Quadratic equation is a 2nd degree polynomial euqtion, generally represented as a x ² + b x + c = 0, where a ≠ 0. Let's calculate the roots of the equation and nature of the roots.
a x ² + b x + c = 0; where a ≠ 0
on divinding with 'a' on both sides we get
x ² +

\frac{b}{a}

x +

\frac{c}{a}

= 0
x ² + 2 .

\frac{b}{2a}

x +

\frac{c}{a}

= 0
let's add and subtract

\frac{b^{2}}{{4a}^{2}}

to form a perfect squre with the first two terms
x ² + 2 .

\frac{b}{2a}

x +

\frac{b^{2}}{{4a}^{2}}

-

\frac{b^{2}}{{4a}^{2}}

+

\frac{c}{a}

= 0
using (a+b)² = a ² + 2.a.b. + b ² formula
( x +

\frac{b}{2a}

) ² +

\frac{c}{a}

-

\frac{b^{2}}{{4a}^{2}}

+ = 0
root1 =

\frac{{-b}^{- \sqrt{b^{2} - 4ac}}}{2a}

and root2 =

\frac{{-b}^{+ \sqrt{b^{2} - 4ac}}}{2a}

where D = b ² - 4ac is called discriminant.
If D > 0 then the equationhas two distict real roots mentioned above.
If D=0 then two equal roots.
If D < 0 then roots are complex numbers

This algebraic equation has application in almost all the branches of science in one form or the other, and very basic equation. another way of looking at the roots of the equation is as follows.
Let p and q be the roots of the equation a x ² + b x + c = 0, where a ≠ 0.
Then sum of the roots p + q = - $\frac{b}{a}$ and product of the roots is pq = $\frac{c}{a}$

1.3 - quadratic2 calculator

quadratic2

1.4 - quadratic calculator

Quadratic-calculator

a:

b:

c:

x1: 0

x2: 0

1.5 - Trigonometry Calulator!

Trigonometrio

2 - Contribute

How to contribute to the docs

These basic sample guidelines assume that your Docsy site is deployed using Netlify and your files are stored in GitHub. You can use the guidelines “as is” or adapt them with your own instructions: for example, other deployment options, information about your doc project’s file structure, project-specific review guidelines, versioning guidelines, or any other information your users might find useful when updating your site. Kubeflow has a great example.

Don’t forget to link to your own doc repo rather than our example site! Also make sure users can find these guidelines from your doc repo README: either add them there and link to them from this page, add them here and link to them from the README, or include them in both locations.

We use Hugo to format and generate our website, the Docsy theme for styling and site structure, and Netlify to manage the deployment of the site. Hugo is an open-source static site generator that provides us with templates, content organisation in a standard directory structure, and a website generation engine. You write the pages in Markdown (or HTML if you want), and Hugo wraps them up into a website.

All submissions, including submissions by project members, require review. We use GitHub pull requests for this purpose. Consult GitHub Help for more information on using pull requests.

Quick start with Netlify

Here’s a quick guide to updating the docs. It assumes you’re familiar with the GitHub workflow and you’re happy to use the automated preview of your doc updates:

Fork the Goldydocs repo on GitHub.
Make your changes and send a pull request (PR).
If you’re not yet ready for a review, add “WIP” to the PR name to indicate it’s a work in progress. (Don’t add the Hugo property “draft = true” to the page front matter, because that prevents the auto-deployment of the content preview described in the next point.)
Wait for the automated PR workflow to do some checks. When it’s ready, you should see a comment like this: deploy/netlify — Deploy preview ready!
Click Details to the right of “Deploy preview ready” to see a preview of your updates.
Continue updating your doc and pushing your changes until you’re happy with the content.
When you’re ready for a review, add a comment to the PR, and remove any “WIP” markers.

Updating a single page

If you’ve just spotted something you’d like to change while using the docs, Docsy has a shortcut for you:

Click Edit this page in the top right hand corner of the page.
If you don’t already have an up to date fork of the project repo, you are prompted to get one - click Fork this repository and propose changes or Update your Fork to get an up to date version of the project to edit. The appropriate page in your fork is displayed in edit mode.
Follow the rest of the Quick start with Netlify process above to make, preview, and propose your changes.

Previewing your changes locally

If you want to run your own local Hugo server to preview your changes as you work:

Follow the instructions in Getting started to install Hugo and any other tools you need. You’ll need at least Hugo version 0.45 (we recommend using the most recent available version), and it must be the extended version, which supports SCSS.
Fork the Goldydocs repo repo into your own project, then create a local copy using git clone. Don’t forget to use --recurse-submodules or you won’t pull down some of the code you need to generate a working site.
```
git clone --recurse-submodules --depth 1 https://github.com/google/docsy-example.git
```
Run hugo server in the site root directory. By default your site will be available at http://localhost:1313/. Now that you’re serving your site locally, Hugo will watch for changes to the content and automatically refresh your site.
Continue with the usual GitHub workflow to edit files, commit them, push the changes up to your fork, and create a pull request.

Creating an issue

If you’ve found a problem in the docs, but you’re not sure how to fix it yourself, please create an issue in the Goldydocs repo. You can also create an issue about a specific page by clicking the Create Issue button in the top right hand corner of the page.

Useful resources

Docsy user guide: All about Docsy, including how it manages navigation, look and feel, and multi-language support.
Hugo documentation: Comprehensive reference for Hugo.
Github Hello World!: A basic introduction to GitHub concepts and workflow.

3 - Physics

3.1 - Electrostaitcs

3.1.1 - Electric charge

What is electric charge?

We have three types of particles in nature on the basis of electrostatics. They are namely positive charge particles, negatively charged particles, and neutral particles. They are determined in such a way based on their interaction i.e.

like charges repel each other,
Unlike charges attract each other, and
neutral bodies don’t interact with any other particles.

I.e. Two positive charges always repel each other, and so do the negative charges. Whereas a positive charge always attracts negative charge and vice versa.

Neutral particles are combinations of both positive and negative charges in equal amounts or mathematically we call 0 charge or no charge. As positive attracts negative and negative attracts positive zero being neutral it does not interact with any charge. For example a neutron will not interact with either an electron or a proton.

Informal definition: From the above information we can understand that charge is a physical quantity of particles which determines the nature of electrical interactions.

SI unit of charge is coulomb and represented as C. An electron has a charge of $-1.6 \times 10^{-19}$ C and a proton has a charge of $+1.6 \times 10^{-19}$ C. Whereas a neutron charge is 0 C

*Note: positive and negative is just a representation of the opposite nature of particles in a mathematical way.

Quantization of charge:

Every particle in the universe is made up of atoms, which are made up of electrons, protons and neutrons. Electron is the smallest negative charge while proton is the smallest positive charge there for any charge should be integral multiple of these particles that is a negatively charged particle should be -n e where and is a positive integer and $e = 1.6 \times 10^{-19}$ C. similarly positive charge should be +n e where and is positive integer and $e = 1.6 \times 10^{-19}$. This means that the fractional charge of e doesn’t exist.

Charge is conserved:

A proton and electron can come by to form a Neutron. Similarly a neutron can be divided into a proton and an electron. And in atoms, protons and neutrons makeup the nucleus while electrons revolve around the nucleus in orbits to form a neutral atom.

By this we can understand that charges can be neutralized and charge can be formed from neutral bodies. But the net charge of a closed system is always the same. i.e. net charge of a closed system can neither be destroyed nor created.

Additivity of electric charge:

When two charges combine together the charge of combination is the algebraic sum of the charges of two bodies. i.e. when a 5 C charge body meets with 6 C charge body and stays together then the resultant charge of the new body is 11 coulombs. Similarly when a 5 C charge body meets with -6 C charge body and stays together then the resultant charge of the new body is -1 coulombs.

A real life example of all the above properties of charges can be observed in beta decay. When a neutron in the nucleus of a heavy atom splits into a proton and an electron the charge is conserved because the charge before splitting is zero and the charge after splitting as a whole is 0. Is an example of conservation of charge and an example of additivity of charge.

Conduction of charge:

When a charged body is made contact with a neutral conducting body, Charge flows to the neutral body so that charge density is equally distributed in the both bodies. this can be understood by following image.

Induction of charge into a neutral body:

Charge can be induced into a body not only by conduction. when a neutral body is brought into the filed of a positively charged body, as shown in the figure, electronss inside the neutral body get attracted towards the positive charge and postive charge get repeled to the otherside as shown in the figure. when we ground the negative charge end, the electrons would floiw to the ground making the body postively charged.

3.1.2 - Coloumb's Law

Coulomb’s law:

The results of many experiments done by Coulomb and many other scientists developed a relation of force exerted by a charged particle on another charge. It is given as:

$$F \propto \frac{q_1 q_2}{r^2}$$

Where $q_1$ and $q_2$ charges of the particles and r is the separation between them. We can write the above proportionality as follows:

$$F = \frac{k q_1 q_2}{r^2}$$

Where k is the proportionality constant. The vectorical representation of coulomb’s law can be written as :

$$\vec{F} = \frac{kq_1 q_2}{r^3} \vec{r}$$

When the constant value was measured experimentally it was found to be $\frac{1}{4\pi \epsilon_0}$. Where $\epsilon_0$ is the permittivity of free space, $\epsilon_0=8.85419\times 10^{-11} C^2 N{-1} m{-2}$. By substituting the constant we get:

$$F = \frac{1}{4\pi \epsilon_0} \frac{ q_1 q_2}{r^2}$$

and vectorically,

$$\vec{F} = \frac{1}{4\pi \epsilon_0} \frac{ q_1 q_2}{r^3} \vec{r}$$

Practice problems

A Charge A of 5 C is placed at origin and another charge B of 5 C is placed at (5,0) on the cordinate plane. Find the electrical force acting on A and it’s direction.
find the force acting on B and its direction in the above question
Three charges of charge q are placed on the vertices of an equilateral triangle of edge a. Find the force acting on one charge because of other two charges.
three charges of charge q are placed on three vertices of square of side length a then find the force acting on test charge Q placed at the centre of the square.
by what distance two charges of charge 1C each must be separated so that force between them is equals to 50 N.
Two charges, 5 C and 6 C are separated by distance 1 m. At what point the third charge can be placed so that it experiences no net force.

3.2 - Statitical Mechanics notes

3.2.1 - Prerequisites and Historical note

This is the notes is the made using lectures of various premium institutes across India.

Prerequisites

Various mathematical concepts such as Advanced calculus which includes multivariable integrations, complex variables, combinatronics and so on.
Basic programming in a language.
Knowledge of Classical Mechanics and, Quantum Mechanics upto undergraduate level. ( you can refer to Classical Mechanics by Herbert Goldstein and Introduction to quantum mechanics by David J. Griffiths)

Historical Note:

Historically, Thermodynamics was the conceptual preceder to Statistical Mechanics.
Since the antiquity to 1600 A.D. notion of heat and temperature were widely accepted. It was believed at that time that heat was associated with the motion of microscopic constituents of matter
Later during 1700s and first half of 1800s,the wrong notion that heat was instead a fluid like substance started becoming popular.
In 1850 experiments of James Joule and others show that heat is a form of energy.
Sadi Carnot had explained the relation between heat and energy. Which was important in the development of steam engines.
In 1850 Rudolf Clausius and William Thomson (Lord Kelvin) formulated the first law which is the idea that total energy is conserved. The second law of thermodynamics which is the idea that heat cannot be completely converted to work was also formulated.
In 1738 Daniel Bernoulli has pointed out that gases consist of molecules in motion. Clausius revived this idea in 1857.
In 1860 James clerk Maxwell derived the expected distribution of molecular speeds in a gas by taking into account molecular conditions.
In 1877 Ludwig Boltzmann constructed an equation that he thought could describe the detail time evolution of a gas regardless of whether it was in equilibrium or not.
In the 1860, Clausius had introduced entropy as a ratio of heat to temperature, and had stated the second law in terms of the increase of this quantity.
Boltzmann then showed that his equation implied the so called H theorem, which states that a quantity equal to entropy in equilibrium must always increase with time. Since molecular collisions were assumed reversible, his derivation could be run in reverse, and would then imply opposite of second law. Boltzmann made the implicit assumption that the motion was uncorrelated before collision but not after which imposes irreversibility.
In 1900, Gibbs introduced the notion of an ensemble - a collection of many possible states of a system, each assigned a certain probability. He argued that if the time evolution of a single state were to visit all other states in the ensemble, the so called ergodic hypothesis - then
averaged over a sufficiency long time a single state would behave in a way that was typical of the ensemble. Gibbs also gave qualitative arguments that entropy would increase if it were measured in a “coarse grained” way in which nearby states were not distributed.

3.3 - Nuclear Physics CSIR-NET prep notes

3.3.1 - Nuclear Radius

As we nucleaus is the core of the atom, where most of the mass of the atom is concentrated, and the nuclear radius is an essential concept in nuclear physics, as it represents the size of the atomic nucleus. It plays a significant role in understanding various nuclear phenomena, such as nuclear reactions, nuclear decay, and nuclear fission. This essay aims to provide a detailed explanation of the nuclear radius, its measurement, and its significance in nuclear physics.

The nuclear radius is the distance between the center of the nucleus and its outermost boundary, or the point at which the nuclear density decreases to a negligible value. The nuclear radius is much smaller than the atomic radius, which is the distance between the nucleus and the outermost electron in the atom. The atomic radius is typically on the order of angstroms (10^-10 meters), while the nuclear radius is on the order of femtometers (10^-15 meters).

Nuclear radius increases with the increase in the nucleons, i.e. atom with higher nucleons has higher radius than those of the lesser ones. However, this increase is not linear but follows a more complex trend due to the strong nuclear force that holds the nucleus together. The strong nuclear force is a short-range force that binds the nucleons together and is much stronger than the electromagnetic force that repels the positively charged protons from each other. As a result, the nuclear radius increases less rapidly than the number of nucleons, and the nuclear density decreases as the nucleus becomes larger.

Measurement of Nuclear Radius:

The measurement of nuclear radius is a challenging task due to the small size of the nucleus and the difficulty in probing its structure. Several methods have been developed to determine the nuclear radius experimentally, each with its advantages and limitations.

One of the most common methods for measuring nuclear radii is through electron scattering experiments. In this technique, high-energy electrons are directed at a thin foil of the element under investigation, and the scattered electrons are detected at different angles. The distribution of scattered electrons provides information about the size and shape of the nucleus. The advantage of this method is that it can be used for a wide range of nuclei and can provide information about the nuclear charge distribution. However, the electron scattering method is limited by the resolution of the detectors and the complexity of the nuclear structure.

R= R₀ A ^1/3 where R₀ = 1.2 fm = 1.2 x 10 ^-15 m The nuclear radius is a fundamental property of the atomic nucleus and plays a significant role in many nuclear phenomena. One of the most important applications of nuclear radius is in the study of nuclear reactions. Nuclear reactions involve the interaction of two or more nuclei, which can result

Further information:

Another method for measuring nuclear radii is through the analysis of nuclear decay processes. In radioactive nuclei, the decay process involves the emission of particles from the nucleus, such as alpha or beta particles. The energy and momentum of these particles depend on the size and structure of the nucleus. By analyzing the decay products, one can determine the nuclear radius indirectly. The advantage of this method is that it can provide information about the excited states of the nucleus and the distribution of nuclear spins. However, the interpretation of the data can be complex, and the method is limited to radioactive nuclei.

A third method for measuring nuclear radii is through the study of nuclear collisions. In this technique, two nuclei are accelerated to high energies and made to collide with each other. The trajectory and energy of the scattered particles can provide information about the nuclear radius and the strength of the nuclear force. The advantage of this method is that it can provide information about the collision dynamics and the nuclear matter distribution. However, the interpretation of the data can be complex, and the method is limited to a small range of nuclei.

3.3.2 - Nuclear Spin and Parity

Spin is a quantum mechanical property that describes the intrinsic angular momentum of a nucleus, while parity is a quantum number that describes the symmetry properties of a nucleus under reflection. The spin and parity of a nucleus play a crucial role in determining its physical and chemical properties and are used extensively in nuclear physics research.

Spin:

The spin of a nucleus is a fundamental property that arises from the intrinsic angular momentum of its constituent particles, such as protons and neutrons. The spin of a nucleus is described using the quantum mechanical concept of angular momentum, which is a vector quantity that represents the rotation of an object around an axis. The magnitude of the angular momentum of a nucleus is proportional to its spin quantum number, denoted by the symbol I.

The spin quantum number can take on half-integer or integer values, depending on the number of protons and neutrons in the nucleus. For example, nuclei with an odd number of protons or neutrons have a half-integer spin, while nuclei with an even number of protons and neutrons have an integer spin. The spin quantum number is important in nuclear physics because it determines the magnetic moment of the nucleus, which is used in various experimental techniques, such as nuclear magnetic resonance (NMR) spectroscopy.

Parity:

The parity of a nucleus is a quantum number that describes its symmetry properties under reflection. A nucleus can be either even or odd parity, depending on whether it is symmetric or asymmetric under reflection. Parity is described using the quantum mechanical concept of parity operator, denoted by the symbol P, which changes the sign of the wave function of a nucleus under reflection.

The parity of a nucleus is related to the number of nucleons with spin up and spin down. For example, if a nucleus has an even number of nucleons with spin up and an even number of nucleons with spin down, it has an even parity. Conversely, if a nucleus has an odd number of nucleons with spin up and an even number of nucleons with spin down, it has an odd parity.

let $\psi$ (x,y,z) be a wave function representing the particle

then if $\psi$ (-x,-y,-z) = $\psi$ (x,y,z) : even parity

and if $\psi$ (-x,-y,-z) = - $\psi$ (x,y,z) : odd parity

4 - How did they do it?

This is about finding out about the journeys of experiments, theories and scientific enquiries that have been done by our scientists. I am interested in the stories of how they have done it I have been trying to find out but currently, I can’t think of what should I find out about. That is why I would write the ideas one by one as I get them.

how did Coulomb define and verify Coulomb’s law?
how did we come to the current understanding of the nuclear model of an atom and the history of the discoveries and models of the nucleus
how did Schrodinger develop his equation and what are the challenges he faced in developing it?
how did Heisenberg discover the uncertainty principle?
How did Max Plank give his hypothesis and give the fundamental constant h to the world?
How did the idea of vectors develop? how did they develop the basis for vectors that helped so much in the scientific study? what is the history of vectors?

5 -

5.1 - Analytical R&D : Scientist

Job title: Analytical R&D : Scientist

Job Description:

Performing analytical method validation/ raw material/ finished product analysis
Performing assay by using HPLC
To perform analysis related to Method Transfer and Analytical Method Verification
Preparing method validation protocols & reporting.

Required Candidate profile:

Education : M.Sc / M. Pharma / Ph.D

Experience : 2 - 8 years

Position : Research Associate to Research Scientist

Location : Hyderabad

Industry Type : Pharmaceutical

Functional Area : R&D, Pharmaceuticals

Employment Type : Full Time

Forward your CV / Resume : careers@cellixbio.com

About Cellix bio:

Cellix Bio employees and teams apply sound science to develop and deliver medicines that focus on unmet medical needs around the world. Our current prescription products at various stages of development target to serve patients across therapeutic areas including metabolism, immunology, infectious diseases, neuroscience, ophthalmology, and oncology.

Here, everyone matters and you will be a vital contributor to our inspiring, bold mission. You will help make an impact on people’s lives and change futures every day. We are equally committed to bringing out your best and fostering a collaborative workplace built on trust and respect.

We empower our people to take charge of their potential. We offer training and leadership development programs tailored to help you build valuable skills and succeed in your career. As you discover your potential, your professional and personal future will become brighter.

What do we do?

Founded in 2014, we are the leading drug discovery and development company. We generate our own drug design data before combining the critical power of AI with the creativity of involving proprietary platform technology, Synergix AITM and expertise of our world-class scientists. This allows us to shorten the pre-clinical drug discovery stage and, in turn, substantially accelerate the delivery of new treatments to patients worldwide. By reducing the number of compounds analysed, we aim to vividly reduce the time and cost of discovering and developing new medicines. Cellix Bio balances Synergix AITM plus human creativity in the drug discovery.

Our targeted discovery process enables the creation of new markets and solutions to both old and new problems. Driven by the speed and accuracy of our platform, Synergix AITM accomplishes in months what previously took years, saving tens of millions in research and development costs. Each of our lead candidates are unique with powerful therapeutic benefits. The molecular entities that we discover are the first of their kind and patented.

Using our Synergix AITM drug discovery platform and the insights from our successful drug development and clinical data, we can identify synergistic pharmacology between bio-actives, combining them to address disease related biochemical pathways and targets. However the rubrics of drug discovery are very complex - not all are known and some are not readily describable as a restricted set of moves.

5.2 - EXECUTIVE PURCHASE

Job title: EXECUTIVE PURCHASE

Location: Hyderabad, India - 1 opening

Qualification

Master’s or Bachelor’s degree in Science / Pharmacy / Commerce
At least 5 years’ experience in similar function in pharmaceutical industry (Especially Finished dosage forms)

Skills Required

Strong initiative skills, working both independently and as part of a team
Clear-cut proficiency in English communication
Advanced Microsoft office skills (Word, Excel, PowerPoint and Outlook)

Change more than just your career:

People at PharmSol are thinkers and doers. That’s why we recruit people with more than just a passion for pharmaceutical solutions – they must also have the courage to bring them to life. There are so many opportunities available – from providing advice, information and solutions to implementing it in pharma plants that make a positive difference for our clients. We build our firm with an entrepreneurial spirit, fueled by dynamic teamwork and collaboration, and founded on trust and respect for each other. The opportunities are endless – the choice is yours.

The Company Overview:

PharmSol, founded in 2004, has grown to become a global pharmaceutical enterprise, offering smart products and providing bespoke solutions throughout the world. PharmSol’s successful track record is credited to its team’s all round technical competence, its drive to be the leading choice for clients, its focus on building long term relationships and its longstanding dedication to prosper the pharma industry and community.

Since its inception 18 years ago, PharmSol has treaded a path of perpetual growth, expanding not only its global reach, but also the products and solutions we offer, the diverse yet detailed expertise of our team and the number of clients whose expectations we have exceeded.

Today, PharmSol has earned a worldwide reputation for developing and registering many first to file products and backward integrating. As a result, we possess a vast portfolio of products and technologies, which are either developed in-house or are co-developments. We endeavour to bring new products to the market and implement new ideas & technologies to meet needs of the future.

PharmSol is well known for offering integrated solutions with ‘single window’ access to all its clients, ensuring seamless and optimal deliverance on all assignments. In key markets and through our experience and operational flexibility, we address challenges of our customers in areas of pharmaceutical regulations, product development, registration, engineering, market affairs and supply of products.

PharmSol has widened its niche, by providing on a supplemental or comprehensive basis, fully tailored solutions for meeting the Product Development, IPR, Facility Design, Compliance, Registrations, Market Access, Audits and Supply Chain requirements, with a focus on EU, US, TGA, WHO and other International Regulatory Requirements.

PharmSol continues to strive for excellence and achieve its goal to forever be the leader in pharma and the foremost choice for its customers. Its visionary leadership, with nearly 40 years of success, continues to innovate and grow, through cementing its presence in Africa and far east Asia as well as through new ventures, such as Eminus eGlobal Institute, which are destined to transform the pharma industry.

explore our site for more jobs in Hyderabad

5.3 - Executive Quality Affairs

Job title: Executive Quality Affairs

Location: Hyderabad, India - 2 openings

Qualification:

Master’s or Bachelor’s degree in Science / Pharmacy
At least 5 years’ experience in similar function in pharmaceutical industry (Especially Finished dosage forms)

Skills Required

Candidate should be from QA / QC background
Sound and deep knowledge in to current GMP requirements (US FDA, EU GMP, PIC/s)
Clear understanding of QA system and principles
Strong initiative skills, working both independently and as part of a team
Clear-cut proficiency in English communication
Advanced Microsoft office skills (Word, Excel, PowerPoint and Outlook)

Change more than just your career:

People at PharmSol are thinkers and doers. That’s why we recruit people with more than just a passion for pharmaceutical solutions – they must also have the courage to bring them to life. There are so many opportunities available – from providing advice, information and solutions to implementing it in pharma plants that make a positive difference for our clients. We build our firm with an entrepreneurial spirit, fueled by dynamic teamwork and collaboration, and founded on trust and respect for each other. The opportunities are endless – the choice is yours.

The Company Overview:

PharmSol, founded in 2004, has grown to become a global pharmaceutical enterprise, offering smart products and providing bespoke solutions throughout the world. PharmSol’s successful track record is credited to its team’s all round technical competence, its drive to be the leading choice for clients, its focus on building long term relationships and its longstanding dedication to prosper the pharma industry and community.

Since its inception 18 years ago, PharmSol has treaded a path of perpetual growth, expanding not only its global reach, but also the products and solutions we offer, the diverse yet detailed expertise of our team and the number of clients whose expectations we have exceeded.

Today, PharmSol has earned a worldwide reputation for developing and registering many first to file products and backward integrating. As a result, we possess a vast portfolio of products and technologies, which are either developed in-house or are co-developments. We endeavour to bring new products to the market and implement new ideas & technologies to meet needs of the future.

PharmSol is well known for offering integrated solutions with ‘single window’ access to all its clients, ensuring seamless and optimal deliverance on all assignments. In key markets and through our experience and operational flexibility, we address challenges of our customers in areas of pharmaceutical regulations, product development, registration, engineering, market affairs and supply of products.

PharmSol has widened its niche, by providing on a supplemental or comprehensive basis, fully tailored solutions for meeting the Product Development, IPR, Facility Design, Compliance, Registrations, Market Access, Audits and Supply Chain requirements, with a focus on EU, US, TGA, WHO and other International Regulatory Requirements.

PharmSol continues to strive for excellence and achieve its goal to forever be the leader in pharma and the foremost choice for its customers. Its visionary leadership, with nearly 40 years of success, continues to innovate and grow, through cementing its presence in Africa and far east Asia as well as through new ventures, such as Eminus eGlobal Institute, which are destined to transform the pharma industry.

explore our site for more jobs in Hyderabad

5.4 - EXECUTIVE QUALITY CONTROL

Job title: EXECUTIVE QUALITY CONTROL

Location: Hyderabad, India - 1 opening

Qualification:

Master’s or Bachelor’s degree in Science / Pharmacy
At least 5 years’ experience in similar function in pharmaceutical industry (Especially Finished dosage forms) Skills Required:
Candidate should be from ARD/QC background
Sound and deep knowledge in to current GMP requirements (US FDA, EU GMP, PIC/S)
Clear understanding of GLP system and principles
Strong initiative skills, working both independently and as part of a team
Clear-cut proficiency in English communication
Advanced Microsoft office skills (Word, Excel, PowerPoint and Outlook)

Change more than just your career:

People at PharmSol are thinkers and doers. That’s why we recruit people with more than just a passion for pharmaceutical solutions – they must also have the courage to bring them to life. There are so many opportunities available – from providing advice, information and solutions to implementing it in pharma plants that make a positive difference for our clients. We build our firm with an entrepreneurial spirit, fueled by dynamic teamwork and collaboration, and founded on trust and respect for each other. The opportunities are endless – the choice is yours.

The Company Overview:

PharmSol, founded in 2004, has grown to become a global pharmaceutical enterprise, offering smart products and providing bespoke solutions throughout the world. PharmSol’s successful track record is credited to its team’s all round technical competence, its drive to be the leading choice for clients, its focus on building long term relationships and its longstanding dedication to prosper the pharma industry and community.

Since its inception 18 years ago, PharmSol has treaded a path of perpetual growth, expanding not only its global reach, but also the products and solutions we offer, the diverse yet detailed expertise of our team and the number of clients whose expectations we have exceeded.

Today, PharmSol has earned a worldwide reputation for developing and registering many first to file products and backward integrating. As a result, we possess a vast portfolio of products and technologies, which are either developed in-house or are co-developments. We endeavour to bring new products to the market and implement new ideas & technologies to meet needs of the future.

PharmSol is well known for offering integrated solutions with ‘single window’ access to all its clients, ensuring seamless and optimal deliverance on all assignments. In key markets and through our experience and operational flexibility, we address challenges of our customers in areas of pharmaceutical regulations, product development, registration, engineering, market affairs and supply of products.

PharmSol has widened its niche, by providing on a supplemental or comprehensive basis, fully tailored solutions for meeting the Product Development, IPR, Facility Design, Compliance, Registrations, Market Access, Audits and Supply Chain requirements, with a focus on EU, US, TGA, WHO and other International Regulatory Requirements.

PharmSol continues to strive for excellence and achieve its goal to forever be the leader in pharma and the foremost choice for its customers. Its visionary leadership, with nearly 40 years of success, continues to innovate and grow, through cementing its presence in Africa and far east Asia as well as through new ventures, such as Eminus eGlobal Institute, which are destined to transform the pharma industry.

explore our site for more jobs in Hyderabad

5.5 - EXECUTIVE REGULATORY AFFAIRS

Job title: EXECUTIVE REGULATORY AFFAIRS

Location: Hyderabad, India - 1 opening

Qualification:

Master’s or Bachelor’s degree in Science / Pharmacy
At least 7-10 years’ experience in similar function in pharmaceutical industry Skills Required:
Candidate should be from RA having supported regulatory affairs function on finished dosage form
Sound and deep knowledge in to current Regulatory / GMP requirements (US FDA, EU GMP, PIC/s )
Strong initiative skills, working both independently and as part of a team
Clear-cut proficiency in English communication
Advanced Microsoft office skills (Word, Excel, PowerPoint and Outlook)

Change more than just your career:

People at PharmSol are thinkers and doers. That’s why we recruit people with more than just a passion for pharmaceutical solutions – they must also have the courage to bring them to life. There are so many opportunities available – from providing advice, information and solutions to implementing it in pharma plants that make a positive difference for our clients. We build our firm with an entrepreneurial spirit, fueled by dynamic teamwork and collaboration, and founded on trust and respect for each other. The opportunities are endless – the choice is yours.

The Company Overview:

PharmSol, founded in 2004, has grown to become a global pharmaceutical enterprise, offering smart products and providing bespoke solutions throughout the world. PharmSol’s successful track record is credited to its team’s all round technical competence, its drive to be the leading choice for clients, its focus on building long term relationships and its longstanding dedication to prosper the pharma industry and community.

Since its inception 18 years ago, PharmSol has treaded a path of perpetual growth, expanding not only its global reach, but also the products and solutions we offer, the diverse yet detailed expertise of our team and the number of clients whose expectations we have exceeded.

Today, PharmSol has earned a worldwide reputation for developing and registering many first to file products and backward integrating. As a result, we possess a vast portfolio of products and technologies, which are either developed in-house or are co-developments. We endeavour to bring new products to the market and implement new ideas & technologies to meet needs of the future.

PharmSol is well known for offering integrated solutions with ‘single window’ access to all its clients, ensuring seamless and optimal deliverance on all assignments. In key markets and through our experience and operational flexibility, we address challenges of our customers in areas of pharmaceutical regulations, product development, registration, engineering, market affairs and supply of products.

PharmSol has widened its niche, by providing on a supplemental or comprehensive basis, fully tailored solutions for meeting the Product Development, IPR, Facility Design, Compliance, Registrations, Market Access, Audits and Supply Chain requirements, with a focus on EU, US, TGA, WHO and other International Regulatory Requirements.

PharmSol continues to strive for excellence and achieve its goal to forever be the leader in pharma and the foremost choice for its customers. Its visionary leadership, with nearly 40 years of success, continues to innovate and grow, through cementing its presence in Africa and far east Asia as well as through new ventures, such as Eminus eGlobal Institute, which are destined to transform the pharma industry.

explore our site for more jobs in Hyderabad

5.6 - Formulation R&D Scientist : Pharma

Job title: Formulation R&D Scientist : Pharma

Forward your CV / Resume : careers@cellixbio.com

Job Description:

R&D Scientist - Formulation Scientist

Pharmaceuticals/ Biotechnology/ Clinical Research

Required Candidate profile:

Education : Doctorate / Masters in Pharmaceutics, Pharmaceutical Chemistry, Industrial Pharmacy, Analytical Chemistry, Pharmaceutical Chemistry, Organic Chemistry or Medicinal Chemistry.

Experience : 5 - 10 Years

Position : Research Associate to Research Scientist

Location : Hyderabad

EXPERIENCE :

In-depth understanding of physical pharmacy, physical chemistry, thermodynamics, chemical reaction kinetics is a must.
In-depth understanding of various pre- formulation and formulation studies including excipient compatibility, forced degradation studies and unit steps in formulation manufacture is a must.
Experience with pilot-scale and manufacturing-scale equipment is a plus.
Experience with process analytical technologies (PAT) is a plus.
Conduct API, excipient, packaging or device characterization; excipient compatibility and forced degradation studies.
Develop robust lab-scale, pilot-scale and commercial-scale manufacturing process based on Quality by Design (QbD) principles using design of experiments (DOE). Recommend stage appropriate acceptance criteria. Utilize PAT where feasible. Utilize relevant statistical tools, as required.
Review and analyze relevant stability data from lab-scale, pilot-scale and commercial-scale, R&D and GMP stability studies by utilizing industry standard modeling and analysis tools including regression analysis for shelf-life prediction. Develop and implement strategies to control related substances, where applicable.
Draft high-quality industry standard technical documents (protocols, reports, technical memorandums, position papers etc.).
Independently draft and/or review standard operating procedures (SOPs).
Maintain effective and pro-active communication and coordination of activities with multi-functional stakeholders.

About Cellix bio:

Cellix Bio employees and teams apply sound science to develop and deliver medicines that focus on unmet medical needs around the world. Our current prescription products at various stages of development target to serve patients across therapeutic areas including metabolism, immunology, infectious diseases, neuroscience, ophthalmology, and oncology.

Here, everyone matters and you will be a vital contributor to our inspiring, bold mission. You will help make an impact on people’s lives and change futures every day. We are equally committed to bringing out your best and fostering a collaborative workplace built on trust and respect.

We empower our people to take charge of their potential. We offer training and leadership development programs tailored to help you build valuable skills and succeed in your career. As you discover your potential, your professional and personal future will become brighter.

What do we do?

Founded in 2014, we are the leading drug discovery and development company. We generate our own drug design data before combining the critical power of AI with the creativity of involving proprietary platform technology, Synergix AITM and expertise of our world-class scientists. This allows us to shorten the pre-clinical drug discovery stage and, in turn, substantially accelerate the delivery of new treatments to patients worldwide. By reducing the number of compounds analysed, we aim to vividly reduce the time and cost of discovering and developing new medicines. Cellix Bio balances Synergix AITM plus human creativity in the drug discovery.

Our targeted discovery process enables the creation of new markets and solutions to both old and new problems. Driven by the speed and accuracy of our platform, Synergix AITM accomplishes in months what previously took years, saving tens of millions in research and development costs. Each of our lead candidates are unique with powerful therapeutic benefits. The molecular entities that we discover are the first of their kind and patented.

Using our Synergix AITM drug discovery platform and the insights from our successful drug development and clinical data, we can identify synergistic pharmacology between bio-actives, combining them to address disease related biochemical pathways and targets. However the rubrics of drug discovery are very complex - not all are known and some are not readily describable as a restricted set of moves.

5.7 - Microbiology Manager

If interested, please mail your resume to meenakshi@horizonbiolabs.com, admin@horizonbiolabs.com

Location: Hyderabad, Telangana

Job Summary:

We are looking for a Microbiology Manager to join our team at Horizon Biolabs Pvt Ltd in Hyderabad, Telangana. The ideal candidate will have a Master’s degree in Microbiology and a minimum of 8 years of experience in any reputed Pharma Testing Lab/CRO/Pharmaceutical Industry. As a Microbiology Manager, you will be responsible for managing and overseeing the day-to-day operations of the microbiology department, including planning and executing microbiology testing, analyzing and interpreting results, and maintaining laboratory compliance with regulatory standards.

Responsibilities:

Manage and oversee the day-to-day operations of the microbiology department
Plan and execute microbiology testing, including sample preparation, culture and media preparation, and analysis of results
Analyze and interpret microbiology test results and prepare reports for internal and external stakeholders
Maintain laboratory compliance with regulatory standards, including GLP, GMP, and ISO

5)Train and supervise microbiology technicians and ensure their compliance with laboratory protocols and safety guidelines

6)Collaborate with other departments and teams to ensure timely delivery of testing results and reports

Qualifications:

Master’s degree in Microbiology or related field
Minimum of 8 years of experience in any reputed Pharma Testing Lab/CRO/Pharmaceutical Industry
Knowledge of microbiology testing methods, techniques, and instrumentation
Experience with laboratory compliance and regulatory standards, including GLP, GMP, and ISO
Strong communication and interpersonal skills

6)Ability to lead and manage a team of technicians

If you are interested in this opportunity and meet the qualifications, please apply through our website. We offer a competitive salary and benefits package, and opportunities for professional growth and development. Horizon Biolabs Pvt Ltd is an equal opportunity employer and values diversity at our company.

If interested, please mail your resume to meenakshi@horizonbiolabs.com, admin@horizonbiolabs.com

About the company:

Horizon Biolabs Pvt Ltd. is a GLP compliance Analytical Contract Research Organization located in Hyderabad , India . We are industry-leading global provider of laboratory testing services in Chemical and Microbiological segments across the pharmaceutical , Biotech , medical devices and packaging Industries .

As a customer focused company , we are committed to provide quality services to meet the client requirements and are compliance with all relevant industry standards and regulations .

We have a team of dedicated scientists working on various research projects outsourced by renowned organizations for their submissions to various regulatory bodies .

Our state of the art analytical laboratory with built up area of 10,000 sq.ft . We use combinations of instruments, analytical techniques and methods to meet service needs .

5.8 - Product Application Specialist - Biopharma Industry

Location: Hyderabad, Telangana

Company: Avantor

Job Type: Full-time

Education: Minimum education requirement: Bachelor’s or Master’s degree in Life Sciences (Biology, Molecular Biology, Microbiology, Immunology, Biochemistry, Biotechnology, Engineering).

Experience:

Minimum of 5 to 10 years industry experience in life sciences – Biopharma industry, including relevant knowledge across the workflow.
Experience working with external customers within the Biotech, Biologics, Life Sciences, and/or Biopharmaceutical industries in a technical, engineering, or manufacturing capacity.

Responsibilities:

Develop and execute a technical engagement strategy that is globally and cross-functionally aligned to support the positioning of Masterflex pump and tubing in customer’s process workflows for biopharma businesses.
Lead as a technical expert in the biopharma industry, providing guidance and support to customers and internal teams.
Collaborate with customers and internal teams to identify opportunities for product and service improvements.
Deliver technical presentations and training to customers and internal teams.
Provide on-site and remote technical support to customers as needed.
Stay up-to-date with industry trends and developments.

Requirements:

A minimum of 5 to 10 years of industry experience in the life sciences - Biopharma industry.
Strong knowledge of upstream process and chromatography purification downstream process.
Excellent communication, presentation, and interpersonal skills.
Ability to work independently and as part of a team.
Willingness to travel as needed.

If you are passionate about the biopharma industry and have the experience and skills we are looking for, we encourage you to apply for this exciting opportunity with Avantor. Apply now and take the first step towards a rewarding career with a global leader in the industry!

The job is for a Product Application Specialist for Avantor’s Fluid Handling (AFH) division in the biopharmaceutical industry. The position requires a Bachelor’s or Master’s degree in Life Sciences (Biology, Molecular Biology, Microbiology, Immunology, Biochemistry, Biotechnology, Engineering) and a minimum of 5 to 10 years of experience in the biopharma industry. The role involves providing technical guidance and support for the upstream process and chromatography purification downstream process. The specialist will be responsible for developing and executing a technical engagement strategy globally, which is aligned to support the positioning of Masterflex pump and tubing in customer’s process workflows. They will work with external and internal customers to meet business goals and maintain awareness of customers’ evolving technical challenges and material requirements. The role involves managing and prioritizing technical engagements in the field, tracking metrics related to technical customer engagements for the India sales team, and supporting the development of applications data for new and existing products and services. The successful candidate will have excellent communication and interpersonal skills and be able to develop and maintain strong technical relationships with targeted key accounts.

To apply, please click the link below:

5.9 - Product Specialist - Protein & Molecular Biology | Thermo Fisher Scientific | Hyderabad

Title: Product Specialist - Protein & Molecular Biology | Thermo Fisher Scientific | Hyderabad

Location: Hyderabad, Telangana

Salary: Competitive

About the Company:

Thermo Fisher Scientific is a global leader in scientific research and innovation, offering a wide range of products and services to help scientists and researchers advance their work. With a focus on protein and molecular biology, Thermo Fisher Scientific is committed to providing cutting-edge solutions to meet the needs of customers worldwide.

Job Description:

We are seeking a talented and driven Product Specialist to join our team in Hyderabad. As a Product Specialist, you will be responsible for maintaining our current customer base while actively developing new process opportunities to build a strong pipeline of projects that will fuel future growth. You will work closely with customers to understand their needs and provide solutions that meet or exceed their expectations.

Key Responsibilities:

Maintain and grow existing customer relationships to maximize revenue and profitability
Develop new business opportunities by identifying and targeting potential customers
Provide technical support and training to customers on the use of our products
Collaborate with cross-functional teams to develop and execute marketing strategies
Stay up-to-date with industry trends and new product developments

Qualifications:

Bachelor’s or Master’s degree in Life Sciences or a related field
3+ years of experience in a sales or business development role, preferably in the life sciences industry
Strong understanding of protein and molecular biology applications and workflows
Excellent communication and interpersonal skills
Ability to work independently and as part of a team

To Apply:

If you are passionate about science and innovation and want to make a difference in the world, we encourage you to apply for this exciting opportunity. Please click on the link provided to submit your application. Thermo Fisher Scientific is an equal opportunity employer and welcomes diversity in the workplace.

About Thermo Fisher Scientific:

Thermo Fisher Scientific Inc. (NYSE: TMO) is the world leader in serving science, with annual revenue of approximately $40 billion. Our Mission is to enable our customers to make the world healthier, cleaner and safer. Whether our customers are accelerating life sciences research, solving complex analytical challenges, increasing productivity in their laboratories, improving patient health through diagnostics or the development and manufacture of life-changing therapies, we are here to support them. Our global team of more than 100,000colleagues delivers an unrivaled combination of innovative technologies, purchasing convenience and pharmaceutical services through our industry-leading brands, including Thermo Scientific, Applied Biosystems, Invitrogen, Fisher Scientific, Unity Lab Services, Patheon and PPD. For more information, please visit www.thermofisher.com

5.10 - Scientific Writer - HEVA

Location: Hyderabad, full-time

Mission statements:

Create HEVA (Health Economics and Value Assessment) Communications deliverables (including manuscripts, posters, abstracts, slide decks) aligned with HEVA strategy and global HEVA communication plan across relevant business units and product teams.
Manage core HEVA communication processes, templates, and products across the portfolio
Ensure Core Value Decks for key products are established and maintained, making available a regularly updated synthesis of critical HEVA evidence on the value of products
Maintain accountability for adherence to the publication SOP and other compliance expectations relevant to HEVA communication processes
Seek opportunities to innovate HEVA value communications to increase the relevance and impact of HEVA evidence and inform optimal access and reimbursement decisions
Develops and maintains TA expertise
Manage end to end process through Datavision & Matrix

Collaboration:

HEVA Global and Local teams
RWE Global and local teams
Scientific communication global or local teams
Medical Information global or local teams

Duties & Responsibilities

People

Maintain effectiveness relationships with the end stakeholders (Medical scientific community) within the allocated Global business unit and product – with an end objective to develop education and communication content as per requirement for HEVA (Health Economics and Value Assessment) Communications
Interact effectively with healthcare professionals on publications content
Constantly assist peer writers in developing knowledge and sharing learning

Performance:

Create HEVA (Health Economics and Value Assessment) Communications deliverables (including manuscripts, posters, abstracts, slide decks) aligned with HEVA strategy and global HEVA communication plan across relevant business units and product teams as per agreed timelines and quality Process

Develop complex publications material:

Act as an expert in the field of medical communication or medical information for the assigned Therapeutic area.
Assist the assigned scientific communication team in conducting comprehensive publication-needs analysis
Manage core HEVA communication processes, templates, and products across the portfolio in accordance to the scientific and value messages aligned with Core Value Dossier, the US AMCP dossier, and HEVA contributions as appropriate to other submissions
Ensure Core Value Decks for key products are established and maintained, making available a regularly updated synthesis of critical HEVA evidence on the value of products
Maintain accountability for adherence to the publication SOP and other compliance expectations relevant to HEVA communication processes
Maintain accountability for adherence to the publication SOP and other compliance expectations relevant to HEVA communication processes
Implement relevant element of publication plan and associated activities for the year identified for the region
Work with selected vendors within the region to deliver the required deliverables as per defined process
Design an overall plan of action basis end-customers feedback & improve course content and delivery Customer

Work closely with HEVA Global and Local teams, RWE Global and local teams & scientific communication teams in regions/areas to identify publications needs and assist in developing assigned deliverables Liaise with Medical, HEVA Global and Local teams to prepare relevant & customized deliverables Knowledge, Skills & Competencies / Language

Therapy Area Exposure:

Diabetes
Familial hypercholesterolemia
Cardiovascular disease
Multiple sclerosis,
Immunology
Oncology
Market access
Emerging markets
Generics
Vaccines
Vitamins and supplements
Digestive
Allergies
Parkinson
Haemophilia
Rare diseases
Rare blood diseases
Stakeholder management

Project management

Publications submission

Procedures

Qualifications

Advanced degree in life sciences/ pharmacy/ similar discipline or medical degree Excellent English language knowledge Relevant training/ experience in health economics, public health, epidemiology, or other relevant health-related scientific discipline preferred Requirements of the job

2 years of experience in content creation for the pharmaceuticals / healthcare industry, or academia

At Sanofi diversity and inclusion is foundational to how we operate and embedded in our Core Values. We recognize to truly tap into the richness diversity brings we must lead with inclusion and have a workplace where those differences can thrive and be leveraged to empower the lives of our colleagues, patients and customers. We respect and celebrate the diversity of our people, their backgrounds and experiences and provide equal opportunity for all.

5.11 - Senior Manager Products

Job title: Senior Manager Products

Location : Hyderabad, India

Qualification:

Academic Qualification (minimum): B.E. / B. Tech in Mechanical / Chemical Engg. (Chem. Engg. preferred)
Years of relevant professional experience: 12+ years

Skills Required:

Experience as a Project Manager in pharma formulation projects involving hands on exposure to facility design & essentially spanning the FDF manufacturing categories (Solid Dosage Forms, Semi-Solid Dosage Forms, Liquid Dosage Forms, Injectable Form)
In-depth knowledge w.r.t. the following project engineering disciplines for formulation projects:
1. Process: Plant capacity calculations, Process & Utility equipment sizing; GMP requirements
2. Mechanical: Process equipment & piping (incl. layouts) / Utilities (black & clean) / HVAC / Fire Protection (FP) & Public Health Engg. (PHE)
3. Building Management Systems & ELV Systems
4. Project Cost Estimation (upto + 10% accuracy level)
Basic knowledge w.r.t. the following project engineering disciplines for formulation projects (Electricals [other than ELV systems], Plant Instrumentation & Automation [other than BMS], Civil Structure & Architecture)
Strong exposure to project capital items procurement
Exposure to & conversant with the requirements of international regulatory body (preferably EDQM, EUGMP, WHO) facility audits.
Advanced knowledge on software skills: MS Office (Word, Excel & Power Point), MS Projects, AutoCad
Possess profound data base w.r.t. indigenous & imported formulation process & utility equipment machinery vendors & cost.

Change more than just your career:

People at PharmSol are thinkers and doers. That’s why we recruit people with more than just a passion for pharmaceutical solutions – they must also have the courage to bring them to life. There are so many opportunities available – from providing advice, information and solutions to implementing it in pharma plants that make a positive difference for our clients. We build our firm with an entrepreneurial spirit, fueled by dynamic teamwork and collaboration, and founded on trust and respect for each other. The opportunities are endless – the choice is yours.

The Company Overview:

PharmSol, founded in 2004, has grown to become a global pharmaceutical enterprise, offering smart products and providing bespoke solutions throughout the world. PharmSol’s successful track record is credited to its team’s all round technical competence, its drive to be the leading choice for clients, its focus on building long term relationships and its longstanding dedication to prosper the pharma industry and community.

Since its inception 18 years ago, PharmSol has treaded a path of perpetual growth, expanding not only its global reach, but also the products and solutions we offer, the diverse yet detailed expertise of our team and the number of clients whose expectations we have exceeded.

Today, PharmSol has earned a worldwide reputation for developing and registering many first to file products and backward integrating. As a result, we possess a vast portfolio of products and technologies, which are either developed in-house or are co-developments. We endeavour to bring new products to the market and implement new ideas & technologies to meet needs of the future.

PharmSol is well known for offering integrated solutions with ‘single window’ access to all its clients, ensuring seamless and optimal deliverance on all assignments. In key markets and through our experience and operational flexibility, we address challenges of our customers in areas of pharmaceutical regulations, product development, registration, engineering, market affairs and supply of products.

PharmSol has widened its niche, by providing on a supplemental or comprehensive basis, fully tailored solutions for meeting the Product Development, IPR, Facility Design, Compliance, Registrations, Market Access, Audits and Supply Chain requirements, with a focus on EU, US, TGA, WHO and other International Regulatory Requirements.

PharmSol continues to strive for excellence and achieve its goal to forever be the leader in pharma and the foremost choice for its customers. Its visionary leadership, with nearly 40 years of success, continues to innovate and grow, through cementing its presence in Africa and far east Asia as well as through new ventures, such as Eminus eGlobal Institute, which are destined to transform the pharma industry.

explore our site for more jobs in Hyderabad

6 - Computational Physics, Python notes

websites that would help for further study: Matplotlib.org
Gallery of different plots using matplotlib

Books: Chapra nd Canale, Numerical MethodS for Engineers, Chapters 2-4

Lecture2 part1
Lecture2 part 2 - setting password
Lecture2 part 3 - setting password 2
Difference between while loop and for loop
Lecture 2 part4 Lists and average of the list containing numbers
lecture 3 part1 - loading and doing operations on a file with single column
lecture3 part2 lading a file with two columns
lecture3 part3 for loop
lecture3 part4 functions
lecture4 part1 basic plotting and saving it in a pdf.html
lecture4 part2 learning more about plotting.html
lecture4 part3 scaling of x and y .html
……

7 - Electrodynamics temporary folder

8 - Maths

This is a Math’s page which include all Maths theories and Problems.

8.1 - Factorial of 100

factoial

Factorial of hundred:

The factorial of a positive integer n is the product of all the positive integers from 1 to n. For example, the factorial of 4 is 4! = 4 * 3 * 2 * 1 = 24. The factorial of 100 is 100! = 100 * 99 * 98 * … * 2 * 1. This is a very large number, equal to approximately 9.3 x 10^157. It has 158 digits! Here is the first few digits of 100!:

eg: $4! = 4 \times 3 \times 2\times 1 = 24$

100!

$100! = 100 \times 99 \times 98 \times 97 \times …….. 4 \times 3 \times 2\times 1 $ = $93326215443944152681699238856266700490715968264381621468592963895217599993229915$ $608941463976156518286253697920827223758251185210916864000000000000000000000000$

The factorial of a positive integer n is the product of all the positive integers from 1 to n. For example, the factorial of 4 is 4! = 4 * 3 * 2 * 1 = 24. The factorial of 100 is 100! = 100 * 99 * 98 * … * 2 * 1. This is a very large number, equal to approximately 9.3 x 10^157. It has 158 digits! Here is the first few digits of 100!:

Factorial of 0! = 1 Proof

The factorial of 0, written as 0!, is defined to be equal to 1. This may seem counterintuitive at first, but there are a few different ways to justify this definition.

One way to think about it is to consider the factorial function as a way of counting the number of ways to arrange a certain number of objects. For example, if you have 3 objects, there are 3! = 6 ways to arrange them: (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), and (3, 2, 1). If you have no objects, there is only 1 way to arrange them: the empty set, or “nothing.”

Another way to justify the definition of 0! as 1 is to consider the definition of the factorial function as a product of all the positive integers from 1 to n. When n = 0, there are no positive integers to include in the product, so the product is equal to 1.

It is also possible to define 0! as the limiting value of n! as n approaches 0. This can be shown using mathematical techniques, such as L’Hopital’s rule. However, this definition relies on advanced mathematical concepts and may not be as intuitive as the other explanations.

8.2 - Logarithms

8.2.1 - Definition and properties

Logarithm: For a number and base, the power the base should have so that it is equal to the given number is called a logarithm.For example, the logarithm of 8 with base 2 means what power should 2 have so that it is equal to 8. 2³=8, the answer is 3.

logarithm is represented as log_b a where a and b are non-zero positive numbers.

Where a is the number inside the logarithmic function and b is the base. And we read it as log a base b.

Symbolically the above example can be represented as: Log₂ 8 =3

I) find the values for the following logarithms:

log₃ 27
log₁₀ 100
log₄ 2
log₁₀ (0.1)

Solutions are at the end of the article.

Logarithm in fact is an inverse of exponential function. Let’s see how logarithm and exponents are interconvertible:

a^b=x ------> b= log_a x

Look at the example 2³=8, this is represented as log₂ 8=3

This is how we interconvert exponents into logarithms and vice versa.

II) convert the following logarithms into exponents

log_a b = x
log₂₅ 5= 1/2
log₃ 9= 2

Solutions are at the end of the article.

Properties:

For a, b, x, y being postive non-zero numbers.

Product rule: Log_a xy = log_a x+ log_a y

Quotient rule: Log_a $\frac{x}{y}$ = log_a x- log_a y

Power rule: (i) log_b $a^m$ = m log_b a

Power rule: (ii) log_bⁿ a= $\frac{1}{n}$ log_b a

log_b a= $\frac{1}{log_a b}$

log_b a= $\frac{log_x a}{log_x b}$

log_a 1 = 0

Other properties:

log_a a= 1

If log_b a= log_b c then a=c

log_b $\frac{1}{a}$=log_{$\frac{1}{b}$} a= -log_a a

a^{log_a x} = x

log_a 0 is undefined

III) Solve the following logarithms:

log₅ x = log₅ 2 + log₅ 7
log_b x+log_b 1/x
log₅ 15 - log₅ 3
log₅ 125
log₁₆ 4

Solutions are at the end of the article.

Sometimes we drop bases in logarithms for base 10 and natural logarithms. Log₁₀ a can be represented dropping 10 i.e. log a. And logarithm with natural base ’e’ can be represented as ln x (=log_e x)

Solutions:

I

1) ans=3

Sol: log₃ 27 = log₃ $3^3$

Using the property $log_b$ $a^n$ = n $log_b$ a

= 3log₃ 3

Using the property log_a a=1

= 3

2) ans=2

Sol:

log₁₀ 100 = log₁₀ $10^2$

Using the property log_b $a^n$ = n log_b a

= 2log₁₀ 10

Using the property log_a a=1

=2

3) ans=½

Sol:

log₄ 2 = log_2² 2

Using the property log_bⁿ a = $\frac{1}{n}$ log_b a

= $\frac{1}{2}$ log₂ 2

Using the property log_a a=1

= $\frac{1}{2}$

4) ans=-1

Sol:

Log₁₀ (0.1) = log₁₀ $\frac{1}{10}$

Using the property Log_a $\frac{x}{y}$ =log_a x- log_a y

= - log₁₀ 10

Using the property log_a a=1 =-1

II)

$a^x$
5= 25^{$\frac{1}{2}$}
9= $3^2$

III)

1) x=14

Sol:

log₅ x = log₅ 2 + log₅ 7

Using the property Log_a xy=log_a x+ log_a y

log₅ x = log₅ 2*7

Using the property if log_a b=log_a c then b=c

x=14

2) ans=0

Sol:

log_b x+log_b 1/x

Using the property Log_a xy =log_a x+ log_a y

=log_b x* $\frac{1}{x}$

=log_b 1

Using the property Log_a 1 = 0

=0

3) ans= 0

Sol:

log₅ 15 - log₅ 3

Using the property Log_a $\frac{x}{y}$ =log_a x- log_a y

=Log₅ $\frac{15}{3}$

= log₅ 5

Using the property Log_a 1 = 0

=0

4) ans =3

Sol:

log₅ 125

=log₅ $5^3$

Using the property log_b aⁿ = n log_b a

=3 log₅ 5

Using the property log_a a=1

=3

5) ans =½

Sol:

log₁₆ 4

=log_4² 4

Using the property log_bⁿ a = $\frac{1}{n}$ log_b a

=½ log₄ 4

Using the property log_a a=1

=½

8.2.2 - Proofs of Logarithmic properties

1) Product Rule: Log_a xy = log_a x+ log_a y

proof:

let log_a x = m, log_a y = n

convert these logarithms into exponents

x=a^m and y=aⁿ

multiply the above equations

xy=a^m * aⁿ

Using the property b^p * b^q=b^p+q from exponents we can write the above equation as follows

xy=a^m+n

Convert back into logarithm

log_a xy = m+n

on substituting the values m= log_a x and n =log_a y we get

log_a xy = log_a x + log_a y

2) Quotient Rule: Log_a $\frac{x}{y}$ =log_a x- log_a y

proof:

let log_a x = m, log_a y = n

convert these logarithms into exponents

x=a^m and y=aⁿ

divide the above equations

$\frac{x}{y}$ = $\frac{a^m}{a^n}$

using the property b^p/b^q=b^p-q from exponents we can write the above equation as follows

$\frac{x}{y}$=a^m-n

convert into logarithm

log_a $\frac{x}{y}$=m-n

on substituting the m= log_a x & n =log_a y we get

log_a (xy)=log_a x - log_a y

3) Power Rules:

i) log_b $a^m$ = m log_b a

Proof:

let x=log_b a

convert logarithm into exponent

b^x=a

taking both sides of the equation to m’th power

(b^x)^m=a^m

using the property (a^m)ⁿ=a^mn from exponents we can write the above equation as follows

b^xm=a^m

convert back to logarithm

xm=log_b a^m

substitute x=log_b a back in the equation and interchanging

log_b a^m = m log_b a

ii) log_bⁿ a= $\frac{1}{n}$ log_b a

proof:

let log_bⁿ a=x

convert this into exponential form

a=(bⁿ)^x

use exponential property

(a^m)ⁿ=a^mn=(aⁿ)^m

a=(b^x)ⁿ

use the property if a^m=b then a=b^{$\frac{1}{m}$} from exponents we can write the above equation as follows

a^{$\frac{1}{n}$}=b^x

interchanging and converting into logarithmic form

b^x=a^{$\frac{1}{n}$}

x=log^b a^{$\frac{1}{n}$}

substitute the value of x=log_bⁿ a and use logarithmic property log_b a^m= m log_b a

log_bⁿ a= $\frac{1}{n}$ log_b a

4) log of 1 rule: log_b 1=0

Proof:

from exponents we have:

b⁰=1

convert this into logarithmic form

0=_b 1

5) Change of Base Rule: log_b a= $\frac{log_x a}{log_x b}$

Proof:

let log_x a=p, log_x b=q

Convert these logarithms into exponential forms

a=x^p and b=x^q

Substituting in the Left hand side of the to solve into right hand side of the Rule

L.H.S. : log_b a= log_xⁿ x^m

Using the power rule: log_n^q a^p= $\frac{p}{q}$ $log_b$ a

log_b a= $\frac{p}{q}$ log_x x

Using the Rule log_a a=1

Log_b a= $\frac{p}{q}$

Substituting the assumed p= $log_x$ a, q= $log_x$ b

Log_b a= $\frac{log_x a}{log_x b}$

6) log_b $\frac{1}{a}$ =log_{$\frac{1}{b}$} a= -log_b a

Proof:

i) Log_b (1/a)=log_b a^-1

Using the property log_a b^m=m log_a b =-log_b a

ii) log_{$\frac{1}{b}$} a=log_$b^{-1}$ a

log_b $\frac{1}{a}$ =log_{$\frac{1}{b}$} a= -log_b a

7) a^{log_a x} = x

Proof:

Let p=a^{log_a x}

Converting the above exponent into logarithm

Log_a p = log_a x

Using he property if log_b a= log_b c then a=c

p=x

But we assumed p=a^{log_a x}

Therefore a^{log_a x} = x

8.2.3 - Practice Problems on logarithms

I) write the following in the logarithmic format

$5^3$ = 125
$2^4$ = 16
$4^{-2}=\frac{1}{16}$
$(\frac{1}{2})^{-3}=8$
$\sqrt {81}$ = 9
$x^{2a}=y$

II) convert the following logarithms into exponents

x= $log_3\ y$
$log_x\ 1$ =0
ln x = 5
$log\ 100=2$
$log_8\ 2=\frac{1}{3}$

III) Find the values of x

$log_5\ 25$ = x
$log_{100}\ 1000$ = x
$log_{23}\ 1$ = x
$log_3\ \frac{1}{27}$ = x
$log_x\ 49$ = 7
$log_9\ x$ = -3
$log_5\ \frac{1}{5}$ = x
$log_8\ x=1- \frac{1}{2}$
$log_{12}\ 1$ = x
$log_{32}\ 2$ = x
$log_4\ \frac{1}{32}$ = x
$log_{12}\ \frac{1}{144}$ = x

IV) Write the following expressions in terms of logs of x, y and z

log $x^3y$
log $xyz$
log $\sqrt[3]x y^2 z$
log $\frac{x}{z^3}y$
log $\sqrt[5]{x^3yz^2}$
log $x \sqrt{ \frac{\sqrt{x^3y^2}}{z^4}}$
log $(x^3y)^{\frac{1}{2}}$
log $\frac{x^3}{y}$

V) Solve the following logarithmic equations

ln x =-2
log (7x+2) = 2
log x + log (x+1) = log 4x
2log x = log 2 + log(3x-4)
$log_3\ (x+3) + log_3\ (x+2) - log_3\ 10 = log_3\ x$
$log_2\ x + log_2\ (x+3) = 1$

VI) If log 2= x, log 3 =y then express the following in x and y

log 24
log 225
log 150
$log_7\ 980$
log 343
log 12.5
log 0.2
log 15
log 2.5
$log_7\ 3.5$

VII) Solve the following equations

$3^x$ - 1 =8
$2^{2x}-2^x-6$=0
$3^{1-x}=5^x$
$3^{2x-1}+3{x+2}-18$=0
$e^{2x}-2e^{x}-15$=0

8.3 - progression

8.3.1 - Arithmetic progression

A progression in which the difference between any two consecutive terms his always as same fixed quantity, such a progression is called arithmetic progression or simply AP. And the fixed difference is called common difference.

If a is the first time and t is the common difference then general AP form is:

a, a+t, a+2t, a+3t ……..

Where we denote terms as follows:

$a_1$ = a

$a_2$ = a+t

$a_3$ = a+ 2t and so on.

General term $a_n$ = a+(n-1)t . That is $a_n$ is the $n^th$ term with a being the first term and t is the common difference.

Examples of AP

1,2,3,4,5,6,7……..
1,3,5,7,9….
9,6,3,0,-3….
1,11,21,31,….

Sum of first n terms of AP

statment: Let $S_n$ be the sum of first n terms of AP $a_1,a_2,a_3,…….a_n$ then $S_n$ = $\frac{n}{2}[2a+(n-1)t]$ = $\frac{n}{2}[a_1+a_n]$

Proof:

Given the AP, $a_1,a_2,a_3,…….a_n$

$$S_n = a_1 + a_2 + a_3 + …… + a_n$$

$$S_n = a + (a+t) + (a+2t) + ….. + [a+ (n-1)t]$$

$$S_n = \sum_{r=1}^n[a+(n-1)t]$$

$$S_n=\sum_{r=1}^{n} a + \sum_{r=1}^{n}(n-1)t $$

As a and t are constants we can take them out of summation

$$S_n = a \sum_{r=1}^{n} 1 + t \sum_{r=1}^{n} (n-1)$$

$$Here \sum_{r=1}^{n} 1 = n \ because\ 1\ is\ added\ n\ times$$

$$S_n = a n + t [\sum_{r=1}^{n} n - \sum_{r=1}^{n} 1]$$

$$And\ we\ know\ that\ sum\ of\ first\ n\ natural\ numbers\ i.e. \sum_{r=1}^{n} n = \frac{n(n+1)}{2}$$

$$S_n = a n + t [ \frac{n(n+1)}{2} - n]$$

$$S_n = n [ a + t (\frac{(n+1)}{2} -1)]$$

$$S_n = n [ a + t \frac{(n-1)}{2} ]$$

$$S_n = \frac{n}{2} [ 2a + t (n-1)]$$

Points to be noted

If every term in an AP is multiplied by a constant k then new series will also be in AP with common difference kt
If every term in an AP is added with k then the new series will be in AP with common difference (k+t)

Arithmetic Mean (AM)

Definition: If $a_1, a_2, a_3, …..a_n$ are real numbers then $\frac{a_1 + a_2 + a_3 + ….. + a_n}{n}$ is called arithmetic mean of $a_1, a_2, a_3, …..a_n$.

In such AM we can observe something interesting. AM of any two numbers make another AP

i.e. if A is the AM of a and b then a, A, b are in AP

If b is the AM of a and c then a, b, c are in AP

Proof:

Given b is the arithematic mean of a and c that means

b = $\frac{a+c}{2}$

for a, b ,c to be in AP the difference between adjacent terms must be equal i.e.

b-a = c - b

as b = $\frac{a+c}{2}$

b-a = $\frac{a+c}{2}$ - a = $\frac{c-a}{2}$

and c-b = c - $\frac{a+c}{2}$ = $\frac{c-a}{2}$

we got b-a = c-b

therefore a,b,c are in AP.

8.3.2 - Geometric progression

Statement: A progression in which the ratio of two consecutive terms is always same constant, such a progression is called geometric progression or simply called GP. and the constant ratio is called common ratio

i.e. if $a_1,a_2,a_3,…….a_n$ are in GP then

$$\frac{a_2}{a_1}=\frac{a_3}{a_2}=\frac{a_4}{a_3}=\frac{a_5}{a_4}=………$$

General form of GP is $a, ar, ar^2, ar^3, ar^4 ……….$ where

first term is $a_1=a$

second term is $a_2=ar$

third term is $a_3=ar^2$ ……..

$n^{th}$ term is $a_n=ar^{n-1}$

Sum of n terms of a GP

Statement:

sum of first n terms of a geometric progression is $S_n= \frac{a(r^n-1)}{r-1}$ when r $\neq$ 1
if r=1 then $S_n$ = a + a + a + a + a + a +. . . . (n terms) = na
if |r|<1 and n = $\infty$ then $S_n = \frac{a}{1-r}$

Proof: Case-1)

$S_n= a+ar+ar^2+…..+ar^{n-1}$—-(1)

now multiply common ration r on both sides of the equation

$rS_n =ar+ar^2+ar^3+……+ar^n$—–(2)

let’s subtract equation (1) from equation (2)

$S_n = a+ar+ar^2+…..+ar^{n-1}$

$\underline{- rS_n = -(ar+ar^2+ar^3+……+ar^n) } $

$(1- r) S_n= [a- ar^n]$

$S_n= \frac{a(1- r^n)}{1-r}$

Case-3)

given |r|<1 and n = $\infty$

from the above proof we can write

$$ S_n= \frac{a(1- r^n)}{1-r}$$

as |r|<1 and n = $\infty$ then $r^n = 0$

$$ S_n= \frac{a(1- 0)}{1-r}$$

$$ S_n= \frac{a}{1-r}$$

Geometric mean (GM):

If $a_1,a_2,a_3,…….a_n$ are positive numbers then $(a_1a_2a_3…….a_n)^{1/n}$ is called geometric mean of $a_1,a_2,a_3,…….a_n$.

therefore GM of a and b is

$$GM=\sqrt{ab}$$

8.3.3 - Relationship between Arithematic and Geometric Means

Relation: For A being the arithematic mean and G being the Geometric mean of positive real numbers a and b then we habe A= $\frac{a+b}{2}$ and G= $\sqrt {ab}$ then $$A \geq G$$

Proof: Given two positive numbers a and b and A and G are arithematic and geometric means respectively.

$$A-G=\frac{a+b}{2}-\sqrt {ab}$$

$$A-G=\frac{a+b-2\sqrt {ab}}{2}$$

$$A-G=\frac{(\sqrt a)^2+(\sqrt b)^2-2\sqrt {ab}}{2}$$

using the algebraic formula $a^2+b^2-2ab=(a+b)^2$ in the equation

$$A-G=\frac{(\sqrt a-\sqrt b)^2}{2}$$

for two positive numbers a and b $(\sqrt a-\sqrt b)^2$ is always positve or zero, which can be represented as

$$A-G=\frac{(\sqrt a-\sqrt b)^2}{2} \geq 0$$

that implies

$$A \geq G$$

hence proved for two numbers

here $A = G$ would mean that a=b. that is for the same number both AM and GM will be same.

Similarly this can be extrapolated and can be expressed more generally as for $a_1, a_2, a_3, …..a_n$ where all being positive numbers then arithematic mean A = $\frac{a_1 + a_2 + a_3 + ….. + a_n}{n}$ and geometric mean $(a_1a_2a_3…….a_n)^{1/n}$. Then it can be shown that

$$A \geq G$$

and A=G when $a_1= a_2= a_3= …..=a_n$

8.4 - Proof of Pythagoras theorem

(There are thousands of independent ways of proving the pythagoras theorem. It was discovered by many mathematicians before and after pythagoras independently. where as proofs are done in various methods. People are so fascinated about this theorem becasue it is so important theorem, because it is useful in almost every subject directly or indirectly)

In this article we used one method to prove the theorem. hope yu find it intresting.

Statement: square of length hypotenuse of a right angled triangle is equal to the sum of squares of lengths other two sides

$$c^2 = a^2+b^2$$

Where c is the length of hypotenuse of a right angled triangle, while a and b are lengths of other two sides

Proof:

Consider the triangle ABC where $\angle$ B = $90^0$ as shown in the figure. Let’s draw BD line segment such that BD is perpendicular to side AC creating two right angled triangles BDA and BDC

$$Let’s\ compare\ \triangle ABC\ with\ \triangle\ ABD$$

$$\angle BAC = \angle BAD$$

$$\angle ABC = \angle ADB = 90^0$$

Therefore by Angle-Angle axiom we can say that $\triangle$ ABC $\sim$ $\triangle$ ADB

$$\triangle ABC \sim \triangle ADB$$

$$\frac{AB}{AD}=\frac{AC}{AB}$$

$$AB^2 = AC AD ————(1)$$

Let’s compare \triangle ABC with $\triangle$ BDC

$$\angle BCA = \angle BCD$$

$$\angle ABC = \angle BDC = 90^0$$

Therefore by Angle-Angle axiom we can say that $\triangle$ ABC $\sim$ $\triangle$ BDC

$$\triangle ABC \sim \triangle BDC$$

$$\frac{BC}{DC}=\frac{AC}{BC}$$

$$BC^2 = DC AC ————(2)$$

From (1) and (2), Add (1) and (2)

$$AB^2 + BC^2 = AC AD + DC AC$$

$$AB^2 + BC^2 = AC ( AD + AC)$$

From the diagram we can see that AC = AD + AC

$$AB ^2 + BC^2 = AC ^2$$

Hence proved that in a right angled triangle square of length of hypotenuse is equal to the sum of squares of the lengths of other two sides.

8.5 - triangle

8.5.1 - Area of triangle proof

There are many formulae for the area of triangle, in this article we would like to prove the following formula of area of triangle. which is more commonly used in mathematics.

$1)\Delta = \frac{1}{2} base * height$ where base is one side of the triangle and height is the altitude on the considered side from the opposite vertex.

1) Proof of $\Delta\ = 1/2\ base * height$

Proof:

Let’s prove the formula using normal geometry. For that let’s consider a paralellogram ABCD as shown in the figure. As we can see in the diagram there is a triangle ABC in paralelllogram ABCD. since $\triangle ABC$ and $\triangle CDA$ have equal lenghts of sides both the triangles, that means both triangles are congruent, which means areas of $\triangle ABC$ and $\triangle CDA$ must be equal. that means area of paralellogram is twice the area of $\triangle ABC$ or $\triangle ADC$.

therefore area of $\triangle$ ABC = $\frac{1}{2}$ area of ABCD

Let’s not use the area of paralellogram formula blindly instead let’s understand how that formula is built. In order to find the area of paralellogram let’s draw DE and CF line segments perpendicular the side AB as shown below.

Now look at the diagram we have produced two triangles $\triangle$ ADE and $\triangle$ BCF, let’s prove that these two triangles are congruent.

$$In\ \triangle\ ADE\ and\ \triangle\ BCF$$

since AD and BC are opposite sides of paralellogram $$AD=BC$$

since AD $\parallel$ BC and AF is transversal corresponding angles must be equal, $$\angle DAE= \angle CBF$$

since $\angle DEA$= $\angle CFB$= $90^0$ and $\angle DAE= \angle CBF$ then $$\angle ADE=\angle BCF$$

by Angle Side Angle axiom $\triangle ADE$ and $\triangle BCF$ are congruent. By this we can say that:

$$area\ of\ paralellogram\ ABCD = area\ of\ rectangle\ EFCD$$

therefore, $$area\ of\ paralellogram\ ABCD = length * breadth$$

from the diagram we can understand that length of rectangle EFCD is base of $\Delta$ ABC and breadth of rectangle EFCD is altitude(height) of $\Delta$ ABC

therefore area of triangle ABC = $\frac{1}{2}$ area of paralellogram ABCD = $\frac{1}{2} base * height$

8.6 - Trigonometry

this is a Trigononometry Docs page

8.6.1 - Trigonometric Identities

1) $sin^2$ x + $cos^2$ x = 1

proof:

To prove this identity let us use pythagoras theorem. consider a right angled triangle as shown in the figure. It is a triangle ABC with right angle at the vertex B and angle x at vertex C.

According to pythagoras theorem we can write

triangle ABC

$AB^2$ + $BC^2$ = $AC^2$

Divide with $AC^2$ on both sides.

$\frac{AB^2+BC^2}{AC^2}$ = 1

$\frac{AB^2}{AC^2}$ + $\frac{BC^2}{AC^2}$ = 1

by definition sin x = $\frac{opposite\ side\ to \ angle\ x}{hypotenuse}$ = $\frac{AB}{AC}$ and cos x = $\frac{adjacent \ side \ to\ angle \ x}{hypotenuse}$ = $\frac{CB}{AC}$

Therefore $sin^2$ x + $cos^2$ x = 1

I) Practice problems

If sin x = $\frac{4}{5}$ find the value of cos x provided x is acute angle.
find the value of $sin^2$ $50^0$ + $sin^2$ $40^0$
find th value of $sin^2$ $18^0$ + $sin^2$ $72^0$ + $sin^2$ $108^0$ + $sin^2$ $162^0$

Solutions at the end of the article.

2) $sec^2$ x - $tan^2$ x =1

proof:

To prove this identity let us use pythagoras theorem. consider a right angled triangle as shown in the figure. It is a triangle ABC with right angle at the vertex B and angle x at vertex C.

triangle ABC

According to pythagoras theorem we can write

$AB^2$ + $BC^2$ = $AC^2$

Divide with $BC^2$ on both sides, so that we can get tan x and sec x functions in the equation.

$\frac{AB^2}{BC^2}$ + 1 = $\frac{AC^2}{BC^2}$

by definition tan x = $\frac{opposite\ side\ to\ angle\ x}{adjacent\ side\ to\ angle\ x}$ = $\frac{AB}{BC}$ and sec x = $\frac{hypotenuse}{adjacent\ side\ to\ angle\ x}$ = $\frac{AC}{BC}$

Therefore $tan^2$ x + 1 = $sec^2$ x

II) solve the following problem

If sec A + tan A = 4, find the value of sin A.
Prove that $\frac{tan\ A+sec\ A -1}{tan\ A-sec\ A +1}$ = $\frac{1+sin\ A}{cos\ A}$

Solutions at the end of the article.

3) $cosec^2$ x - $cot^2$ x =1

proof:

To prove this identity let us use pythagoras theorem. consider a right angled triangle as shown in the figure. It is a triangle ABC with right angle at the vertex B and angle x at vertex C.

triangle ABC

According to pythagoras theorem we can write

$AB^2$ + $BC^2$ = $AC^2$

Divide with $AB^2$ on both sides, so that we can get tan x and sec x functions in the equation.

$\frac{BC^2}{AB^2}$ + 1 = $\frac{BC^2}{AC^2}$

by definition cot x = $\frac{adjacent\ side\ to\ angle\ x}{opposite\ side\ to\ angle\ x}$ = $\frac{BC}{AB}$ and cosec x = $\frac{hypotenuse}{opposite\ side\ to\ angle\ x}$ = $\frac{BC}{AC}$

Therefore $cosec^2$ x - $cot^2$ x =1

III) solve the following

find the value of (sin x + cosec x) $^2$ + (cos x + sex x) $^2$ - (tan x + cot x) $^2$
If Cosex A + Cot A = $\frac{2}{3}$ find the value of Sin A

Solutions at the end of the article.

Solutions

I)

ans= $\frac{3}{5}$

sol: Given Sin x = $\frac{4}{5}$

using the identity of trigonometry

$sin^2$ x + $cos^2$ x = 1

$\frac{4^2}{5^2}$ + $cos^2$ x = 1

$cos^2$ x = $\frac{9}{25}$

cos x = + $\frac{3}{5}$ or - $\frac{3}{5}$

ans= 1

sol:

$sin^2$ $50^0$ + $sin^2$ $40^0$

= $sin^2$ $50^0$ + $sin^2$ $(90^0-50^0)$

= $sin^2$ $50^0$ + $cos^2$ $50^0$

use the identity of trigonometry $sin^2$ x + $cos^2$ x = 1

= 1

ans= 2

sol:

$sin^2$ $18^0$ + $sin^2$ $72^0$ + $sin^2$ $108^0$ + $sin^2$ $162^0$

= $sin^2$ $18^0$ + $sin^2$ $72^0$ + $sin^2$ $(90^0+18^0)$ + $sin^2$ $(90^0+72^0)$

= $sin^2$ $18^0$ + $sin^2$ $72^0$ + $cos^2$ $18^0$ + $cos^2$ $72^0$

= ($sin^2$ $18^0$ + $cos^2$ $18^0$ ) + ( $sin^2$ $72^0$ + $cos^2$ $72^0$ )

use the identity of trigonometry $sin^2$ x + $cos^2$ x = 1

= 2

II)

Ans= $\frac{15}{17}$

Sol:

given sec A+tan A=4

let’s use the trigonometric identity

$sec^2$ x - $tan^2$ x =1

use the algebraic equation $a^2-b^2$=(a+b)(a-b)

(sec A + tan A)(Sec A - tan A) =1

4*(Sec A - tan A) =1

Sec A - tan A = $\frac{1}{4}$

adding the two equations

(Sec A + tan A)+(Sec A - tan A)= 4+ $\frac{1}{4}$

sec A = $\frac{17}{8}$

similarly subtracting the two equations

(Sec A + tan A)-(Sec A - tan A) = 4- $\frac{1}{4}$

tan A = $\frac{15}{8}$

on divinding tan A with sec A we get sin A

sin A = $\frac{15}{8}$ * $\frac{8}{17}$

sin A = $\frac{15}{17}$

Sol:

Consider LHS

$\frac{tan\ A+sec\ A -1}{tan\ A-sec\ A +1}$

multiply and divide the fraction with $tan\ A+sec\ A +1$

= $\frac{ [ (tan\ A+sec\ A) -1 ][(tan\ A+sec\ A) +1 ]}{[(tan\ A+1)-sec\ A][(tan\ A+1) +sec\ A]}$

= $\frac{ (tan\ A+sec\ A)^2 -1}{(tan\ A+1)^2-sec^2\ A}$

= $\frac{tan^2\ A+ sec^2\ A+2tan\ A sec\ A -1}{tan^2\ A+2tan\ A+1-sec^2\ A}$

use the trigonometric identity $sec^2$ x - $tan^2$ x =1 which can also be written as $sec^2$ x -1= $tan^2$ x

= $\frac{2tan^2\ A +2tan\ A sec\ A}{2tan\ A+1-1}$

= tan A + sec A

= $\frac{sin\ A}{cos\ A}$ + $\frac{1}{cos\ A}$

= $\frac{1+sin\ A}{cos\ A}$

III)

ans = 5 Sol:

(sin x + cosec x) $^2$ + (cos x + sex x) $^2$ - (tan x + cot x) $^2$

= $sin^2$ x + $cosec^2$ x + 2+ $cos^2$ x+ $sec^2$ x+ 2 - $tan^2$ x - $cot^2$ x-2

= ( $sin^2$ x + $cos^2$ x)+( $cosec^2$ x- $cot^2$ x) + ( $sec^2$ x- $tan^2$ x) + 2

= 5

ans= $\frac{3}{13}$

Sol:

Cosex A + Cot A = $\frac{2}{3}$

let’s use the trigonometric identity

$cosec^2$ x - $cot^2$ x =1

use the algebraic equation $a^2-b^2$=(a+b)(a-b)

(Cosex A + Cot A)(Cosex A - Cot A)=1

$\frac{2}{3}$ (Cosex A - Cot A)=1

(Cosex A - Cot A)= $\frac{3}{2}$

add both the equations

(Cosex A + Cot A)+(Cosex A - Cot A)= $\frac{2}{3}$ + $\frac{3}{2}$

cosec A = $\frac{13}{3}$

sin A = $\frac{3}{13}$

8.6.2 - Trigonometric Identities (problem solving)

1) If cos 2A = $tan^2$ B then find the value of $tan^2$ A in terms of B.

Sol:

Given, cos 2A = $tan^2$ B

Use multiple angle formula of cos, i.e. cos 2A = $\frac{1- tan^2\ A}{1+ tan^2\ A}$

$\frac{1- tan^2\ A}{1+ tan^2\ A}$ = $tan^2$ B

1- $tan^2$ A = $tan^2$ B (1+ $tan^2$ A)

1- $tan^2$ A = $tan^2$ B + $tan^2$ B $tan^2$ A

1- $tan^2$ B = $tan^2$ A + $tan^2$ B $tan^2$ A

1- $tan^2$ B = $tan^2$ A(1- $tan^2$ B)

$\frac{1- tan^2\ B}{1+ tan^2\ B}$ = $tan^2$ A

Again use the formula cos 2A = $\frac{1- tan^2\ A}{1+ tan^2\ A}$

cos 2B = $tan^2$ A

Therefore $tan^2$ A = cos 2B

2) If a cos A - b sin A = c then find the value of a sin A + b cos A

Sol:

Given, a cos A - b sin A = c

Squaring on both sides

$(a cos A - b sin A)^2$ = $c^2$

$a^2\ cos^2$ A + $b^2\ sin^2$ A -2ab sin A cos A = $c^2$

$a^2\ (1-sin^2$ A) + $b^2\ (1-sin^2$ A) -2ab sin A cos A = $c^2$

$a^2$ - $a^2\ sin^2$ A + $b^2$ - $b^2\ sin^2$ A -2ab sin A cos A = $c^2$

$a^2$ + $b^2$ - ( $a^2\ sin^2$ A + $b^2\ sin^2$ A + 2ab sin A cos A) = $c^2$

(a sin A + b cos A $)^2$ = $a^2$ + $b^2$ - $c^2$

a sin A + b cos A = $\pm \sqrt{a^2 + b^2 - c^2}$

3) If sin A + cosec A = 2 then find the value of sin $^20$ A + cosec $^20$ A

Sol:

sin A + cosec A = 2

sin A + $\frac{1}{sin\ A}$ = 2

Sin $^2$ A + 1 = 2 sin A

Sin $^2$ A + 1 - 2 sin A = 0

(sin A - 1 $)^2$ = 0

Sin A = 1

So cosec A = 1

sin $^20$ A + cosec $^20$ A =1+1

sin $^20$ A + cosec $^20$ A = 2

4) If sin A + sin $^2$ A + sin $^3$ A = 1 find the value of cos $^6$ A - 4cos $^4$ A + 8cos $^2$

Sol:

Given, sin A + sin $^2$ A + sin $^3$ A = 1

sin A + sin $^3$ A = 1- sin $^2$ A

sin A + sin $^3$ A = cos $^2$ A

Square on both sides

( sin A + sin $^3$ A $)^2$ = cos $^4$ A

sin $^2$ A + 2sin $^4$ A + sin $^6$ A = cos $^4$ A

1-cos $^2$ A + 2(1-cos $^2$ A $)^2$ + (1-cos $^2$ A $)^3$ = cos $^4$ A

1-cos$^2$ A + 2-4cos $^2$ A + 2cos $^4$ A + 1- 3cos $^2$ A + 3cos $^4$ A - cos $^6$ A = cos $^4$ A

4 - cos $^6$ A + 4cos $^4$ A - 8cos $^2$ = 0

Therefore cos $^6$ A - 4cos $^4$ A + 8cos $^2$= 0

5) if sin A + cos A = m and sin $^3$ A + cos $^3$ A = n, then prove that m$^3$ -3m+2n=0

sol:

Given,

sin A + cos A = m and sin $^3$ A + cos $^3$ A = n

Consider the second equation

sin $^3$ A + cos $^3$ A = n

Use the algebraic formula a $^3$ A + b $^3$ A = (a+b)( a $^2$ - ab + b $^2$ )

(sin A + cos A)( sin $^2$ A - sin A cos A + cos $^2$ A) = n

But it is given that sin A + cos A = m and we know that sin $^2$ A + cos $^2$ A = 1

m(1 - sin A cos A) = n ————–(1)

Now let’s consider the given equation sin A + cos A = m and lets square on both sides of the equation

(sin A + cos A $)^2$ = m $^2$

sin $^2$ A + 2sin A cos A + cos $^2$ A = m $^2$

we know that sin $^2$ A + cos $^2$ A = 1

1 + 2sin A cos A = m $^2$ sin A cos A = $\frac{m^2 -1}{2}$ ————(2)

Now from (1) and (2)

m(1 - $\frac{m^2 -1}{2}$) = n

2m - m $^3$ +m = 2n

m$^3$ -3m+2n=0

8.6.3 - Trigonometric ratios of multiple of Angle-2A

If A an angle, then its integral multiples 2A, 3A, 4A….. are called multiple angles of A and the multiple angles of A

Trigonometric ratios of Multiple angles of A Formulae

Sin 2A = 2Sin A cos A
cos 2A = $cos^2$ A - $sin^2$ A = 2 $cos^2$ A-1 = 1- $sin^2$ A
tan 2A = $\frac{2tan\ A}{1-tan^2\ A}$
cot 2A = $\frac{cot^2\ A-1}{2cot\ A}$
Sin 2A = $\frac{2tan\ A}{1+tan^2\ A}$
cos 2A = $\frac{1-tan^2\ A}{1+tan^2\ A}$

1) Sin 2A = 2Sin A cos A

Proof:

We have a compound angle formula:

Sin(A+B)= sin A cos B + cos A sin B

If A=B

Sin 2A = sin A cos A + cos A sin A

Sin 2A = 2sin A cos A

2) cos 2A = $cos^2$ A - $sin^2$ A = 2$cos^2$ A-1 = 1- $sin^2$ A

Proof:

i) We have a compound angle formula:

Cos(A+B)= Cos A cos B- Sin A Sin B

If A= B

Cos 2A = $cos^2$ A - $sin^2$ A

ii) Using the trigonometric identity $cos^2$ A + $sin^2$ A =1 in Cos 2A = $cos^2$ A - $sin^2$ A

Cos 2A = (1-$sin^2$ A)-$sin^2$ A

Cos 2A = 1-2$sin^2$ A

iii) Similarly Using the trigonometric identity $cos^2$ A + $sin^2$ A =1 in Cos 2A = $cos^2$ A - $sin^2$ A

Cos 2A = $cos^2$ A - $sin^2$ A

Cos 2A = $cos^2$ A - (1-$cos^2$ A)

Cos 2A = 2 $cos^2$ A -1

Therefore cos 2A = $cos^2$ A - $sin^2$ A = 2$cos^2$ A-1 = 1- $sin^2$ A

3) tan 2A = $\frac{2tan\ A}{1-tan^2\ A}$

Proof:

tan 2A = $\frac{sin\ 2A}{cos\ 2A}$

Using the multiple angle formulae Sin 2A=2Sin A cos A and cos 2A = $cos^2$ A - $sin^2$ A

tan 2A = $\frac{2Sin\ A cos\ A}{cos^2\ A - sin^2\ A }$

Divide $cos^2$ A on both numerator and denominator.

tan 2A = $\frac{\frac{2Sin\ A cos\ A}{cos^2\ A}}{\frac{cos^2\ A - sin^2\ A}{cos^2\ A }}$

tan 2A = $\frac{2tan\ A}{1-tan^2\ A}$

4) cot 2A = $\frac{cot^2\ A-1}{2cot\ A}$

Proof:

Cot 2A = $\frac{1}{tan\ 2A}$

Using tan 2A = $\frac{2tan\ A}{1-tan^2\ A}$ formula

Cot 2A = $\frac{1-tan^2\ A}{2tan\ A}$

= $\frac{1-\frac{1}{cot^2\ A}}{2\frac{1}{cot\ A}}$

cot 2A = $\frac{cot^2\ A-1}{2cot\ A}$

5) Sin 2A = $\frac{2tan\ A}{1+tan^2\ A}$

Proof:*

We have a trigonometric multiple angle formula:

Sin 2A=2Sin A cos A

Use trigonometric identity $cos^2$ A + $sin^2$ A =1

Sin 2A = $\frac{2Sin\ A cos\ A}{cos^2\ A + sin^2\ A}$

Divide with $cos^2$ A on numerator and denominator

Sin 2A = $\frac{\frac{2Sin\ A cos\ A}{cos^2\ A}}{\frac{cos^2\ A + sin^2\ A}{cos^2\ A} }$

Sin 2A = $\frac{2tan\ A}{1+tan^2\ A}$

6) cos 2A = $\frac{1-tan^2\ A}{1+tan^2\ A}$

Proof:

We have a trigonometric multiple angle formula:

cos 2A = $cos^2$ A - $sin^2$ A

Use trigonometric identity $cos^2$ A + $sin^2$ A =1

Cos 2A = $\frac{cos^2\ A-sin^2\ A}{cos^2\ A+sin^2\ A}$

Divide with $cos^2$ A on numerator and denominator

Cos 2A = $\frac{\frac{cos^2\ A-sin^2\ A}{cos^2\ A}}{\frac{cos^2\ A+sin^2\ A}{cos^\ A}}$

Cos 2A = $\frac{1-tan^2\ A}{1+tan^2\ A}$

8.6.4 - Trigonometric ratios of multiple of Angle-3A

Page contents:

proofs of sin 3A, cos 3A, tan 3A, cot 3A

Formulae:

Sin 3A = 3Sin A-4 $sin^3$ A
cos 3A = 4 $cos^3$ A - 3cos A
tan 3A = $\frac{3tan\ A-tan^3\ A}{1-3tan^2\ A}$
cot 3A = $\frac{3cot\ A-cot^3\ A}{1-3cot^\ A}$

1) Sin 3A = 3Sin A-4 $sin^3$ A

Proof:

Sin 3A = sin(A +2A)

use compund angle formula of sine function, i.e. Sin(A+B)= sinA cosB + cosA sinB

Sin 3A = sin 2A cos A + cos 2A sin A

use Sin 2A = 2sin A cos A; Cos 2A = 1-2 $sin^2 A$

Sin 3A = 2sin A $cos^2$ A + (1-2 $sin^2 A$) sin A

using the trigonometric identity $sin^2$ A + $cos^2$ A =1

sin 3A = 2sin A(1- $sin^2$ A) + sin A - 2 $sin^3 A$

Sin 3A = 3Sin A-4 $sin^3$ A

2) Cos 3A = 4 $cos^3$ A - 3cos A

Proof:

cos 3A = cos(A+2A)

use compund angle formula of sine function, i.e. cos(A+B)= cosA cosB + sinA sinB

cos(A+2A)= cosA cos2A + sinA sin 2A

use Sin 2A = 2sin A cos A; Cos 2A = 2 $cos^2 A$-1

cos 3A = cos A(2 $cos^2 A$-1) + sin A 2sin Acos A

cos 3A = 2 $cos^3 A$ -cos A +2 $sin^2$ A cos A

using the trigonometric identity $sin^2$ A + $cos^2$ A =1

cos 3A = 2 $cos^3 A$ -cos A +2(1- $cos^2$ A) cos A

Cos 3A = 4 $cos^3$ A - 3cos A

3) tan 3A = $\frac{3tan\ A-tan^3\ A}{1-3tan^2\ A}$

Proof:

tan 3A = tan(A+2A)

use compund angle formula of sine function, i.e. tan(A+B) = $\frac{tan\ A+tan\ B}{1-tan\ Atan\ B}$

tan 3A = $\frac{tan\ A+tan\ 2A}{1-tan\ Atan\ 2A}$

use the formula tan 2A = $\frac{2tan\ A}{1-tan^2\ A}$

tan 3A = $\frac{tan\ A+\frac{2tan\ A}{1-tan^2\ A}}{1-tan\ A\frac{2tan\ A}{1-tan^2\ A}}$

tan 3A = $\frac{3tan\ A-tan^3\ A}{1-3tan^2\ A}$

4) cot 3A = $\frac{3cot\ A-cot^3\ A}{1-3cot^\ A}$

Proof:

cot 3A = cot(A+2A)

use compund angle formula of sine function, i.e. cot(A+B) = $\frac{cot\ A cot\ B-1}{cot\ A+cot\ B}$

cot 3A = $\frac{cot\ A cot\ 2A-1}{cot\ A+cot\ 2A}$

use the formula cot 2A = $\frac{cot^2\ A-1}{2cot\ A}$

cot 3A = $\frac{cot\ A \frac{cot^2\ A-1}{2cot\ A}-1}{cot\ A+\frac{cot^2\ A-1}{2cot\ A}}$

cot 3A = $\frac{3cot\ A-cot^3\ A}{1-3cot^\ A}$

8.6.5 -

8.6.6 - Trigonometric transformations from sum( or difference) to product and vice-versa

Formulae

sum into product transformations:

2sin A cos B = sin(A+B) + sin(A-B)

2cos A sin B = sin(A+B) - sin(A-B)

2cos A cos B = cos(A-B) + cos(A+B)

2sin A sin B = cos(A-B) - cos(A+B)

product into sum transformations:

sin C + sin D = 2sin( $\frac{C+D}{2}$ ) cos( $\frac{C-D}{2}$ )

sin C - sin D = 2cos( $\frac{C+D}{2}$ ) sin( $\frac{C-D}{2}$ )

cos C + cos D = 2cos( $\frac{C+D}{2}$ ) cos( $\frac{C-D}{2}$ )

cos C - cos D = -2sin( $\frac{C+D}{2}$ ) sin( $\frac{C-D}{2}$ )

Proofs

2sin A cos B = sin(A+B) + sin(A-B)

Proof:

consider Right hand sideo of the statement

sin(A+B) + sin(A-B)

use the coumpund angle formulae i.e. sin(A+B) = sin A cos B + cos A sin B and sin(A-B) = sin A cos B - cos A sin B

sin(A+B) + sin(A-B) = (sin A cos B + cos A sin B) + (sin A cos B - cos A sin B)

sin(A+B) + sin(A-B) =2sin A cos B

hence 2sin A cos B = sin(A+B) + sin(A-B)

sin C + sin D = 2sin( $\frac{C+D}{2}$ ) cos( $\frac{C-D}{2}$ )

Proof:

consider Right hand sideo of the statement

2sin( $\frac{C+D}{2}$ ) cos( $\frac{C-D}{2}$ )

this looks liks the previous proof statement i.e 2sin A cos B = sin(A+B) + sin(A-B) where A = $\frac{C+D}{2}$ and B = $\frac{C-D}{2}$

2sin( $\frac{C+D}{2}$ ) cos( $\frac{C-D}{2}$ ) = sin( $\frac{C+D}{2}$ + $\frac{C-D}{2}$ ) cos ( $\frac{C+D}{2}$ - $\frac{C-D}{2}$ )

2sin( $\frac{C+D}{2}$ ) cos( $\frac{C-D}{2}$ ) = sin C + sin D

hence sin C + sin D = 2sin( $\frac{C+D}{2}$ ) cos( $\frac{C-D}{2}$ )

similalry rest of the proofs can be done

2cos A sin B = sin(A+B) - sin(A-B)

Proof:

consider Right hand sideo of the statement

sin(A+B) - sin(A-B)

use the coumpund angle formulae i.e. sin(A+B) = sin A cos B + cos A sin B and sin(A-B) = sin A cos B - cos A sin B

sin(A+B) sin(A-B) = (sin A cos B + cos A sin B) (sin A cos B - cos A sin B)

sin(A+B) + sin(A-B) = 2cos A sin B

hence 2cos A sin B = sin(A+B) - sin(A-B)

2cos A cos B = cos(A-B) + cos(A+B)

proof:

consider Right hand sideo of the statement

cos(A-B) + cos(A+B)

use the coumpund angle formulae i.e. cos(A+B) = cos A cos B - sin A sin B and cos(A-B) = cos A cos B + sin A sin B

cos(A-B) + cos(A+B) = (cos A cos B - sin A sin B) + (cos A cos B - sin A sin B)

cos(A-B) + cos(A+B) = 2cos A cos B

hence 2cos A cos B = cos(A-B) + cos(A+B)

2sin A sin B = cos(A-B) - cos(A+B)

proof:

consider Right hand sideo of the statement

cos(A-B) - cos(A+B)

use the coumpund angle formulae i.e. cos(A+B) = cos A cos B - sin A sin B and cos(A-B) = cos A cos B + sin A sin B

cos(A-B) - cos(A+B) = (cos A cos B - sin A sin B) - (cos A cos B - sin A sin B)

cos(A-B) - cos(A+B) = 2sin A sin B

hence 2sin A sin B = cos(A-B) - cos(A+B)

8.7 - Trigonometric Properties of triangle

8.7.1 - cos rules or Laws of cos

Cos rule or law of cos states that in A $\triangle$ ABC a, b and c are lengths of sides BC CA and AB. Then

$$a^2=b^2+c^2-2bc \ Cos \ A$$

$$b^2=a^2+c^2-2ac\ Cos\ B$$

$$c^2=a^2+b^2-2ab\ Cos\ C$$

$$Or$$

$$cos\ B= \frac{a^2+c^2-b^2}{2ac}$$

$$cos\ C= \frac{b^2+a^2-c^2}{2ba}$$

$$cos\ A= \frac{b^2+c^2-a^2}{2bc}$$

It seems like a different form of Pythagoras theorem for every triangle doesn’t it?

Proof:

Let’s consider a $\triangle$ ABC as shown in the figure with a, b and c being the lengths of the sides BC, CA and AB. we can represent the triangle vectorically as shown in the diagram. That means $\vec{AB}$, $\vec{BC}$ and $\vec{AC}$ are forming the triangle so we can understand that | $\vec{AB}$ | = c, | $\vec{BC}$ | = a and | $\vec{AC}$ | = b.

From the triangle we can write that

$$\vec{AC}=\vec{AB}+\vec{BC}$$

Apply mod on both sides

$$| \vec{AC} |=| \vec{AB} + \vec{BC} |$$

$$ b = | \vec{AB} + \vec{BC} |$$

From vectors we have $| \vec{x} + \vec{y} | = \sqrt {x^2+y^2+2xy\ cos(\vec{x},\vec{y})}$

To find the angle between $\vec{AB}$ and $\vec{BC}$ lets extend the $\vec{AB}$ so that both vector tails meet. Let $\alpha$ be the angle between those two vectors.

(note: we extended the vector to find the angle between two vectors by not considering B as angle, because angle between two vectors is angle made by vectors when both heads meet or tails meet but not one tail and one head)

From the above figure we can find $\alpha$

$$\alpha + B=180^0$$

$$\alpha =180^0 - B $$

Therefore,

$$ b = | \vec{AB} + \vec{BC} |$$

$$ b =\sqrt{ | \vec{AB}|^2 +| \vec{BC} |^2 +2 | \vec{AB} | |\vec{BC}| cos(\vec{AB} , \vec{BC})}$$

$$ b =\sqrt{ c^2 + a^2 +2 ca\ cos(\alpha)}$$

$$ b =\sqrt{ c^2 + a^2 +2 ca\ cos(180^0 - B)}$$

$$ b =\sqrt{ c^2 + a^2 -2 ca\ cos\ B}$$

$$b^2=a^2+c^2-2ac\ Cos\ B$$

Hence proved.

Now we can extend this proof by simplification and we can write

$$cos\ B= \frac{a^2+c^2-b^2}{2ac}$$

Similarly we can do the same for the other two angles and we can prove the other two equations of cos rule

8.7.2 - Sine rule or Laws of sine

Statement: In a $\triangle$ ABC with a, b and c are the lengths of sides BC, CA and AB respectively then

$$\frac{a}{Sin A}=\frac{b}{Sin B}=\frac{c}{Sin C}= 2R$$

Where R is the circum radius of the $\triangle$ ABC.

Therefore, $$a= 2R\ Sin\ A, b= 2R\ sin\ B, c= 2R\ sin\ C$$

Proof: Let’s consider the $\triangle$ ABC as shown in the figure circumscribed. O is the circum centre.

We know that a, b and c are lengths of sides of $\triangle$ ABC and R is the circum radius.

From the diagram we can see that BC is the chord of the circle. From circle properties we know that

$$\angle BOC = 2A$$

Let’s draw altitude to OM in $\triangle$OBC perpendicular to side BC. As OB and OC are also radii of the circle we can understand that $\triangle$OBC is an isosceles triangle. In an isosceles triangle altitude and median coincides. Therefore OM is not just altitude but also median. Therefore BM =$\frac{a}{2}$

Now let’s consider $\triangle$ OMB, since OBC is isosceles OM bisects $\angle$ BOC. And $\angle$ OMB =$90^0$

$$sin \ \angle BOM = \frac{MB}{OB}$$

$$sin \ A= \frac{a}{2R}$$

$$a=2R\ sin\ A$$

Similarly we can prove that c = 2R sin C and b= 2R sin B.

And we can write

$$\frac{a}{Sin A}=\frac{b}{Sin B}=\frac{c}{Sin C}= 2R$$

x²+	x+

Result
Root 1	Root 2