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Trigonometry

1: Trigonometric Identities
2: Trigonometric Identities (problem solving)
3: Trigonometric ratios of multiple of Angle-2A
4: Trigonometric ratios of multiple of Angle-3A
5:
6: Trigonometric transformations from sum( or difference) to product and vice-versa

this is a Trigononometry Docs page

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

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 = $\frac{3}{13}$

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

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

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

5 -

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)