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

1. Sin 2A = 2Sin A cos A

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

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

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

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

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