# sin 2A in Terms of A

We will learn to express trigonometric function of sin 2A in terms of A. We know if A is a given angle then 2A is known as multiple angles.

How to proof the formula of sin 2A is equals 2 sin A cos A?

We know that for two real numbers or angles A and B,

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

Now, putting B = A on both sides of the above formula we get,

sin (A + A) = sin A cos A + sin A cos A

⇒ sin 2A = 2 sin A cos A

Note: In the above formula we should note that the angle on the R.H.S. is half of the angle on L.H.S. Therefore, sin 60° = 2 sin 30° cos 30°.

The above formula is also known as double angle formulae for sin 2A.

Now, we will apply the formula of multiple angle of sin 2A in terms of A to solve the below problems.

1. Express sin 8A in terms of sin 4A and cos 4A

Solution:

sin 8A

= sin (2 ∙ 4A)

= 2 sin 4A cos 4A, [Since, we know sin 2A = 2 sin A cos A]

2. If sin A = $$\frac{3}{5}$$ find the values of sin 2A.

Solution:

Given, sin A = $$\frac{3}{5}$$

We know that, sin$$^{2}$$ A + cos$$^{2}$$ A = 1

cos$$^{2}$$ A = 1 - sin$$^{2}$$ A

cos$$^{2}$$ A = 1 - ($$\frac{3}{5}$$)$$^{2}$$

cos$$^{2}$$ A = 1 - $$\frac{9}{25}$$

cos$$^{2}$$ A = $$\frac{25 - 9}{25}$$

cos$$^{2}$$ A = $$\frac{16}{25}$$

cos A = √$$\frac{16}{25}$$

cos A = $$\frac{4}{5}$$

sin 2A

= 2 sin A cos A

= 2 ∙ $$\frac{3}{5}$$ ∙ $$\frac{4}{5}$$

= $$\frac{24}{25}$$

3. Prove that,16 cos $$\frac{2π}{15}$$  cos $$\frac{4π}{15}$$  cos $$\frac{8π}{15}$$   $$\frac{16π}{15}$$ = 1.

Solution:

Let, $$\frac{2π}{15}$$ = θ

L.H.S = 16 cos $$\frac{2π}{15}$$  cos $$\frac{4π}{15}$$  cos $$\frac{8π}{15}$$   $$\frac{16π}{15}$$ = 1.

= 16  cos θ cos 2θ cos 4θ cos 8θ, [Since, θ = $$\frac{2π}{15}$$]

= $$\frac{8}{sin θ}$$ (2 sin θ cos θ) cos 2θ cos 4θ cos 8θ

= $$\frac{4}{sin θ}$$ (2 sin 2θ cos 2θ) cos 4θ cos 8θ

= $$\frac{2}{sin θ}$$ (2 sin 4θ cos 4θ) cos 8θ

= $$\frac{1}{sin θ}$$ (2 sin 8θ cos 8θ)

= $$\frac{1}{sin θ}$$ ∙ sin 16θ

= $$\frac{1}{sin θ}$$ ∙ sin (15θ + θ)

= $$\frac{1}{sin θ}$$ ∙ sin (2π + θ), [Since, $$\frac{2π}{15}$$ = θ 15θ = 2π]

= $$\frac{1}{sin θ}$$ ∙ sin (θ), [Since, sin (2π + θ) = sin θ]

= 1 = R.H.S.                Proved

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