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

● Multiple Angles

• sin 2A in Terms of A
• cos 2A in Terms of A
• tan 2A in Terms of A
• sin 2A in Terms of tan A
• cos 2A in Terms of tan A
• Trigonometric Functions of A in Terms of cos 2A
  
• sin 3A in Terms of A
• cos 3A in Terms of A
• tan 3A in Terms of A
• Multiple Angle Formulae

11 and 12 Grade Math

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