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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 = 35 find the values of sin 2A.

Solution:

Given, sin A = 35

We know that, sin2 A + cos2 A = 1

                                  cos2 A = 1 - sin2 A

                                  cos2 A = 1 - (35)2

                                  cos2 A = 1 - 925

                                  cos2 A = 25925

                                  cos2 A = 1625

                                  cos A = √1625

                                  cos A = 45

   sin 2A

= 2 sin A cos A

= 2 ∙ 3545

= 2425 


3. Prove that,16 cos 2π15  cos 4π15  cos 8π15   16π15 = 1.

Solution: 

Let, 2π15 = θ

L.H.S = 16 cos 2π15  cos 4π15  cos 8π15   16π15 = 1.

= 16  cos θ cos 2θ cos 4θ cos 8θ, [Since, θ = 2π15]

= 8sinθ (2 sin θ cos θ) cos 2θ cos 4θ cos 8θ 

= 4sinθ (2 sin 2θ cos 2θ) cos 4θ cos 8θ 

= 2sinθ (2 sin 4θ cos 4θ) cos 8θ 

= 1sinθ (2 sin 8θ cos 8θ)

= 1sinθ ∙ sin 16θ

= 1sinθ ∙ sin (15θ + θ)

= 1sinθ ∙ sin (2π + θ), [Since, 2π15 = θ 15θ = 2π]

= 1sinθ ∙ sin (θ), [Since, sin (2π + θ) = sin θ]

= 1 = R.H.S.                Proved

 Multiple Angles






11 and 12 Grade Math

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