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






11 and 12 Grade Math

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