Question

302.Show that : 4 sinthetha cos³thetha - 4 costhetha sin³theta = sin 4thetha

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Answer 1

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Answer 2

Step-by-step explanation:

Given : We have a trigonometric equation .

To prove :  $4sin\theta*cos^{3} \theta-4cos\theta*sin^{3} \theta=sin(4\theta)$

Trigonometric identities used :

1. $cos^{2}\theta-sin^2\theta=cos(2\theta)$
2. $2*sin\theta*cos\theta=sin(2\theta)$

• Proof :

We have to prove ;

⇒  $4sin\theta*cos^{3} \theta-4cos\theta*sin^{3} \theta=sin(4\theta)$

Let's take L.H.S separately,

⇒  $4sin\theta*cos^{3} \theta-4cos\theta*sin^{3} \theta$

taking $4*sin\theta*cos\theta$ common from L.H.S,

⇒  $4*sin\theta*cos\theta[cos^2\theta-sin^2\theta]$

by using identity (1),

⇒  $4*sin\theta*cos\theta *cos(2\theta)$

rearranging equation so that we can apply identity (2),

⇒  $2*(2*sin\theta*cos\theta )*cos(2\theta)$

⇒  $2*sin(2\theta)*cos(2\theta)$

again using identity (2) we get L.H.S as,

⇒  $sin(4\theta)$

Hence proved that $L.H.S =R.H.S$ or  $4sin\theta*cos^{3} \theta-4cos\theta*sin^{3} \theta=sin(4\theta)$ .