Question

2 ( 1- sin^2theta) cos4theta =

[ Hint : write 1-sin^2theta=cos2 theta]​

Answer 1

Answer:

3

Step-by-step explanation:

$2(1-2sin^2\theta)cos4\theta \\= 2cos2\theta cos4\theta$since $1-sin^2x = cos2x$

$= cos6\theta + cos 2\theta$ since $cos x + cos y = 2cos\frac{x+y}{2}cos\frac{x-y}{2}$

Ignore the 2 before $sin^2 \theta$ in the question, may be a printing error.

Mark as brainliest if it helps :D

Answer 2

$\green{\large\underline{\sf{Given \:Question - }}}$

Evaluate

$\rm :\longmapsto\:2(1 - 2 {sin}^{2} \theta)cos4\theta$

$\rm \: \: \: \: \: \: \: \:(1) \: sin6\theta + cos2\theta$

$\rm \: \: \: \: \: \: \: \:(2) \: sin6\theta + sin2\theta$

$\rm \: \: \: \: \: \: \: \:(3) \: cos6\theta + cos2\theta$

$\rm \: \: \: \: \: \: \: \:(3) \: sin6\theta + sin2\theta$

$\red{\large\underline{\sf{Solution-}}}$

Given expression is

$\rm :\longmapsto\:2(1 - 2 {sin}^{2} \theta)cos4\theta$

We know,

$\red{ \boxed{ \sf{ \:1 - {2sin}^{2}\theta = cos2\theta }}}$

So, using this identity, we get

$\rm \: = \: 2 \: cos2\theta \: cos4\theta$

can be rewritten as

$\rm \: = \: 2 \: cos4\theta \: cos2\theta$

We know,

$\red{ \boxed{ \sf{ \:2cosx \: cosy = cos(x + y) + cos(x - y)}}}$

So, using this identity, we get

$\rm \: = \: cos(4\theta + 2\theta ) + cos(4\theta - 2\theta )$

$\rm \: = \: cos6\theta \: + \: cos2\theta$

Hence,

$\rm :\longmapsto\:\red{ \boxed{ \sf{ \:2(1 - 2 {sin}^{2} \theta)cos4\theta = cos6\theta +cos2 \theta}}}$

• So, option (3) is correct

Additional Information :-

$\red{ \boxed{ \sf{ \:2sinxcosy = sin(x + y) + sin(x - y)}}}$

$\red{ \boxed{ \sf{ \:2sinxsiny = cos(x - y) - cos(x + y)}}}$

$\red{ \boxed{ \sf{ \:sinx + siny = 2sin\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg]}}}$

$\red{ \boxed{ \sf{ \:sinx - siny = 2cos\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{x - y}{2} \bigg]}}}$

$\red{ \boxed{ \sf{ \:cosx - cosy = - \: 2sin\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{x - y}{2} \bigg]}}}$

$\red{ \boxed{ \sf{ \:cosx + cosy = 2cos\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg]}}}$