Question

The value of $2cos\theta -cos3\theta -cos5\theta -16co{s}^3\theta si{n}^2$ is

• A. 2
• B. 1
• C. 0
• D. -1

Answer 1

The correct option is C

0

We have,

$2cos\theta -cos3\theta -cos5\theta -16co{s}^3\theta si{n}^2\theta =2cos\theta -3\theta -cos5\theta -16[\frac{cos3\theta +3cos\theta }{4}\times \frac{(1-cos2\theta )}{2}]=2cos\theta -cos3\theta -cos5\theta -2\left[\right(cos3\theta +3cos2\theta \left)\right]=2cos\theta -cos3\theta -cos5\theta -2[cos3\theta -cos3\theta cos2\theta +3cos\theta -3cos\theta cos2\theta ]=2cos\theta -cos3\theta -cos5\theta -2[cos3\theta +cos3\theta ]+2cos3\theta cos2\theta +3\left[2cos\theta cos2\theta \right]=2cos\theta -cos3\theta -cos5\theta -2[cos3\theta +cos3\theta ]+cos5\theta +cos\theta +3cos3\theta +3cos\theta [2cosAcosB=cos(A+B)+cos(A-B\left)\right]$

$=2cos\theta -cos3\theta -cos5\theta -2cos3\theta -6cos\theta +cos5\theta +cos\theta +3cos3\theta +3cos\theta$

= 0