Question

If $cos 5\theta =5cos\theta ⁡-20cos^{3}⁡\theta +acos^{5}⁡\theta +b$ , then the value of $a+b$ is equal to

• A. $20$
• B. $16$
• C. $-16$
• D. $15$

Answer 1

Answer: (B)

$cos 5 \theta =cos ⁡ \left(4 \theta + \theta \right)$
$=cos 4\theta cos⁡\theta -sin⁡4\theta sin⁡\theta $
$=\left(2 \left(\left[2 c o s^{2} \theta - 1\right]\right)^{2} - 1\right)cos \theta -2.2sin⁡\theta cos⁡\theta \left[2 c o s^{2} \theta - 1\right]sin⁡\theta $
$=cos \theta \left[2 \left(4 \left(cos\right)^{4} ⁡ \theta + 1 - 4 \left(cos\right)^{2} ⁡ \theta \right) - 1\right]$ $-4sin^{2} \theta cos ⁡ \theta \left[2 cos^{2} ⁡ \theta - 1\right]$
$=8cos^{5} \theta - 8 cos^{3} ⁡ \theta + 2 cos ⁡ \theta - cos ⁡ \theta $ $-4\left(1 - \left(cos\right)^{2} \theta \right)cos \theta \left(2 \left(cos\right)^{2} ⁡ \theta - 1\right)$
$=8cos^{5} \theta -8cos^{3}⁡\theta +cos⁡\theta -4cos⁡\theta $ $\left[2 cos^{2} \theta - 1 - 2 cos^{4} ⁡ \theta + cos^{2} ⁡ \theta \right]$
$=16cos^{5} \theta - 20 cos^{3} ⁡ \theta + 5 cos ⁡ \theta $
Clearly, $a=16,b=0$