Question

The value of $2\cos \mathrm{\theta }-\cos 3\mathrm{\theta }-\cos 5\mathrm{\theta }-16{\cos }^3\mathrm{\theta }{\sin }^2\mathrm{\theta }$
(a) 2
(b) 1
(c) 0
(d) −1

Answer 1

(c) 0

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ 2\mathrm{cos\theta }-\cos 3\mathrm{\theta }-\cos 5\mathrm{\theta }-16{\cos }^3\theta {\sin }^2\mathrm{\theta } \\ =2\mathrm{cos\theta }-\cos 3\mathrm{\theta }-\cos 5\mathrm{\theta }-16\left[\frac{\cos 3\mathrm{\theta }+3\mathrm{cos\theta }}{4}\times \frac{\left(1-\cos 2\mathrm{\theta }\right)}{2}\right] \\ =2\mathrm{cos\theta }-\cos 3\mathrm{\theta }-\cos 5\mathrm{\theta }-2\left[\left(\cos 3\mathrm{\theta }+3\mathrm{cos\theta }\right)\left(1-\cos 2\mathrm{\theta }\right)\right] \\ =2\mathrm{cos\theta }-\cos 3\mathrm{\theta }-\cos 5\mathrm{\theta }-2\left[\cos 3\mathrm{\theta }-\cos 3\mathrm{\theta cos}2\mathrm{\theta }+3\mathrm{cos\theta }-3\mathrm{cos\theta cos}2\mathrm{\theta }\right] \\ =2\mathrm{cos\theta }-\cos 3\mathrm{\theta }-\cos 5\mathrm{\theta }-2\left[\cos 3\mathrm{\theta }+3\mathrm{cos\theta }\right]+2\cos 3\mathrm{\theta cos}2\mathrm{\theta }+3\left[2\mathrm{cos\theta cos}2\mathrm{\theta }\right] \\ =2\mathrm{cos\theta }-\cos 3\mathrm{\theta }-\cos 5\mathrm{\theta }-2\left[\cos 3\mathrm{\theta }+3\mathrm{cos\theta }\right]+\cos 5\mathrm{\theta }+\mathrm{cos\theta }+3\cos 3\mathrm{\theta }+3\mathrm{cos\theta } \\ \left[2\mathrm{cosAcosB}=\cos \left(\mathrm{A}+\mathrm{B}\right)+\cos \left(\mathrm{A}-\mathrm{B}\right)\right] \\ =2\mathrm{cos\theta }-\cos 3\mathrm{\theta }-\cos 5\mathrm{\theta }-2\cos 3\mathrm{\theta }-6\mathrm{cos\theta }+\cos 5\mathrm{\theta }+\mathrm{cos\theta }+3\cos 3\mathrm{\theta }+3\mathrm{cos\theta } \\ =0 \end{gathered}$$