Question

If $\left(2^n+1\right)\mathrm{\theta }=\pi ,$ then $2^n\cos \mathrm{\theta }\cos 2\mathrm{\theta }\cos 2^2\mathrm{\theta }...\cos 2^{n-1}\mathrm{\theta }=1$
(a) $-1$
(b) 1
(c) 1/2
(d) None of these

Answer 1

(b) 1

$$\begin{gathered} \left(2^n+1\right)\mathrm{\theta }=\mathrm{\pi }\left(\mathrm{Given}\right) \\ \Rightarrow 2^n\mathrm{\theta }+\mathrm{\theta }=\mathrm{\pi } \\ \Rightarrow 2^n\mathrm{\theta }=\mathrm{\pi }-\mathrm{\theta } \\ \Rightarrow \sin 2^n\mathrm{\theta }=\sin \left(\mathrm{\pi }-\mathrm{\theta }\right) \\ \Rightarrow \sin 2^n\mathrm{\theta }=\sin \mathrm{\theta }...\left(1\right) \end{gathered}$$


$$\begin{gathered} 2^n\cos \mathrm{\theta }\cos 2\mathrm{\theta }\cos 2^2\mathrm{\theta }...\cos 2^{n-1}\mathrm{\theta }=2^{\mathrm{n}}\times \frac{\sin 2^n\mathrm{\theta }}{2^n\sin \mathrm{\theta }} \\ =\frac{\sin 2^n\mathrm{\theta }}{\sin \mathrm{\theta }} \\ =\frac{\sin \mathrm{\theta }}{\sin \mathrm{\theta }}\left[\mathrm{From}\left(1\right)\right] \\ =1 \end{gathered}$$