Q61 / 160 · TS EAMCET 2018 (05 May Shift 2) · Mathematics

If $\cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15}$ $\cos \frac{5 \pi}{15} \cos \frac{7 \pi}{15} \cos \frac{30 \pi}{15}=x$, then $\frac{1}{8 x}=$

  • Correct answer
    A
    4
  • B
    $\frac{1}{4}$
  • C
    8
  • D
    $\frac{4}{3}$

Solution

We have, $$ \begin{aligned} & \Rightarrow \cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{5 \pi}{15} \cos \frac{7 \pi}{15} \cos \frac{30 \pi}{15}=x \\ & \Rightarrow \cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{\pi}{3} \cos \frac{7 \pi}{15} \cos 2 \pi=x \\ & \Rightarrow \quad \cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{7 \pi}{15}=2 x \\ & {\left[\because \cos \frac{\pi}{3}=\frac{1}{2} \text { and } \cos 2 \pi=1\right]} \end{aligned} $$ $$ \begin{aligned} & \Rightarrow \cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \left(\pi-\frac{8 \pi}{15}\right)=2 x \\ & \Rightarrow \quad-\cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15}=2 x \\ & \Rightarrow-\cos \frac{\pi}{15} \cos \frac{2 \pi}{15} \cos 2^2 \frac{\pi}{15} \cdot \cos 2^3 \cdot \frac{\pi}{15}=2 x \\ & \Rightarrow \quad-\frac{\sin 2^4 \frac{\pi}{15}}{24} \sin \frac{\pi}{15} \\ & \Rightarrow \quad\left[\because \cos A \cos 2 A \cos 2^2 A \ldots \cos 2^{14-1} A=\frac{\sin 2^{14} A}{2^n \sin A}\right] \\ & \Rightarrow 2 x=-\frac{\sin \frac{16 \pi}{15}}{16 \sin \frac{\pi}{15}} \Rightarrow 2 x=-\frac{\sin \left(\pi+\frac{\pi}{15}\right)}{16 \sin \frac{\pi}{15}}=\frac{\sin \frac{\pi}{15}}{16 \sin \frac{\pi}{15}} \\ & \Rightarrow 2 x=\frac{1}{16} \Rightarrow x=\frac{1}{32} \Rightarrow \frac{1}{x}=32 \Rightarrow \frac{1}{8 x}=4 \end{aligned} $$

Hence, the correct answer is option: 'A'.

Relevant Questions