Q62 / 160 · TS EAMCET 2018 (05 May Shift 1) · Mathematics

$\cos \left(\frac{\pi}{7}\right) \cos \left(\frac{2 \pi}{7}\right) \cos \left(\frac{4 \pi}{7}\right)=$

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

Solution

We have, $$ \cos \frac{\pi}{7} \cdot \cos \frac{2 \pi}{7} \cdot \cos \frac{4 \pi}{7} $$ Let $A=\frac{\pi}{7} \Rightarrow \pi=7 A$ $$ =\cos A \cos 2 A \cos 4 A=\cos A \cos 2 A \cos 2^2 A $$ Now, we will use the formula $$ \begin{aligned} & \cos A \cos 2 A \cos 2^2 A \cos 2^3 A \ldots \ldots \cos ^{n-1} A=\frac{\sin 2^n A}{2^n \sin A} \\ & \therefore \cos A \cos 2 A \cos 2^2 A=\frac{\sin 2^3 A}{2^3 \sin A} \\ & =\frac{\sin 8 A}{8 \sin A}=\frac{\sin (7 A+A)}{8 \sin A}=\frac{\sin (\pi+A)}{8 \sin A} \quad[\because 7 A=\pi] \\ & =\frac{-\sin A}{8 \sin A} \quad[\because \sin (\pi+\theta)=-\sin \theta] \\ & =-\frac{1}{8} \end{aligned} $$

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

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