$\begin{array}{r} \left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{2 \pi}{8}\right) \\ \left(1+\cos…

$\begin{array}{r} \left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{2 \pi}{8}\right) \\ \left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{4 \pi}{8}\right) \\ \left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{6 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)= \end{array}$
  1. $\frac{1}{8}$
  2. $\frac{1}{16}$
  3. $\frac{1}{32}$
  4. $\frac{1}{64}$

Solution

Given that, $ \begin{aligned} & \left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{2 \pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right) \\ & \left(1+\cos \frac{4 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{6 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right) \\ & =\left[1+\cos \frac{\pi}{8}\right]\left[1+\cos \frac{\pi}{4}\right]\left[1+\cos \frac{3 \pi}{8}\right]\left[1+\cos \frac{\pi}{2}\right] \\ & {\left[1+\cos \left(\pi-\frac{3 \pi}{8}\right)\right]\left[1+\cos \frac{3 \pi}{4}\right]\left[1+\cos \left(\pi-\frac{\pi}{8}\right)\right]} \end{aligned} $ $\begin{aligned} & =\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{\pi}{4}\right)\left(1+\cos \frac{3 \pi}{8}\right) \\ & =\left(1+\cos \frac{\pi}{8}\right)\left(1-\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{3 \pi}{4}\right)\left(1-\cos \frac{\pi}{8}\right) \\ & =\left(1-\cos ^2 \frac{\pi}{8}\right)\left(1-\cos ^2 \frac{3 \pi}{8}\right)\left(1-\frac{1}{2}\right) \\ & =\sin ^2 \frac{\pi}{8} \cdot \sin ^2 \frac{3 \pi}{8} \cdot \frac{1}{2}=\frac{1}{4}\left(2 \sin ^2 \frac{\pi}{8}\right)\left(2 \sin ^2 \frac{3 \pi}{8}\right) \cdot \frac{1}{2} \\ & =\frac{1}{4}\left(1-\cos \frac{\pi}{4}\right)\left(1-\cos \frac{3 \pi}{4}\right) \cdot \frac{1}{2} \\ & =\frac{1}{4} \cdot\left(1-\frac{1}{\sqrt{2}}\right)\left(1+\frac{1}{\sqrt{2}}\right) \cdot \frac{1}{2} \\ & =\frac{1}{4}\left(1-\frac{1}{2}\right) \cdot \frac{1}{2}=\frac{1}{4} \cdot \frac{1}{2} \cdot \frac{1}{2}=\frac{1}{16} .\end{aligned}$

Asked in: AP EAMCET 2018 (24 Apr Shift 1)

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