Question

$\left(1+\cos \frac{\pi }{8}\right)\left(1+\cos \frac{3\pi }{8}\right)\left(1+\cos \frac{5\pi }{8}\right)\left(1+\cos \frac{7\pi }{8}\right)$

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{1}{8}$
• D. $\frac{1}{16}$

Answer 1

The correct option is C

$\frac{1}{8}$

Explanation for the correct option:

Step 1. Solve the given expression:

$\left(1+\cos \frac{\mathrm{\pi }}{8}\right)\left(1+\cos \frac{3\mathrm{\pi }}{8}\right)\left(1+\cos \frac{5\mathrm{\pi }}{8}\right)\left(1+\cos \frac{7\mathrm{\pi }}{8}\right)$

$=\left(1+\cos \frac{\mathrm{\pi }}{8}\right)\left(1+\cos \frac{3\mathrm{\pi }}{8}\right)\left[1+\cos \left(\pi -\frac{3\mathrm{\pi }}{8}\right)\right]\left[1+\cos \left(\pi -\frac{\mathrm{\pi }}{8}\right)\right]$

$=\left(1+\cos \frac{\mathrm{\pi }}{8}\right)\left(1+\cos \frac{3\mathrm{\pi }}{8}\right)\left(1-\cos \frac{3\mathrm{\pi }}{8}\right)\left(1-\cos \frac{\mathrm{\pi }}{8}\right)$ $\left[\mathbf{\because }\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathbf{\pi }\mathbf{-}\mathbf{\theta }\mathbf{)}\mathbf{=}\mathbf{-}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\right]$

$=\left(1+{\cos }^2\frac{\mathrm{\pi }}{8}\right)\left(1+{\cos }^2\frac{3\mathrm{\pi }}{8}\right)$

$=\left({\sin }^2\frac{\mathrm{\pi }}{8}\right)\left[1-{\cos }^2\left(\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{8}\right)\right]$

$=\left({\sin }^2\frac{\mathrm{\pi }}{8}\right)\left({\cos }^2\frac{\mathrm{\pi }}{8}\right)$ $\left[\mathbf{\because }\mathbf{1}\mathbf{-}\mathbf{}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathbf{(}\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{-}\mathbf{\theta }\mathbf{)}\mathbf{=}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{\theta }\right]$

Step 2. Multiply and divide it by 4:

$=\frac{1}{4}\times \left(2\sin \frac{\mathrm{\pi }}{8}\cos \frac{\mathrm{\pi }}{8}\right)\left(2\sin \frac{\mathrm{\pi }}{8}\cos \frac{\mathrm{\pi }}{8}\right)$

$=\frac{1}{4}\times \left(\sin \frac{\mathrm{\pi }}{4}\right)\left(\sin \frac{\mathrm{\pi }}{4}\right)$ $\left[\mathbf{\because }\mathbf{2}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathbf{2}\mathit{\theta }\right]$

$=\frac{1}{4}\times {\sin }^2\frac{\mathrm{\pi }}{4}$

$$\begin{gathered} =\frac{1}{4}\times {\left(\frac{1}{\sqrt{2}}\right)}^2 \\ =\frac{\mathbf{1}}{\mathbf{8}} \end{gathered}$$ $\left[\mathbf{\because }\mathit{s}\mathit{i}\mathit{n}\frac{\mathbf{\pi }}{\mathbf{4}}\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\right]$

Hence, Option ‘C' is Correct.