Question

The value of $\sin \left(\frac{\mathrm{\pi }}{14}\right)\sin \left(\frac{3\mathrm{\pi }}{14}\right)\sin \left(\frac{5\mathrm{\pi }}{14}\right)\sin \left(\frac{7\mathrm{\pi }}{14}\right)\sin \left(\frac{9\mathrm{\pi }}{14}\right)\sin \left(\frac{11\mathrm{\pi }}{14}\right)\sin \left(\frac{13\mathrm{\pi }}{14}\right)$ is equal to

• A. $\frac{1}{8}$
• B. $\frac{1}{16}$
• C. $\frac{1}{32}$
• D. $\frac{1}{64}$

Answer 1

The correct option is D

$\frac{1}{64}$

Step 1: Simplify the expression by the formula $\sin \left(\mathrm{\pi }-\mathrm{\theta }\right)=\sin \left(\theta \right)$

The given trigonometric expression: $\sin \left(\frac{\mathrm{\pi }}{14}\right)\sin \left(\frac{3\mathrm{\pi }}{14}\right)\sin \left(\frac{5\mathrm{\pi }}{14}\right)\sin \left(\frac{7\mathrm{\pi }}{14}\right)\sin \left(\frac{9\mathrm{\pi }}{14}\right)\sin \left(\frac{11\mathrm{\pi }}{14}\right)\sin \left(\frac{13\mathrm{\pi }}{14}\right)$.

$$\begin{gathered} \sin \left(\frac{\mathrm{\pi }}{14}\right)\sin \left(\frac{3\mathrm{\pi }}{14}\right)\sin \left(\frac{5\mathrm{\pi }}{14}\right)\sin \left(\frac{7\mathrm{\pi }}{14}\right)\sin \left(\frac{9\mathrm{\pi }}{14}\right)\sin \left(\frac{11\mathrm{\pi }}{14}\right)\sin \left(\frac{13\mathrm{\pi }}{14}\right) \\ =\sin \left(\frac{\mathrm{\pi }}{14}\right)\sin \left(\frac{3\mathrm{\pi }}{14}\right)\sin \left(\frac{5\mathrm{\pi }}{14}\right)\sin \left(\frac{\mathrm{\pi }}{2}\right)\sin \left(\mathrm{\pi }-\frac{5\mathrm{\pi }}{14}\right)\sin \left(\mathrm{\pi }-\frac{3\mathrm{\pi }}{14}\right)\sin \left(\mathrm{\pi }-\frac{\mathrm{\pi }}{14}\right) \\ =\sin \left(\frac{\mathrm{\pi }}{14}\right)\sin \left(\frac{3\mathrm{\pi }}{14}\right)\sin \left(\frac{5\mathrm{\pi }}{14}\right)\left(1\right)\sin \left(\frac{5\mathrm{\pi }}{14}\right)\sin \left(\frac{3\mathrm{\pi }}{14}\right)\sin \left(\frac{\mathrm{\pi }}{14}\right)\left[\because \sin \left(\mathrm{\pi }-\theta \right)=\sin \left(\theta \right)\right] \\ ={\left[\sin \left(\frac{\mathrm{\pi }}{14}\right)\sin \left(\frac{3\mathrm{\pi }}{14}\right)\sin \left(\frac{5\mathrm{\pi }}{14}\right)\right]}^2 \end{gathered}$$

Step 2: Simplify the expression by the formula $\sin \left(\frac{\mathrm{\pi }}{2}-\mathrm{\theta }\right)=\cos \left(\theta \right)$

$$\begin{gathered} \sin \left(\frac{\mathrm{\pi }}{14}\right)\sin \left(\frac{3\mathrm{\pi }}{14}\right)\sin \left(\frac{5\mathrm{\pi }}{14}\right)\sin \left(\frac{7\mathrm{\pi }}{14}\right)\sin \left(\frac{9\mathrm{\pi }}{14}\right)\sin \left(\frac{11\mathrm{\pi }}{14}\right)\sin \left(\frac{13\mathrm{\pi }}{14}\right)={\left[\sin \left(\frac{\mathrm{\pi }}{14}\right)\sin \left(\frac{3\mathrm{\pi }}{14}\right)\sin \left(\frac{5\mathrm{\pi }}{14}\right)\right]}^2 \\ ={\left[\sin \left(\frac{\mathrm{\pi }}{2}-\frac{3\mathrm{\pi }}{7}\right)\sin \left(\frac{\mathrm{\pi }}{2}-\frac{2\mathrm{\pi }}{7}\right)\sin \left(\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{7}\right)\right]}^2 \\ ={\left[\cos \left(\frac{3\mathrm{\pi }}{7}\right)\cos \left(\frac{2\mathrm{\pi }}{7}\right)\cos \left(\frac{\mathrm{\pi }}{7}\right)\right]}^2\left[\because \sin \left(\frac{\mathrm{\pi }}{2}-\theta \right)=\cos \left(\theta \right)\right] \\ ={\left[\cos \left(\mathrm{\pi }-\frac{4\mathrm{\pi }}{7}\right)\cos \left(\frac{2\mathrm{\pi }}{7}\right)\cos \left(\frac{\mathrm{\pi }}{7}\right)\right]}^2 \\ ={\left[-\cos \left(\frac{4\mathrm{\pi }}{7}\right)\cos \left(\frac{2\mathrm{\pi }}{7}\right)\cos \left(\frac{\mathrm{\pi }}{7}\right)\right]}^2\left[\because \cos \left(\mathrm{\pi }-\theta \right)=-\cos \left(\theta \right)\right] \\ ={\left[\cos \left(\frac{4\mathrm{\pi }}{7}\right)\cos \left(\frac{2\mathrm{\pi }}{7}\right)\cos \left(\frac{\mathrm{\pi }}{7}\right)\right]}^2 \end{gathered}$$

Step 3: Simplify the expression by the formula $\sin \left(2A\right)=2\sin \left(A\right)\cos \left(A\right)$

$$\begin{gathered} \sin \left(\frac{\mathrm{\pi }}{14}\right)\sin \left(\frac{3\mathrm{\pi }}{14}\right)\sin \left(\frac{5\mathrm{\pi }}{14}\right)\sin \left(\frac{7\mathrm{\pi }}{14}\right)\sin \left(\frac{9\mathrm{\pi }}{14}\right)\sin \left(\frac{11\mathrm{\pi }}{14}\right)\sin \left(\frac{13\mathrm{\pi }}{14}\right)={\left[\cos \left(\frac{4\mathrm{\pi }}{7}\right)\cos \left(\frac{2\mathrm{\pi }}{7}\right)\cos \left(\frac{\mathrm{\pi }}{7}\right)\right]}^2 \\ =\frac{1}{4{\sin }^2\left({\displaystyle \frac{\mathrm{\pi }}{7}}\right)}{\left[2\sin \left(\frac{\mathrm{\pi }}{7}\right)\cos \left(\frac{\mathrm{\pi }}{7}\right)\cos \left(\frac{2\mathrm{\pi }}{7}\right)\cos \left(\frac{4\mathrm{\pi }}{7}\right)\right]}^2 \\ =\frac{1}{4{\sin }^2\left({\displaystyle \frac{\mathrm{\pi }}{7}}\right)}{\left[\sin \left(\frac{2\mathrm{\pi }}{7}\right)\cos \left(\frac{2\mathrm{\pi }}{7}\right)\cos \left(\frac{4\mathrm{\pi }}{7}\right)\right]}^2\left[\because \sin \left(2A\right)=2\sin \left(A\right)\cos \left(A\right)\right] \\ =\frac{1}{16{\sin }^2\left({\displaystyle \frac{\mathrm{\pi }}{7}}\right)}{\left[2\sin \left(\frac{2\mathrm{\pi }}{7}\right)\cos \left(\frac{2\mathrm{\pi }}{7}\right)\cos \left(\frac{4\mathrm{\pi }}{7}\right)\right]}^2 \\ =\frac{1}{16{\sin }^2\left({\displaystyle \frac{\mathrm{\pi }}{7}}\right)}{\left[\sin \left(\frac{4\mathrm{\pi }}{7}\right)\cos \left(\frac{4\mathrm{\pi }}{7}\right)\right]}^2 \\ =\frac{1}{64{\sin }^2\left({\displaystyle \frac{\mathrm{\pi }}{7}}\right)}{\left[2\sin \left(\frac{4\mathrm{\pi }}{7}\right)\cos \left(\frac{4\mathrm{\pi }}{7}\right)\right]}^2 \\ =\frac{1}{64{\sin }^2\left({\displaystyle \frac{\mathrm{\pi }}{7}}\right)}{\left[\sin \left(\frac{8\mathrm{\pi }}{7}\right)\right]}^2 \\ =\frac{1}{64{\sin }^2\left({\displaystyle \frac{\mathrm{\pi }}{7}}\right)}{\left[\sin \left(\mathrm{\pi }+\frac{\mathrm{\pi }}{7}\right)\right]}^2 \\ =\frac{1}{64{\sin }^2\left({\displaystyle \frac{\mathrm{\pi }}{7}}\right)}{\left[-\sin \left(\frac{\mathrm{\pi }}{7}\right)\right]}^2\left[\because \sin \left(\mathrm{\pi }+\mathrm{\theta }\right)=-\sin \left(\theta \right)\right] \\ =\frac{1}{64} \end{gathered}$$

Therefore, the value of $\sin \left(\frac{\mathrm{\pi }}{14}\right)\sin \left(\frac{3\mathrm{\pi }}{14}\right)\sin \left(\frac{5\mathrm{\pi }}{14}\right)\sin \left(\frac{7\mathrm{\pi }}{14}\right)\sin \left(\frac{9\mathrm{\pi }}{14}\right)\sin \left(\frac{11\mathrm{\pi }}{14}\right)\sin \left(\frac{13\mathrm{\pi }}{14}\right)$ is equal to $\frac{1}{64}$.

Hence, the correct option is (D).