Question

The value of $\sin \left(\frac{\mathrm{\pi }}{16}\right)\sin \left(\frac{3\mathrm{\pi }}{16}\right)\sin \left(\frac{5\mathrm{\pi }}{16}\right)\sin \left(\frac{7\mathrm{\pi }}{16}\right)$ is

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

Answer 1

The correct option is B

$\frac{\sqrt{2}}{16}$

The explanation for the correct option

Step 1 . Reduction of the trigonometric function

The given trigonometric expression: $\sin \left(\frac{\mathrm{\pi }}{16}\right)\sin \left(\frac{3\mathrm{\pi }}{16}\right)\sin \left(\frac{5\mathrm{\pi }}{16}\right)\sin \left(\frac{7\mathrm{\pi }}{16}\right)$.

$$\begin{gathered} \sin \left(\frac{\mathrm{\pi }}{16}\right)\sin \left(\frac{3\mathrm{\pi }}{16}\right)\sin \left(\frac{5\mathrm{\pi }}{16}\right)\sin \left(\frac{7\mathrm{\pi }}{16}\right) \\ =\frac{1}{4}\left[2\sin \left(\frac{\mathrm{\pi }}{16}\right)\sin \left(\frac{7\mathrm{\pi }}{16}\right)\right]\left[2\sin \left(\frac{3\mathrm{\pi }}{16}\right)\sin \left(\frac{5\mathrm{\pi }}{16}\right)\right] \\ =\frac{1}{4}\left[\cos \left(\frac{7\mathrm{\pi }}{16}-\frac{\mathrm{\pi }}{16}\right)-\cos \left(\frac{7\mathrm{\pi }}{16}+\frac{\mathrm{\pi }}{16}\right)\right]\left[\cos \left(\frac{5\mathrm{\pi }}{16}-\frac{3\mathrm{\pi }}{16}\right)-\cos \left(\frac{5\mathrm{\pi }}{16}+\frac{3\mathrm{\pi }}{16}\right)\right]\left[\because \cos \left(A-B\right)-\cos \left(A+B\right)=2\sin \left(A\right)\sin \left(B\right)\right] \\ =\frac{1}{4}\left[\cos \left(\frac{6\mathrm{\pi }}{16}\right)-\cos \left(\frac{8\mathrm{\pi }}{16}\right)\right]\left[\cos \left(\frac{2\mathrm{\pi }}{16}\right)-\cos \left(\frac{8\mathrm{\pi }}{16}\right)\right] \\ =\frac{1}{4}\left[\cos \left(\frac{3\mathrm{\pi }}{8}\right)-\cos \left(\frac{\mathrm{\pi }}{2}\right)\right]\left[\cos \left(\frac{\mathrm{\pi }}{8}\right)-\cos \left(\frac{\mathrm{\pi }}{2}\right)\right] \end{gathered}$$

Step 2. Further reduction

$$\begin{gathered} \therefore \frac{1}{4}\left[\cos \left(\frac{3\mathrm{\pi }}{8}\right)-\cos \left(\frac{\mathrm{\pi }}{2}\right)\right]\left[\cos \left(\frac{\mathrm{\pi }}{8}\right)-\cos \left(\frac{\mathrm{\pi }}{2}\right)\right] \\ =\frac{1}{4}\left[\cos \left(\frac{3\mathrm{\pi }}{8}\right)-0\right]\left[\cos \left(\frac{\mathrm{\pi }}{8}\right)-0\right] \\ =\frac{1}{4}\cos \left(\frac{3\mathrm{\pi }}{8}\right)\cos \left(\frac{\mathrm{\pi }}{8}\right) \\ =\frac{1}{8}\times 2\cos \left(\frac{3\mathrm{\pi }}{8}\right)\cos \left(\frac{\mathrm{\pi }}{8}\right) \\ =\frac{1}{8}\left[\cos \left(\frac{3\mathrm{\pi }}{8}+\frac{\mathrm{\pi }}{8}\right)+\cos \left(\frac{3\mathrm{\pi }}{8}-\frac{\mathrm{\pi }}{8}\right)\right]\left[\because \cos \left(A+B\right)+\cos \left(A-B\right)=2\cos \left(A\right)\cos \left(B\right)\right] \\ =\frac{1}{8}\left[\cos \left(\frac{\mathrm{\pi }}{2}\right)+\cos \left(\frac{\mathrm{\pi }}{4}\right)\right] \\ =\frac{1}{8}\left[0+\frac{1}{\sqrt{2}}\right] \\ =\frac{1}{8\sqrt{2}} \\ =\frac{\sqrt{2}}{8\times 2} \\ =\frac{\sqrt{2}}{16} \end{gathered}$$

Therefore, the value of $\sin \left(\frac{\mathrm{\pi }}{16}\right)\sin \left(\frac{3\mathrm{\pi }}{16}\right)\sin \left(\frac{5\mathrm{\pi }}{16}\right)\sin \left(\frac{7\mathrm{\pi }}{16}\right)$ is equal to $\frac{\sqrt{2}}{16}$.

Hence, the correct option is (B).