Question

If $k=\sin \frac{\pi }{18}\sin \frac{5\pi }{18}\sin \frac{7\pi }{18}$, then the numerical value of $k$ is

• A. $\frac{1}{4}$
• B. $\frac{1}{8}$
• C. $\frac{1}{16}$
• D. None of these

Answer 1

The correct option is B

$\frac{1}{8}$

Explanation for the correct option.

Step 1: Change it in form of $2\sin A\sin B$.

Multiplying and dividing $\sin \frac{\pi }{18}\sin \frac{5\pi }{18}\sin \frac{7\pi }{18}$ by $2$, we get

$$\begin{gathered} \frac{1}{2}\left[\left(2\sin \frac{\pi }{18}\sin \frac{5\pi }{18}\right)\sin \frac{7\pi }{18}\right] \\ =\frac{1}{2}\left[\left\{\cos \left(\frac{\pi }{18}-\frac{5\pi }{18}\right)-\cos \left(\frac{\pi }{18}+\frac{5\pi }{18}\right)\right\}\sin \frac{7\pi }{18}\right]\left[by2\sin A\sin B=\cos \left(A-B\right)-\cos \left(A+B\right)\right] \\ =\frac{1}{2}\left[\left(\cos \frac{4\pi }{18}-\cos \frac{6\pi }{18}\right)\sin \frac{7\pi }{18}\right]\left[by\cos \left(-\theta \right)=\cos \theta \right] \\ =\frac{1}{2}\left[\left(\cos \frac{4\pi }{18}·\sin \frac{7\pi }{18}\right)-\left(\cos \frac{\pi }{3}·sin\frac{7\pi }{18}\right)\right] \end{gathered}$$

Step 2: Change it in form of $2\cos A\cos B$.

$$\begin{gathered} =\frac{1}{2}\left[\frac{1}{2}\left(2\cos \frac{4\pi }{18}·\sin \frac{7\pi }{18}\right)-\left(\frac{1}{2}·sin\frac{7\pi }{18}\right)\right] \\ =\frac{1}{2}\left[\frac{1}{2}\left\{\sin \left(\frac{7\pi }{18}+\frac{4\pi }{18}\right)+\sin \left(\frac{7\pi }{18}-\frac{4\pi }{18}\right)\right\}-\left(\frac{1}{2}·sin\frac{7\pi }{18}\right)\right];\left[by2\sin A\cos B=\sin \left(A+B\right)+\sin s\left(A-B\right)\right] \\ =\frac{1}{2}\left[\frac{1}{2}\left\{\sin \left(\frac{11\pi }{18}\right)+\sin \left(\frac{3\pi }{18}\right)\right\}-\left(\frac{1}{2}·sin\frac{7\pi }{18}\right)\right] \\ =\frac{1}{2}\left[\frac{1}{2}\left\{\sin \left(\mathrm{\pi }-\frac{7\pi }{18}\right)+\sin \left(\frac{\pi }{6}\right)\right\}-\left(\frac{1}{2}·sin\frac{7\pi }{18}\right)\right] \\ =\frac{1}{2}\left[\frac{1}{2}\left\{\sin \frac{7\pi }{18}+\frac{1}{2}\right\}-\left(\frac{1}{2}·sin\frac{7\pi }{18}\right)\right]\left[by\sin \left(\mathrm{\pi }-\mathrm{\theta }\right)=\sin \theta \right] \\ =\frac{1}{2}\left[\frac{1}{2}\sin \frac{7\pi }{18}+\frac{1}{4}-\frac{1}{2}sin\frac{7\pi }{18}\right] \\ =\frac{1}{8} \end{gathered}$$

Therefore, $k=\frac{1}{8}$

Hence, option B is correct.