\mathrm{The}\mathrm{ value}\mathrm{of}\sin \frac{\pi }{18}·\sin \frac{5}{\pi }·\sin \frac{7}{\mathrm{\pi /}}is\; equal\; to\frac{\mathrm{}}{}

The correct answer is 1/8.

To find the value of the given expression:

$P=\sin \left(\frac{\pi }{18}\right)·\sin \left(\frac{5\pi }{18}\right)·\sin \left(\frac{7\pi }{18}\right)$

We can use the co-function trigonometric identity:

$\sin \theta =\cos \left(\frac{\pi }{2},-,\theta \right)$

Let us express each sine term in terms of cosine:

$\sin \left(\frac{\pi }{18}\right)=\cos \left(\frac{\pi }{2},-,\frac{\pi }{18}\right)=\cos \left(\frac{9\pi -\pi }{18}\right)=\cos \left(\frac{8\pi }{18}\right)=\cos \left(\frac{4\pi }{9}\right)$

$\sin \left(\frac{5\pi }{18}\right)=\cos \left(\frac{\pi }{2},-,\frac{5\pi }{18}\right)=\cos \left(\frac{9\pi -5\pi }{18}\right)=\cos \left(\frac{4\pi }{18}\right)=\cos \left(\frac{2\mathrm{��}}{9}\right)$

$\sin \left(\frac{7\pi }{18}\right)=\cos \left(\frac{\pi }{2},-,\frac{7\pi }{18}\right)=\cos \left(\frac{9\pi -7\pi }{18}\right)=\cos \left(\frac{2\pi }{18}\right)=\cos \left(\frac{\pi }{9}\right)$

Substituting these values back into the expression gives:

$P=\cos \left(\frac{\pi }{9}\right)·\cos \left(\frac{2\pi }{9}\right)·\cos \left(\frac{4\pi }{9}\right)$

We can now apply the product formula for cosines:

$\cos A·\cos 2A·\cos 2^{2}A\cdots \cos 2^{n-1}A=\frac{\sin \left(2^{n}A\right)}{2^n\sin A}$

For this problem, let $A=\frac{\pi }{9}$ and $n=3$. Substituting these parameters:

$P=\frac{\sin \left(2^3·\frac{\pi }{9}\right)}{2^3\sin \left(\frac{\pi }{9}\right)}$

$P=\frac{\sin \left(\frac{8\pi }{9}\right)}{8\sin \left(\frac{\pi }{9}\right)}$

Using the identity $\sin \left(\pi -\theta \right)=\sin \theta$, we find that $\sin \left(\frac{8\pi }{9}\right)=\sin \left(\pi -\frac{\pi }{9}\right)=\sin \left(\frac{\pi }{9}\right)$. Thus:

$P=\frac{\sin \left(\frac{\pi }{9}\right)}{8\sin \left(\frac{\pi }{9}\right)}=\frac{1}{8}$