Given,

Expression $96·\cos \frac{\mathrm{\pi }}{33}·\cos \frac{2\mathrm{\pi }}{33}·\cos \frac{4\mathrm{\pi }}{33}.........\cos \frac{16\mathrm{\pi }}{33}$

Now we know that,

$\cos A·\cos 2A·\cos 2^{2}A·\cos 2^{3}A.....·\cos 2^{n-1}A=\frac{\sin 2^{n}A}{2^n\sin A}$

Now using the above formula in given expression we get,

$96·\cos \frac{\mathrm{\pi }}{33}·\cos \frac{2\mathrm{\pi }}{33}·\cos \frac{4\mathrm{\pi }}{33}.........\cos \frac{16\mathrm{\pi }}{33}$

$=96\times \frac{\sin {\displaystyle \frac{32\mathrm{\pi }}{33}}}{2^5\sin {\displaystyle \frac{\mathrm{\pi }}{33}}}$

$=96\times \frac{\sin {\displaystyle \left(\mathrm{\pi }-\frac{\mathrm{\pi }}{33}\right)}}{2^5\sin {\displaystyle \frac{\mathrm{\pi }}{33}}}$

$=96\times \frac{\sin {\displaystyle \left(\frac{\mathrm{\pi }}{33}\right)}}{2^5\sin {\displaystyle \frac{\mathrm{\pi }}{33}}} \left\{\text{as} \sin \left(\mathrm{\pi }-\mathrm{\alpha }\right)=\sin \alpha \right\}$

$=96\times \frac{1}{32}=3$