Question

If the value of $2^{100}\sin \left(\frac{\pi }{100}\right)\sin \left(\frac{2\pi }{100}\right)...\sin \left(\frac{99}{100}\pi \right)$ is ${10}^k$, then find the value of $k$.

Answer 1

100th roots of unity are ${\alpha }_0,{\alpha }_1,....{\alpha }_{99}$
where ${\alpha }_j=\cos \left(\frac{2\pi j}{100}\right)+i\sin \left(\frac{2\pi j}{100}\right)$
Thus $(x-{\alpha }_1)...(x-{\alpha }_{99})=x^{99}+...+x+1$
Putting $x=1$
$(1-{\alpha }_1)(1-{\alpha }_2)...(1-a_{99})=100\Rightarrow (1-a_1)(1-\overline{{\alpha }_2})...(1-\overline{{\alpha }_{99}})=100\therefore (1-{\alpha }_1)(1-\overline{{\alpha }_1})(1-{\alpha }_2)(1-\overline{{\alpha }_2})...(1-{\alpha }_{99})(1-\overline{a_{99}})={10}^4$
But $(1-a_k)(1-\overline{{\alpha }_k})=2(1-\cos \frac{2k\pi }{100})=2^2{\sin }^2\left(\frac{k\pi }{100}\right)$
Thus, $2^{198}{\sin }^2\left(\frac{\pi }{100}\right)...{\sin }^2\left(\frac{99\pi }{100}\right)={10}^4$
$\Rightarrow 2^{99}\sin \left(\frac{\pi }{100}\right)...\sin \left(\frac{99\pi }{100}\right)=100={10}^2$

Hence, $k=2$