Question

If $1,{\alpha }_1,{\alpha }_2{\alpha }_3..........{\alpha }_n{}_{-}1$ be the $n^{th}$ roots of unity, then the value of $sin\frac{\pi }{n},sin\frac{2\pi }{n},sin\frac{3\pi }{n}..........\frac{(n-1)\pi }{n}$ equals.

• A. $\frac{n}{2^n}$
• B. $\frac{n}{2^{n+1}}$
• C. $\frac{n+1}{2^{n-1}}$
• D. $\frac{n}{2^{n-1}}$

Answer 1

The correct option is D $\frac{n}{2^{n-1}}$

Let ${\alpha }_k=\cos \frac{2k\pi }{\eta }+\sin \frac{2k\pi }{\eta };K=1,2,3,.....,\eta -1$

$\Rightarrow {\alpha }_1,{\alpha }_2,{\alpha }_3.......{\alpha }_{\eta -1}$ are zeros of $1+x+x^2+....+x^{\eta -1}$

$\Rightarrow 1+x+x^2+.......+x^{\eta -1}=\stackrel{\eta -1}{\underset{k=1}{\pi }}(x-{\alpha }_R)$

put $x=1$

$\Rightarrow \eta =\stackrel{\eta -1}{\underset{k=1}{\pi }}(1-{\alpha }_k)$

Apply rod

$\Rightarrow |n|=\stackrel{\eta -1}{\underset{k=1}{\pi }}|(1-{\alpha }_K)|$

$-1{\alpha }_K=1-\cos \frac{2K\pi }{\eta }-i\sin \frac{2K\pi }{\eta }=2\sin \frac{k\pi }{\eta }(\sin \frac{k\pi }{\eta }-1\cos \frac{k\pi }{\eta })$

$\Rightarrow |1-{\alpha }_K|=2\sin \frac{k\pi }{\eta }$

$\Rightarrow |\eta |=2^{\eta -1}\stackrel{\eta -1}{\underset{k=1}{\pi }}\sin \frac{k\pi }{\eta }$

$\therefore \sin \frac{\pi }{\eta }\sin \frac{2\pi }{\eta }......\sin \frac{\eta -1}{\eta }\pi =\frac{\eta }{2^{\eta -1}}$