Question

$\underset{n\to \infty }{\lim }{\left(\sin \frac{\pi }{2n}\cdot \sin \frac{2\pi }{2n}\cdot \sin \frac{3\pi }{2n}\dots \dots .\cdot \sin \frac{(n-1)\pi }{n}\right)}^{1/n}$ is equal to :

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{1}{5}$

Answer 1

Answer: (C)

$A={\left[\underset{n\to \infty }{\lim }\left(\sin \frac{\pi }{2n}\sin \frac{2\pi }{2n}\dots \dots \sin \frac{(n-1)}{2n}\right)\right]}^{1/n}$
$\Rightarrow \ln A=\frac{1}{n}\sum _{r=1}^{2(n-1)}\ln \sin \frac{r\pi }{2n}=\underset{0}{\overset{2}{\int }}\ell n\sin \left(\frac{\pi x}{2}\right)dx$
put $\frac{\pi x}{2}=t$
$\Rightarrow \ln A=\frac{2}{\pi }\underset{0}{\overset{\pi }{\int }}\ln (\sin t)dt=\frac{4}{\pi }\underset{0}{\overset{\pi /2}{\int }}\ln (\sin t)dt$
$\Rightarrow \ln A=-2\ln 2\Rightarrow A=\frac{1}{4}$