Question

The value of $\left(\cos \left(\frac{\pi }{2}\right)+i\sin \left(\frac{\pi }{2}\right)\right)·\left(\cos \left(\frac{\pi }{4}\right)+i\sin \left(\frac{\pi }{4}\right)\right)·\left(\cos \left(\frac{\pi }{8}\right)+i\sin \left(\frac{\pi }{8}\right)\right)······\infty$

• A. $1$
• B. $0$
• C. $-1$
• D. None of these

Answer 1

The correct option is C

$-1$

Explanation for correct option

Given function is $\left(\cos \left(\frac{\pi }{2}\right)+i\sin \left(\frac{\pi }{2}\right)\right)·\left(\cos \left(\frac{\pi }{4}\right)+i\sin \left(\frac{\pi }{4}\right)\right)·\left(\cos \left(\frac{\pi }{8}\right)+i\sin \left(\frac{\pi }{8}\right)\right)······\infty$

We know that $e^{i\theta }=\cos \left(\theta \right)+i\sin \left(\theta \right)$

$$\begin{gathered} \therefore \left(\cos \left(\frac{\pi }{2}\right)+i\sin \left(\frac{\pi }{2}\right)\right)·\left(\cos \left(\frac{\pi }{4}\right)+i\sin \left(\frac{\pi }{4}\right)\right)·\left(\cos \left(\frac{\pi }{8}\right)+i\sin \left(\frac{\pi }{8}\right)\right)······\infty \\ =e^{\frac{i\pi }{2}}.e^{\frac{i\pi }{2^2}}.e^{\frac{i\pi }{2^3}}······\infty \\ =e^{\frac{i\pi }{2}\left(1+\frac{1}{2}+\frac{1}{2^2}+...\infty \right)} \end{gathered}$$

We know that the sum of infinite GP series is $\frac{a}{1-r}$

$$\begin{gathered} \therefore e^{\frac{i\pi }{2}\left(1+\frac{1}{2}+\frac{1}{2^2}+...\infty \right)} \\ =e^{\frac{i\pi }{2}\left(\frac{1}{1-\frac{1}{2}}\right)} \\ =e^{\frac{i\pi }{2}·2} \\ =e^{i\pi } \\ =\cos \left(\pi \right)+i\sin \left(\pi \right) \\ =-1+i·0 \\ =-1 \end{gathered}$$

Hence the correct option is $\mathrm{Option}\left(\mathrm{C}\right)$