Question

If $(cos\theta +isin\theta )(cos2\theta +isin\theta )$ ....
$(cosn\theta +isinn\theta )=1$, then the value of $\theta$ is

• A. $\frac{2m\pi }{n(n+1)}$
• B. $4m\pi$
• C. $\frac{4m\pi }{n(n+1)}$
• D. $\frac{m\pi }{n(n+1)}$

Answer 1

The correct option is C $\frac{4m\pi }{n(n+1)}$

We have, $(cos\theta +isin\theta )(cos2\theta +isin2\theta )...$
$(cosn\theta +isinn\theta )=1$
$\therefore cos(\theta +2\theta +3\theta +....+n\theta )+isin(\theta +2\theta +3\theta +.........+n\theta )=1$
$\Rightarrow cos\frac{nn+1}{2}\theta +isin\frac{nn+1}{2}\theta =1$
On comparing the coefficients of real and imaginary parts on both sides; we get
$cos\frac{n(n+1)}{2}\theta =1$
And $sin\frac{n(n+1)}{2}\theta =0$
$\therefore \frac{nn+1}{2}\therefore =2m\pi$
$\Rightarrow \theta =\frac{4m\pi }{n(n+1)}$