Q56 / 155 · WBJEE 2019 · Mathematics

The general value of the real angle $\theta$, which satisfies the equation. $(\cos \theta+i \sin \theta)(\cos 2 \theta+i \sin 2 \theta) \ldots$$(\cos n \theta+t \sin n \theta)=1$ is given by, (assuming $k$ is an integer)

  • A
    $\frac{2 k \pi}{n+2}$
  • Correct answer
    B
    $\frac{4 k \pi}{n(n+1)}$
  • C
    $\frac{4 x \pi}{n+1}$
  • D
    $\frac{6 k \pi}{n(n+1)}$

Solution

We have, $(\cos\theta+i\sin\theta)(cos2\theta+isin2\theta)...(cosn\theta+isinn\theta) = 1$ $\Rightarrow e^{i\theta}.e^{i(2\theta)}.e^{i(3\theta)}. ...e^{i(n\theta)} = 1$ $\Rightarrow e^{i\theta(1+2+3+...+n}=1$ $\Rightarrow e^{\frac{in(n+1)\theta}{2}}=1$ $\Rightarrow cos(\frac{n(n+1)}{2}\theta)+isin(\frac{n(n+1)}{2}\theta = 1 + 0i$ $\Rightarrow cos(\frac{n(n+1)}{2}\theta)=1$ $\Rightarrow \frac{n(n+1)}{2}\theta = 2k\pi$ $\Rightarrow \theta=\frac{4k}{n(n+1)}\pi$

Hence, the correct answer is option: 'B'.

Relevant Questions