If $$(\cos\Theta+i\sin\Theta)(\cos2\Theta+i\sin2\Theta)(\cos3\Theta+i\sin3\Theta) \dots (\cos n\Theta+i\sin n\Theta)=i $$ then show that general Value of $$\Theta=\left[2r+\frac1{n(n+1)}\right]\pi$$ $$OR$$
$$\Theta = \frac{\pi+4\pi m}{n(n+1)} $$ How to get the value of theta
As stated by john.abraham, the product can be written as an exponential raised to some sum which can then be equated to $\left({\frac{\pi}{2} + 2\pi m}\right)i$ (the exponent of $i = e^{\left(\frac{\pi}{2} + 2\pi m\right)i}$)--the $i$'s cancel:
$$ \prod_{k = 1}^{n}\left(\cos\left(k\theta\right) + i\sin\left(k\theta\right)\right) = \prod_{k = 1}^n e^{k\theta i} = i = e^{\left(\frac{\pi}{2} + 2\pi m\right)i} \\ \prod_{k = 1}^n e^{k\theta i} = \exp\left(\sum\limits_{k = 1}^n \left(k\theta i\right)\right) \\ \sum\limits_{k = 1}^n \left(k\theta i\right) = \left(\theta \left(\sum\limits_{k = 1}^n k\right)\right)i = \left(\theta \frac{n(n + 1)}{2}\right)i = \left(\frac{\pi}{2} + 2\pi m\right)i \\ \theta \frac{n(n + 1)}{2} = \frac{\pi}{2} + 2\pi m \\ \theta = \frac{\pi + 4\pi m}{n(n+1)} \text{, q.e.d.} $$