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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

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  • $\begingroup$ Use $$(\cos A+i\sin A)(\cos B+i\sin B)=\cos(A+B)+i\sin(A+B)$$ $\endgroup$ – lab bhattacharjee Apr 7 '14 at 4:49
  • $\begingroup$ It would be then $cos(\theta+2\theta+3\theta+..n\theta)+isin(\theta+2\theta+3\theta+..n\theta)$ where $$\theta+2\theta+3\theta+..n\theta=\frac{n\theta(n\theta+1)}2$$ $\endgroup$ – Abdullah Sorathia Apr 7 '14 at 5:00
  • $\begingroup$ You have a mistake in $$\theta+2\theta+3\theta+..n\theta=\frac{n\theta(n\theta+1)}2$$ It should be $$\frac{1}{2} \theta n (n+1)$$ $\endgroup$ – Claude Leibovici Apr 7 '14 at 7:26
  • $\begingroup$ Are you sure your formula is correct? Is $r$ a real number or a natural number? $\endgroup$ – john.abraham Apr 7 '14 at 7:58
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    $\begingroup$ Using that $\cos(k\Theta)+i\sin(k\Theta)=e^{ik\Theta}$ we have $\begin{array}{ccc} e^{i\Theta}e^{i2\Theta}e^{i3\Theta}\dots e^{in\Theta} & = & i\\ e^{i\Theta(1+2+3+\dots+n)} & = & i\\ e^{i\Theta\frac{1}{2}n(n+1)} & = & i \end{array} $ Which is only true if (for $n,m\in\mathbb{N}$) $\begin{array}{ccc} \Theta\frac{1}{2}n(n+1) & = & \frac{1}{2}\pi+2\pi m\\ \Theta n(n+1) & = & \pi+4\pi m\\ \Theta & = & \frac{\pi+4\pi m}{n(n+1)} \end{array} $ I don't see how $\Theta$ could be anything else. Hopefully someone else can chime in on this. $\endgroup$ – john.abraham Apr 7 '14 at 11:17
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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:

$$ \prod_{k = 1}^{n}\left(\cos\left(k\theta\right) + i\sin\left(k\theta\right)\right) = \prod_{k = 1}^n e^{k\theta i} = i = e^{\left(\frac{\pi}{2} + 2\pi m\right)i} \\ \prod_{k = 1}^n e^{k\theta i} = \exp\left(\sum\limits_{k = 1}^n \left(k\theta i\right)\right) \\ \sum\limits_{k = 1}^n \left(k\theta i\right) = \left(\theta \left(\sum\limits_{k = 1}^n k\right)\right)i = \left(\theta \frac{n(n + 1)}{2}\right)i = \left(\frac{\pi}{2} + 2\pi m\right)i \\ \theta \frac{n(n + 1)}{2} = \frac{\pi}{2} + 2\pi m \\ \theta = \frac{\pi + 4\pi m}{n(n+1)} \text{, q.e.d.} $$

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    $\begingroup$ This is essentially what john.abraham posted in the comments on 7 April, right? $\endgroup$ – Gerry Myerson May 11 '14 at 4:28
  • $\begingroup$ @GerryMyerson Yes, I would agree he stated the same thing. $\endgroup$ – Jared May 11 '14 at 4:29
  • $\begingroup$ Then may I suggest that the polite thing to do is to acknowledge it in your post. $\endgroup$ – Gerry Myerson May 11 '14 at 4:39
  • $\begingroup$ @GerryMyerson I agree (and there's no way to prove this) except that I wrote this without looking at the comments--I don't know why this came up as a "new" question considering how old the comments are. $\endgroup$ – Jared May 11 '14 at 4:46
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    $\begingroup$ @GerryMyerson I appreciate your criticism of conduct and hopefully my edited acknowledgement alleviates that. It's one of those things where a cursory glance of a comment doesn't seem clear to me so I write my own solution which, after-the-fact, is clear to be the same as the comment. Hopefully the OP can see that the two explanations are equivalent. $\endgroup$ – Jared May 11 '14 at 5:02

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