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# If $\cos\;\alpha + \cos\;\beta + \cos\;\gamma = 0 = \sin \;\alpha + \sin \;\beta + \sin \;\gamma$, prove that $\sin 2\alpha + \sin 2\beta + \sin 2\gamma = 0$

(Hints: Take $a = \textit{$\cos$} \;\alpha, \;b = \textit{$\cos$} \;\beta, c = \textit{$\cos$} \; \gamma$, $a+b+c = 0 => a^{3} + b^{3}+ c^{3} = 3abc$ $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 0 => a^{2}+b^{2}+c^{2}=0$ This is the fourth part of the multi-part Q3.

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Toolbox:
• From De moivre's theorem we have
• (i) $(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta,n\in Q$
• (ii) $(\cos\theta+i\sin\theta)^{-n}=\cos n\theta-i\sin n\theta$
• (iii) $(\cos\theta-i\sin\theta)^n=\cos n\theta-i\sin n\theta$
• (iv) $(\sin \theta+i\cos \theta)^n=[\cos(\large\frac{\pi}{2}$$-\theta)+i\sin(\large\frac{\pi}{2}$$-\theta)]^n=\cos n(\large\frac{\pi}{2}$$-\theta)+i\sin n(\large\frac{\pi}{2}$$-\theta)$
• $e^{i\theta}=\cos\theta+i\sin\theta$
• $e^{-i\theta}=\cos\theta-i\sin\theta$,also written as $\cos\theta$ and $\cos(-\theta)$
Step 1:
Now consider $e^{\large -i\alpha},e^{\large -i\beta},e^{\large -i\gamma}$.
(i.e) $\cos\alpha-i\sin\alpha,\cos\beta-i\sin\beta,\cos\gamma-i\sin\gamma$ which represent $\large\frac{1}{a},\frac{1}{b},\frac{1}{c}$.
We can see $\large\frac{1}{a}$$+\large\frac{1}{b}$$+\large\frac{1}{c}$$=0 \Rightarrow \large\frac{ab+bc+ca}{abc}$$=0$
If $ab+bc+ca=0$ we have $(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca=0$
Step 2:
$\Rightarrow a^2+b^2+c^2=0$
(i.e) $e^{\large i2\alpha}+e^{\large i2\beta}+e^{\large i2\gamma}=0$
We know that $e^{i\theta}=\cos\theta+i\sin\theta$
$\cos 2\alpha+i\sin 2\alpha+\cos 2\beta+i\sin 2\beta+\cos 2\gamma+i\sin 2\gamma=0$
$(\cos 2\alpha+\cos 2\beta+\cos 2\gamma)+i(\sin 2\alpha+\sin 2\beta+\sin 2\gamma)=0$
From this we have
$\sin 2\alpha+\sin 2\beta+\sin 2\gamma=0$
answered Jun 11, 2013