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If $\cos \alpha + cos \beta + cos \gamma = 0 = sin \alpha + sin \beta + sin \gamma$, prove that $cos 3\alpha + cos 3\beta + cos 3\gamma = 3 cos\left ( \alpha +\beta + \gamma \right )$

This is the first part of the multi-part Q3.

1 Answer

  • 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:
Therefore $(\cos\alpha+i\sin\alpha)+(\cos\beta+i\sin\beta)+(\cos\gamma+i\sin\gamma)=0$
This is of the form $a+b+c=0$
Therefore $a^3+b^3+c^3=3abc$
Step 2:
$\cos 3\alpha+i\sin 3\alpha+\cos 3\beta+i\sin 3\beta+\cos 3\gamma+i\sin 3\gamma=(\cos 3\alpha+\cos 3\beta+\cos 3\gamma)+i(\sin 3\alpha+\sin 3\beta+\sin 3\gamma)$
$3e^{i\alpha}e^{i\beta}e^{i\gamma}=3e^{\large i(\alpha+\beta+\gamma)}$
$\Rightarrow 3[\cos(\alpha+\beta+\gamma)+i\sin(\alpha+\beta+\gamma)]$
Step 3:
Equating real parts we get
$\cos 3\alpha+\cos 3\beta+\cos 3\gamma=3\cos(\alpha+\beta+\gamma)$
answered Jun 11, 2013 by sreemathi.v
edited Jun 11, 2013 by sreemathi.v

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