Browse Questions

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

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

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:
$\cos\alpha+\cos\beta+\cos\gamma=0$
$\sin\alpha+\sin\beta+\sin\gamma=0$
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 $a^3+b^3+c^3=3abc$
$(\cos\alpha+i\sin\alpha)^3+(\cos\beta+i\sin\beta)^3+(\cos\gamma+i\sin\gamma)^3=3(\cos\alpha+i\sin\alpha)(\cos\beta+i\sin\beta)(\cos\gamma+i\sin\gamma)$
Step 2:
LHS:
$\cos 3\alpha+i\sin 3\alpha+\cos 3\beta+i\sin 3\beta+\cos 3\gamma+i\sin 3\gamma=(\cos 3\alpha+\cos 3\beta+\cos 3\gamma)+i(\sin 3\alpha+\sin 3\beta+\sin 3\gamma)$
RHS:
$3e^{i\alpha}e^{i\beta}e^{i\gamma}=3e^{\large i(\alpha+\beta+\gamma)}$
$\Rightarrow 3[\cos(\alpha+\beta+\gamma)+i\sin(\alpha+\beta+\gamma)]$
Step 3:
Equating the imaginary parts we get
$\Rightarrow \sin 3\alpha+\sin 3\beta+\sin 3\gamma=3\sin(\alpha+\beta+\gamma)$