# If $\cos\;\alpha + \cos\;\beta + \cos\;\gamma = 0 = \sin \;\alpha + \sin \;\beta + \sin \;\gamma$, prove that $\cos^{2}\;\alpha + \cos^{2}\;\beta + \cos^{2}\;\gamma = \sin^{2} \alpha + \sin^{2}\beta + \sin^{2}\gamma = \large\frac{3}{2}$.

This is the fifth 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:
Consider $(\cos^2\alpha-\sin^2\alpha)+(\cos^2\beta-\sin^2\beta)+(\cos^2\gamma-\sin^2\gamma)=\cos 2\alpha+\cos 2\beta+\cos 2\gamma=0$
We know that $\cos 2\alpha+\cos 2\beta+\cos 2\gamma=0$
Step 2:
Now $\cos^2\alpha+\cos^2\beta+\cos^2\gamma+\sin^2\alpha+\sin^2\beta+\sin^2\gamma=(\cos^2\alpha+\sin^2\alpha)+(\cos^2\beta+\sin^2\beta)+(\cos^2\gamma+\sin^2\gamma)=3$
Therefore $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=\sin^2\alpha+\sin^2\beta+\sin^2\gamma=\large\frac{3}{2}$
Hence proved.