If $\;\cos\alpha+\cos\beta+\cos\gamma= \sin\alpha+\sin\beta+\sin\gamma=0$, then show that
$\cos^2\!\alpha+\cos^2\!\beta+\cos^2\!\gamma=\sin^2\!\alpha+\sin^2\!\beta+\sin^2\!\gamma=\!3/2$
I have done this problem with just trigonometric identities, but I want to know how can I get it with complex numbers.
I have tried and gotten the following results,
$\mathrm{CiS}(\alpha)+\mathrm{CiS}(\beta)+\mathrm{CiS}(\gamma)=0$
$\mathrm{CiS}^2(\alpha) + \mathrm{CiS}^2(\beta) + \mathrm{CiS}^2(\gamma) = 0$
I have seen very similar questions with the same given condition and tried methods from there but I couldn't solve it all the way. Thanks.
Best Answer
$$CiS(\alpha)=x, CiS(\beta)=y, CiS(\gamma)=z$$
$$x+y+z = (\cos(\alpha)+\cos(\beta)+\cos(\gamma))+i(\sin(\alpha)+\sin(\beta)+\sin(\gamma)=0+i(0))=0$$
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z} = (\cos(\alpha)+\cos(\beta)+\cos(\gamma))-i(\sin(\alpha)+\sin(\beta)+\sin(\gamma)) = 0$$ $$xy+yz+zx=0$$
Therefore, $x^2+y^2+z^2 = 0$
$$CiS(\alpha)^2+CiS(\beta)^2+CiS(\gamma)^2 = 0$$ $$(\cos(2\alpha)+i\sin(2\alpha))+(\cos(2\beta)+i\sin(2\beta))+(\cos(2\gamma)+i\sin(2\gamma))=0$$ $$1-2\sin^2(\alpha)+i\sin(2\alpha)+1-2\sin^2(\beta)+i\sin(2\beta)+1-2\sin^2(\gamma)+i\sin(2\gamma)=0$$ $$3-2(\sin^2(\alpha)+\sin^2(\beta)+\sin^2(\gamma))+i(\sin(2\alpha)+\sin(2\beta)+\sin(2\gamma))=0+i(0)$$
Comparing the real part of this equation, $$\sin^2(\alpha)+\sin^2(\beta)+\sin^2(\gamma) = \frac{3}{2}$$
It is now apparent that, $$\cos^2(\alpha)+\cos^2(\beta)+\cos^2(\gamma) = \frac{3}{2}$$