# $\cos(\alpha-\beta)+\cos(\beta-\gamma)+\cos(\gamma-\alpha)=\frac{-3}{2}$,show that $\cos\alpha+\cos\beta+\cos\gamma=\sin\alpha+\sin\beta+\sin\gamma=0$

I think that I've done a major part of the problem but I'm stuck at a point.
Here's what I've done :

It's given to us that $$\cos(\alpha-\beta)+\cos(\beta-\gamma)+\cos(\gamma-\alpha) = \dfrac{-3}{2}$$ Using the identity $$\cos(A-B) = \cos A \cos B + \sin A \sin B$$, we obtain : $$\cos\alpha\cos\beta + \sin\alpha\sin\beta + \cos\beta\cos\gamma + \sin\beta\sin\gamma + \cos\gamma\cos\alpha + \sin\gamma\sin\alpha = \dfrac{-3}{2}$$ Multiplying both sides by $$2$$, we obtain : $$2\cos\alpha\cos\beta + 2\cos\beta\cos\gamma + + 2\cos\gamma\cos\alpha + 2\sin\alpha\sin\beta + 2\sin\beta\sin\gamma + 2\sin\gamma\sin\alpha = -3$$ Adding $$\sin^2\alpha+\sin^2\beta+\sin^2\gamma+\cos^2\alpha+\cos^2\beta+\cos^2\gamma$$ to both sides, we obtain : $$\text{LHS : } (\cos^2\alpha + \cos^2\beta + \cos^2\gamma + 2\cos\alpha\cos\beta + 2\cos\beta\cos\gamma + 2\cos\gamma\cos\alpha)$$ $$+ (\sin^2\alpha + \sin^2\beta + \sin^2\gamma + 2\sin\alpha\sin\beta + 2\sin\beta\sin\gamma + 2\sin\gamma\sin\alpha)$$ $$\text{RHS : } -3 + (\cos^2\alpha + \sin^2\alpha) + (\cos^2\beta + \sin^2\beta) + (\cos^2\gamma + \sin^2\gamma)$$ On simplifying, $$\text {LHS : } (\cos\alpha + \cos\beta + \cos\gamma)^2 + (\sin\alpha + \sin\beta + \sin\gamma)^2$$ $$\text{RHS : } -3+1+1+1 = -3+3 = 0$$ So, we obtain : $$(\cos\alpha + \cos\beta + \cos\gamma)^2 + (\sin\alpha + \sin\beta + \sin\gamma)^2 = 0$$ $$\implies (\cos\alpha + \cos\beta + \cos\gamma)^2 = -(\sin\alpha + \sin\beta + \sin\gamma)^2$$ Now, square rooting both sides would involve $$\iota$$ i.e. $$\sqrt{-1}$$ but I haven't learnt about complex numbers yet and I think that the solution can be continued without using complex numbers but I don't know how.

Any help would be appreciated.
Thanks!

• sum of two squares is $0$ and your are working in real numbers, i.e. $a^2+b^2=0 \implies a=0 \, \& \, b=0$. So from your second to last step, you get the conclusion you are looking for. Jun 19, 2020 at 9:01
• @AnuragA Thanks! But, how do I prove that I'm dealing with real numbers here? Jun 19, 2020 at 9:14
• The presumption of any question of this sort is that the numbers involved are real. If you were to allow complex numbers, then the conclusion cannot be reached. Jun 19, 2020 at 10:20
• Thank You, @JohnBentin... Jun 19, 2020 at 10:23
• Jun 19, 2020 at 18:08

• What do I prove using Reductio-Ad-Absurdum? That the expressions I'm dealing with have real values? Or that if the sum of squares of two real numbers equals $0$, then they both are $0$? Jun 19, 2020 at 9:19