# $\cos(\alpha-\beta)+\cos(\beta-\gamma)+\cos(\gamma-\alpha)=\frac{-3}{2}$,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. – Anurag A Jun 19 at 9:01
• @AnuragA Thanks! But, how do I prove that I'm dealing with real numbers here? – Rajdeep Sindhu Jun 19 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. – John Bentin Jun 19 at 10:20
• Thank You, @JohnBentin... – Rajdeep Sindhu Jun 19 at 10:23
• – lab bhattacharjee Jun 19 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 $0$? – Rajdeep Sindhu Jun 19 at 9:19