3

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!

Rajdeep Sindhu
  • 1,573
  • 3
    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 '20 at 09:01
  • @AnuragA Thanks! But, how do I prove that I'm dealing with real numbers here? – Rajdeep Sindhu Jun 19 '20 at 09:14
  • 1
    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 '20 at 10:20
  • Thank You, @JohnBentin... – Rajdeep Sindhu Jun 19 '20 at 10:23
  • https://math.stackexchange.com/questions/1382047/trigonometry-problem and the linked https://math.stackexchange.com/questions/1397066/clarification-regarding-a-question – lab bhattacharjee Jun 19 '20 at 18:08

1 Answers1

3

If the sum of squares of two real numbers is zero, it implies that both numbers are zero. If you want you can simply prove this using Reductio-Ad-Absurdum.

AxyuS
  • 399
  • 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 '20 at 09:19
  • Reductio-Ad-Absurdum can be used to prove the latter part you mentioned in your comment. To prove that the expression has real values you can use the generalization of sine and cosine function which are as follow: sin(x) = (exp(ix) - exp(-ix))/2i and cos(x) = (exp(ix) + exp(-ix))/2. Here i represents square root of -1. – AxyuS Jun 19 '20 at 09:23
  • Can the former be proved without using the generalization? I'm still working with trigonometric functions with respect to the unit circle. Thanks! – Rajdeep Sindhu Jun 19 '20 at 09:53
  • If you don't want to use generalization you can use the very basic idea of the trigonometry. Sine and cosine are basically the ratios of the length of the sides of a triangle. The length by definition is a real quantity. Hence sine and cosine are the ratios of two real numbers. Now I think you can proceed forward. – AxyuS Jun 19 '20 at 10:28
  • I can, thank you! – Rajdeep Sindhu Jun 19 '20 at 10:33