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Given, $$\cos(\beta-\gamma)+\cos(\gamma-\alpha)+\cos(\alpha-\beta)=-\frac{3}{2}$$ Prove that:

  1. $\cos\alpha+\cos\beta+\cos\gamma=0$
  2. $\sin\alpha+\sin\beta+\sin\gamma=0$

It would be especially helpful if anyone would point out a way to do these quickly, in competitive examinations, such as in True or False questions.

Hungry Blue Dev
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  • Hint: Imagine you have 3 unit vectors $x_1 = (\cos\alpha,\sin\alpha), x_2 = (\cos\beta,\sin\beta), x_3 = (\cos\gamma,\sin\gamma)$. What is $x_1 \cdot x_2$, $x_2\cdot x_3$, $x_3 \cdot x_1$ and $|x_1 + x_2 + x_3|^2$ ? – achille hui Jan 28 '16 at 20:48
  • @achillehui Sorry, I don't know about vectors yet. – Hungry Blue Dev Jan 28 '16 at 20:49
  • Okay, what do you get if you expand $(\cos\alpha + \cos\beta + \cos\gamma)^2 + (\sin\alpha+\sin\beta + \sin\gamma)^2$? Remember you can group some of terms together using identities like $\cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta$. – achille hui Jan 28 '16 at 20:52
  • Related : http://math.stackexchange.com/questions/1397066/clarification-regarding-a-question/1397721 – lab bhattacharjee Jan 29 '16 at 06:31

1 Answers1

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You have $$\cos(\beta-\gamma)+\cos(\gamma-\alpha)+\cos(\alpha-\beta)=-\frac{3}{2}$$ From this we have $$-2(\cos\alpha \cos\beta+\cos\beta\cos\gamma+\cos\gamma\cos\alpha)+2(\sin\alpha \sin\beta+\sin\beta\sin\gamma+\sin\gamma\sin\alpha)=3\left(\sum_{cyclic} \cos^2 \alpha+\sum_{cyclic} \sin^2 \alpha\right)\\\implies\left(\sum_{cyclic}\cos\alpha\right)^2+\left(\sum_{cyclic}\sin\alpha\right)^2=0.$$

So, $$\begin{align} \sum_{cyclic}\cos\alpha&=0,\\ \text{and}\sum_{cyclic}\sin\alpha&=0 \end{align}$$