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If $\cos \left( {\alpha - \beta } \right) + \cos \left( {\beta - \gamma } \right) + \cos \left( {\gamma - \alpha } \right) = - \frac{3}{2}$, where $(α,β,γ ∈ R).$

(A) $\sum {\cos \alpha } = 0$

(B) $\sum {\sin \alpha } = 0$

(C) $\sin \alpha \sin \beta \sin \gamma = 0$

(D) $\sum {\cos \alpha } +\sum {\sin \alpha } = 0$

This is a multiple choice question with one or more options

My approach is bases on Complex number

$\cos \left( {\alpha - \beta } \right) + \cos \left( {\beta - \gamma } \right) + \cos \left( {\gamma - \alpha } \right) = - \frac{3}{2}$

$T = {e^{i\left( {\alpha - \beta } \right)}} + {e^{i\left( {\beta - \gamma } \right)}} + {e^{i\left( {\gamma - \alpha } \right)}}$

${\mathop{\rm Re}\nolimits} \left( T \right) = - \frac{3}{2}$

${e^{i\frac{\alpha }{\beta }}} + {e^{i\frac{\beta }{\gamma }}} + {e^{i\frac{\gamma }{\alpha }}} \Rightarrow {e^{i\frac{\alpha }{\beta }}}\left( {1 + \frac{{{e^{i\frac{\beta }{\gamma }}}}}{{{e^{i\frac{\alpha }{\beta }}}}}} \right) + {e^{i\frac{\gamma }{\alpha }}} \Rightarrow {e^{i\frac{\alpha }{\beta }}}\left( {1 + {e^{i\left( {\frac{\beta }{\gamma } = \frac{\alpha }{\beta }} \right)}}} \right) + {e^{i\frac{\gamma }{\alpha }}}$

How do I proceed from here

amWhy
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Samar Imam Zaidi
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  • $$\sum_{\text{cyc}}((\cos\alpha-\cos\beta)^2+(\sin\alpha-\sin\beta))^2=?$$. See also : https://math.stackexchange.com/questions/1419652/prove-that-in-any-triangle-abc-cos2a-cos2b-cos2c-geq-frac34 $$(\cos\alpha+\cos\beta+\cos\gamma)^2+(\sin\alpha+\sin\beta+\sin\gamma)^2=?$$ – lab bhattacharjee Nov 14 '22 at 18:13
  • $e^{i(\alpha-\beta)} = e^{i \frac{\alpha}{\beta}}$ is not in general true. You were probably thinking of $e^{i(\alpha-\beta)} = \frac{e^{i \alpha}}{e^{i \beta}}$ – aschepler Nov 14 '22 at 18:20
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    No where it is given that $\alpha, \beta , \gamma$ are sides of a triangle – Samar Imam Zaidi Nov 14 '22 at 18:28
  • See https://math.stackexchange.com/questions/2425077/if-cosz-x-cosy-z-cosx-y-frac32-then-sin-x-sin-y/2426242#2426242 – Hari Shankar Nov 17 '22 at 11:38

2 Answers2

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Note that $\cos (\alpha -\beta)=\cos \alpha \cos \beta +\sin \alpha \sin \beta$, therefore:

$$(\sum \sin \alpha)^2+(\sum \cos \alpha)^2=3+2(\sum \sin \alpha \sin \beta+\sum \cos \alpha \cos \beta)=3+2(\frac{-3}{2})=0;$$

$$\implies \sum \sin \alpha=\sum \cos \alpha=0.$$

Reza Rajaei
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Let $x=\alpha-\beta$ and $y=\beta-\gamma$. Then the given expression is $$f(x,y)=\cos x+\cos y+\cos(x+y)=-\frac{3}{2}.$$ The extrema of this function occur at $f_x=-\sin x-\sin(x+y)=0$ and $f_y=-\sin y-\sin(x+y)=0$. Then $\sin x=\sin y$ and $x=y$ is a solution.

The equation is now $2\cos x+\cos2x=-\frac{3}{2}$ with a solution $x=\frac{2\pi}{3}.$

Then, $\beta=\alpha-\frac{2\pi}{3}$ and $\gamma=\alpha-\frac{4\pi}{3}$.

And then: $\cos\alpha+\cos\beta+\cos\gamma=0$. $\implies A$. (It is just a test question. Be fast. If there are two correct answers, this question is cancelled in ÖSS exam. By the way Reza R. is genius.)

Bob Dobbs
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