If $a = \cos{\alpha + i\sin\alpha,\ \ \ \ b = \cos{\beta + i\sin\beta}}$, $c = \cos{\gamma + i\sin\gamma}$ and $b/c + c/a + a/b = 1$, then $\cos{(\beta - \gamma) + \cos{(\gamma - \alpha) + \cos{(\alpha - \beta)}}}$ is equal to

Question

Dr. Abubacker · Accepted Answer

(b)
Given, $\frac{b}{c} + \frac{c}{a} + \frac{a}{b} = 1$
$\Rightarrow \ \ \ \frac{\cos{\beta + i\sin\beta}}{\cos{\gamma + i\sin\gamma}} + \frac{\cos{\gamma + i\sin\gamma}}{\cos{\alpha + i\sin\alpha}} + \frac{\cos{\alpha + i\sin\alpha}}{\cos{\beta + i\sin\beta}} = 1$
$\Rightarrow \ \ \ \cos{(\beta - \gamma) + i\sin{(\beta - \gamma) + \cos{(\gamma - \alpha) + i\sin{(\gamma - \alpha) + \cos{(\alpha - \beta) + i\sin{(\alpha - \beta) = 1}}}}}}$
On equating real part on both sides, we get
$\cos{(\beta - \gamma) + \cos{(\gamma - \alpha) + \cos{(\alpha - \beta) = 1}}}$

Q. If $a = \cos{\alpha + i\sin\alpha,\ \ \ \ b = \cos{\beta + i\sin\beta}}$, $c = \cos{\gamma + i\sin\gamma}$ and $b/c + c/a + a/b = 1$, then $\cos{(\beta - \gamma) + \cos{(\gamma - \alpha) + \cos{(\alpha - \beta)}}}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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