If $\alpha,\ \beta,\ \gamma \in \left\lbrack 0,\ \frac{\pi}{2} \right\rbrack$, then the value of $\frac{\sin{(\alpha + \beta + \gamma)}}{\sin{\alpha + \sin{\beta + \sin\gamma}}}$ is

Question

Dr. Abubacker · Accepted Answer

(a)
$\sin{\alpha + \sin{\beta + \sin{\gamma - \sin{(\alpha + \beta + \gamma)}}}}$
$= \sin{\alpha + \sin{\beta + \sin{\gamma - \sin{\alpha\cos{\beta\cos{\gamma - \cos{\alpha\sin{\beta\cos{\gamma - \cos{\alpha\cos{\beta\sin{\gamma + \sin{\alpha\sin{\beta\sin\gamma}}}}}}}}}}}}}}$
$= \sin{\alpha\left( 1 - \cos{\beta\cos\gamma} ight) + \sin\beta}\left( 1 - \cos{\alpha\cos\gamma} ight) + \sin{\gamma\left( 1 - \cos{\alpha\cos\beta} ight) + \sin{\alpha\sin{\beta\sin\gamma}}}$
$	herefore\sin{\alpha + \sin{\beta + \sin{\gamma > \sin{(\alpha + \beta + \gamma)}}}}$
$\Rightarrow \ \ \ \frac{\sin{(\alpha + \beta + \gamma)}}{\sin{\alpha + \sin{\beta + \sin\gamma}}} < 1$

Q. If $\alpha,\ \beta,\ \gamma \in \left\lbrack 0,\ \frac{\pi}{2} \right\rbrack$, then the value of $\frac{\sin{(\alpha + \beta + \gamma)}}{\sin{\alpha + \sin{\beta + \sin\gamma}}}$ is

Explanation

Reviewed by Dr. Abubacker

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