જો alpha ,,beta ,,gamma ,,delta એ ચડતા ક્રમમ | vedclass

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Question

Given that, $\alpha<\beta<\gamma<\delta$ and 

$\sin \alpha=\sin \beta=\sin \gamma=\sin \delta=\mathrm{k}$

Also $\alpha, \beta, \ga

Vedclass · Accepted Answer

$2\sqrt {\left( {1 + k} \right)} $

Vedclass · Answer

Given that, $\alpha<\beta<\gamma<\delta$ and 

$\sin \alpha=\sin \beta=\sin \gamma=\sin \delta=\mathrm{k}$

Also $\alpha, \beta, \gamma, \delta$ are smallest positive angles, satisfying above two conditions. 

$	herefore $ we can take $\beta=\pi-\alpha, \gamma=2 \pi+\alpha, \delta=3 \pi-\alpha$

The given expression

$=4 \sin \frac{\alpha}{2}+3 \sin \left(\frac{\pi}{2}-\frac{\alpha}{2}ight)$

$+2 \sin \left(\pi+\frac{\alpha}{2}ight)+\sin \left(\frac{3 \pi}{2}-\frac{\alpha}{2}ight)$

$=4 \sin \frac{\alpha}{2}+3 \cos \frac{\alpha}{2}-2 \sin \frac{\alpha}{2}-\cos \frac{\alpha}{2}$

$=2\left(\sin \frac{\alpha}{2}+\cos \frac{\alpha}{2}ight)$

$ = 2\sqrt {\left\{ {{{\left( {\sin \frac{\alpha }{2} + \cos \frac{a}{2}} ight)}^2}} ight\}} $

$=2 \sqrt{(1+\sin \alpha)}=2 \sqrt{(1+\mathrm{k})}$

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