यदि alpha ,,,beta ,gamma ,,,delta परिमाण के | vedclass

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Question

(c) दिया है $\alpha < \beta < \gamma < \delta $
एवं $\sin \alpha = \sin \beta = \sin \gamma = \sin \delta = k$ एवं $\alpha ,\beta ,\gamma ,\

Vedclass · Accepted Answer

$2\,\sqrt {1 + k} $

Vedclass · Answer

(c) दिया है $\alpha < \beta < \gamma < \delta $
एवं $\sin \alpha = \sin \beta = \sin \gamma = \sin \delta = k$ एवं $\alpha ,\beta ,\gamma ,\delta $ सबसे छोटे धनात्मक कोण हैं जो ऊपर दिये गये प्रतिबन्धों को संतुष्ट करते हैं।
$	herefore$  $\beta = \pi - \alpha ,\gamma = 2\pi + \alpha ,\delta = 3\pi - \alpha $.
अत: दिया गया व्यंजक
$ = 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( {\sin \frac{1}{2}\alpha + \cos \frac{1}{2}\alpha } ight)}^2}} = 2\sqrt {1 + \sin \alpha } = 2\sqrt {1 + k} $.

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