If alpha, beta, gamma, delta are the smallest | vedclass

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Question

(C) Given $\alpha < \beta < \gamma < \delta$ and $\sin \alpha = \sin \beta = \sin \gamma = \sin \delta = k$. Since these are the smallest positive ang

Vedclass · Accepted Answer

$2\sqrt{1 + k}$

Vedclass · Answer

(C) Given $\alpha Substituting these into the expression:$E = 4\sin \frac{\alpha}{2} + 3\sin \frac{\pi - \alpha}{2} + 2\sin \frac{2\pi + \alpha}{2} + \sin \frac{3\pi - \alpha}{2}$$E = 4\sin \frac{\alpha}{2} + 3\sin \left( \frac{\pi}{2} - \frac{\alpha}{2} \right) + 2\sin \left( \pi + \frac{\alpha}{2} \right) + \sin \left( \frac{3\pi}{2} - \frac{\alpha}{2} \right)$Using trigonometric identities:$E = 4\sin \frac{\alpha}{2} + 3\cos \frac{\alpha}{2} - 2\sin \frac{\alpha}{2} - \cos \frac{\alpha}{2}$$E = 2\sin \frac{\alpha}{2} + 2\cos \frac{\alpha}{2} = 2 \left( \sin \frac{\alpha}{2} + \cos \frac{\alpha}{2} \right)$Squaring the term inside the square root:$E = 2\sqrt{\left( \sin \frac{\alpha}{2} + \cos \frac{\alpha}{2} \right)^2} = 2\sqrt{\sin^2 \frac{\alpha}{2} + \cos^2 \frac{\alpha}{2} + 2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}$$E = 2\sqrt{1 + \sin \alpha} = 2\sqrt{1 + k}$.

If $\alpha, \beta, \gamma, \delta$ are the smallest positive angles in ascending order of magnitude which have their sines equal to the positive quantity $k$,then the value of $4\sin \frac{\alpha}{2} + 3\sin \frac{\beta}{2} + 2\sin \frac{\gamma}{2} + \sin \frac{\delta}{2}$ is equal to

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