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

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

Vedclass · Accepted Answer

$2\sqrt{1 + k}$

Vedclass · Answer

(C) Given that $\alpha Since these are the smallest positive angles,we have:$\alpha = \alpha$$\beta = \pi - \alpha$$\gamma = 2\pi + \alpha$$\delta = 3\pi - \alpha$Substituting these into the expression:$E = 4\sin \frac{\alpha}{2} + 3\sin \frac{\pi - \alpha}{2} + 2\sin \frac{2\pi + \alpha}{2} + \sin \frac{3\pi - \alpha}{2}$$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)$$E = 4\sin \frac{\alpha}{2} + 3\cos \frac{\alpha}{2} - 2\sin \frac{\alpha}{2} - \cos \frac{\alpha}{2}$$E = 2\sin \frac{\alpha}{2} + 2\cos \frac{\alpha}{2} = 2\left(\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2}\right)$Squaring the term inside:$E^2 = 4\left(\sin^2 \frac{\alpha}{2} + \cos^2 \frac{\alpha}{2} + 2\sin \frac{\alpha}{2}\cos \frac{\alpha}{2}\right) = 4(1 + \sin \alpha) = 4(1 + k)$Therefore,$E = 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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