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(D) Given the equation $\cos 2	heta = \sin \alpha$. Using the identity $\sin \alpha = \cos \left( \frac{\pi}{2} - \alpha ight)$,we have: $\cos 2	h

Vedclass · Accepted Answer

$	heta = n\pi \pm \left( \frac{\pi}{4} - \frac{\alpha}{2} ight)$

Vedclass · Answer

(D) Given the equation $\cos 2	heta = \sin \alpha$. Using the identity $\sin \alpha = \cos \left( \frac{\pi}{2} - \alpha ight)$,we have: $\cos 2	heta = \cos \left( \frac{\pi}{2} - \alpha ight)$. The general solution for $\cos x = \cos y$ is $x = 2n\pi \pm y$. Therefore,$2	heta = 2n\pi \pm \left( \frac{\pi}{2} - \alpha ight)$. Dividing by $2$,we get: $	heta = n\pi \pm \left( \frac{\pi}{4} - \frac{\alpha}{2} ight)$.

The general value of $\theta$ obtained from the equation $\cos 2\theta = \sin \alpha$ is

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