If sin beta is the geometric mean between sin | vedclass

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(D) Since $\sin \beta$ is the geometric mean between $\sin \alpha$ and $\cos \alpha,$$\sin^2 \beta = \sin \alpha \cos \alpha.$Now,$\cos 2\beta = 1 - 2

Vedclass · Accepted Answer

Both $(a)$ and $(b)$

Vedclass · Answer

(D) Since $\sin \beta$ is the geometric mean between $\sin \alpha$ and $\cos \alpha,$$\sin^2 \beta = \sin \alpha \cos \alpha.$Now,$\cos 2\beta = 1 - 2\sin^2 \beta = 1 - 2\sin \alpha \cos \alpha.$Using the identity $\sin 2\alpha = 2\sin \alpha \cos \alpha,$ we get $\cos 2\beta = 1 - \sin 2\alpha.$Since $1 = \sin^2 \alpha + \cos^2 \alpha,$ we have $\cos 2\beta = \cos^2 \alpha + \sin^2 \alpha - 2\sin \alpha \cos \alpha = (\cos \alpha - \sin \alpha)^2.$We can write this as $2\left( \frac{1}{\sqrt{2}}\cos \alpha - \frac{1}{\sqrt{2}}\sin \alpha ight)^2 = 2\left( \sin \frac{\pi}{4} \cos \alpha - \cos \frac{\pi}{4} \sin \alpha ight)^2 = 2\sin^2\left( \frac{\pi}{4} - \alpha ight).$This matches option $(a).$Also,using $\cos^2 	heta = \sin^2(\frac{\pi}{2} - 	heta),$ we have $2\sin^2\left( \frac{\pi}{4} - \alpha ight) = 2\cos^2\left( \frac{\pi}{2} - (\frac{\pi}{4} - \alpha) ight) = 2\cos^2\left( \frac{\pi}{4} + \alpha ight).$This matches option $(b).$Therefore,both $(a)$ and $(b)$ are correct.

If $\sin \beta$ is the geometric mean between $\sin \alpha$ and $\cos \alpha,$ then $\cos 2\beta$ is equal to

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