If sin beta is the geometric mean between sin | vedclass

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(D) Given that $\sin \beta$ is the geometric mean between $\sin \alpha$ and $\cos \alpha.$Therefore,$\sin^2 \beta = \sin \alpha \cos \alpha.$Now,$\cos

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

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(D) Given that $\sin \beta$ is the geometric mean between $\sin \alpha$ and $\cos \alpha.$Therefore,$\sin^2 \beta = \sin \alpha \cos \alpha.$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 have $\cos 2\beta = 1 - \sin 2\alpha.$Since $1 - \sin 2\alpha = (\cos \alpha - \sin \alpha)^2,$ we can write:$\cos 2\beta = (\cos \alpha - \sin \alpha)^2 = 2\left(\frac{1}{\sqrt{2}}\cos \alpha - \frac{1}{\sqrt{2}}\sin \alphaight)^2 = 2\sin^2\left(\frac{\pi}{4} - \alphaight).$Also,using the identity $\sin 	heta = \cos(\frac{\pi}{2} - 	heta),$$\cos 2\beta = 2\cos^2\left(\frac{\pi}{2} - (\frac{\pi}{4} - \alpha)ight) = 2\cos^2\left(\frac{\pi}{4} + \alphaight).$Thus,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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