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Solution

Since $\sin\beta$ is the G.M. of $\sin\alpha$ and $\cos\alpha$, $\sin^2\beta=\sin\alpha\cos\alpha=\dfrac12\sin2\alpha$ So, $\cos2\beta=1-2\sin^2\beta=1-\sin2\alpha$ But $1-\sin2\alpha=2\sin^2\left(\dfrac{\pi}{4}-\alpha\right)$