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If $\cos 2 \theta=\sin \propto, \quad$ then $\theta=$
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The correct answer is:
$n \pi \pm\left(\frac{\pi}{4}-\frac{\alpha}{2}\right), n \in z$
We have $\cos 2 \theta=\sin \alpha \Rightarrow \cos 2 \theta=\cos (90-\alpha)$ When $\cos \theta=\cos \alpha$, we get $\theta=2 n \pi \pm \alpha, n \in Z$ $\therefore \quad 2 \theta=2 \mathrm{n} \pi \pm(90-\alpha)$
$\theta=\frac{2 n \pi}{2} \pm\left(\frac{90-\alpha}{2}\right) \Rightarrow \theta=n \pi \pm\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)$
$\theta=\frac{2 n \pi}{2} \pm\left(\frac{90-\alpha}{2}\right) \Rightarrow \theta=n \pi \pm\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)$
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