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Trả lời:
Ta có:
$0<\alpha<\pi⇒0<\dfrac{\alpha}{2}<\dfrac{\pi}{2}⇒\begin{cases}\cos\dfrac{\alpha}{2}>0\\\cos\dfrac{\alpha}{4}>0\end{cases}$
$Q=\sqrt{2+\sqrt{2+2\cos\alpha}}=\sqrt{2+\sqrt{4\cos^2\dfrac{\alpha}{2}}}=\sqrt{2+2\cos\dfrac{\alpha}{2}}=\sqrt{4\cos^2\dfrac{\alpha}{4}}=2\cos\dfrac{\alpha}{4}$