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Question: If π\< 2θ\<\(\frac{3\pi}{2}\), then \(\sqrt{2 + \sqrt{2 + 2\cos 4\theta}}\) is equal to...

If π< 2θ<3π2\frac{3\pi}{2}, then 2+2+2cos⁡4θ\sqrt{2 + \sqrt{2 + 2\cos 4\theta}} is equal to

A

–2cosθ

B

–2sinθ

C

2cosθ

D

2sinθ

Answer

2sinθ

Explanation

Solution

2+2(1+cos⁡4θ)=2+2∣cos⁡2θ∣\sqrt{2 + \sqrt{2(1 + \cos 4\theta)}} = \sqrt{2 + 2|\cos 2\theta|}

=2(1−cos⁡2θ)\sqrt{2(1 - \cos 2\theta)}

= 2 | sinθ | = 2sinθ as π2<θ<3π4\frac{\pi}{2} < \theta < \frac{3\pi}{4}