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Question Number 135111 by deleteduser12 last updated on 10/Mar/21

Answered by Ñï= last updated on 10/Mar/21

$$2\cos \theta =\sqrt{2+\sqrt{2+\sqrt{3}}}$$$$4\cos {}^2\theta =2+\sqrt{2+\sqrt{3}}$$$$2(2\cos {}^2\theta -1)=2\cos 2\theta =\sqrt{2+\sqrt{3}}$$$$4\cos {}^{2}2\theta =2+\sqrt{3}$$$$2(2\cos {}^{2}2\theta -1)=2\cos 4\theta =\sqrt{3}$$$$\mathrm{co}s4\theta =\frac{\sqrt{3}}{2}$$$$4\theta =\frac{\pi }{6}+2k\pi$$$$\theta =\frac{\pi }{24}+\frac{k\pi }{2}(k\in \mathbb{Z})$$

Answered by mr W last updated on 10/Mar/21

$${\cos }^2\theta =\frac{1}{2}+\frac{1}{4}\sqrt{2+\sqrt{3}}$$$${2\cos }^2\theta -1=\frac{1}{2}\sqrt{\frac{{(\sqrt{3}+1)}^2}{2}}=\frac{(\sqrt{3}+1)}{2\sqrt{2}}$$$$\cos 2\theta =\frac{\sqrt{3}}{2}\times \frac{1}{\sqrt{2}}+\frac{1}{2}\times \frac{1}{\sqrt{2}}$$$$\cos 2\theta =\cos \frac{\pi }{6}\times \cos \frac{\pi }{4}+\sin \frac{\pi }{6}\times \sin \frac{\pi }{4}$$$$\cos 2\theta =\cos (\frac{\pi }{4}-\frac{\pi }{6})$$$$2\theta =\pm (\frac{\pi }{4}-\frac{\pi }{6})=\pm \frac{\pi }{12}$$$$\Rightarrow \theta =\pm \frac{\pi }{24}=\pm 7.5°(+2k\pi )$$

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