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Question Number 172817 by Mikenice last updated on 01/Jul/22

Answered by som(math1967) last updated on 02/Jul/22

$$\sqrt{2+\sqrt{2+\sqrt{2(1+cos8\theta )}}}$$$$\sqrt{2+\sqrt{2+2cos4\theta }}$$$$\sqrt{2+\sqrt{2(1+cos4\theta )}}$$$$\sqrt{2+2cos2\theta }=\sqrt{2(1+cos2\theta )}=2cos\theta$$

Answered by CElcedricjunior last updated on 02/Jul/22

$$\sqrt{2+\sqrt{2+\sqrt{2+2\mathbf{cos}8\mathbf{\Theta }}}}=\mathbf{k}$$$$\mathbf{cos}2\mathbf{x}=2{\mathbf{cos}}^2\mathbf{x}-1=>4{\mathbf{cos}}^2\mathbf{x}=2+2\mathbf{cos}2\mathbf{x}$$$$\mathbf{k}=\sqrt{2+\sqrt{2+\sqrt{4{\mathbf{cos}}^{2}4\mathbf{\Theta }}}}$$$$\mathbf{k}=\sqrt{2+\sqrt{2+2\mathbf{cos}4\mathbf{\Theta }}}\mathbf{si}\mathbf{\Theta }\in ]\frac{-3\mathit{\pi }}{8};\frac{\mathit{\pi }}{8}[$$$$\mathbf{k}=\sqrt{2+2\mathbf{cos}2\mathbf{\Theta }}\mathit{s}\mathit{i}\mathbf{\Theta }\in ]\frac{-3\mathit{\pi }}{4};\frac{\mathit{\pi }}{4}[$$$$\mathbf{k}=2\mathbf{cos}\mathbf{\Theta }\mathit{s}\mathit{i}\mathbf{\Theta }\in ]\frac{-3\mathit{\pi }}{2};\frac{\mathit{\pi }}{2}[$$$$$$$$........\mathcal{L}ec\acute{el}\stackrel{`}{ebre}cedricjunior......$$$$$$

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