Question Number 135111 by faysal last updated on 10/Mar/21
Answered by Ñï= last updated on 10/Mar/21
$$\mathrm{2cos}\:\theta=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}} \\ $$$$\mathrm{4cos}\:^{\mathrm{2}} \theta=\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$$$\mathrm{2}\left(\mathrm{2cos}\:^{\mathrm{2}} \theta−\mathrm{1}\right)=\mathrm{2cos}\:\mathrm{2}\theta=\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$$$\mathrm{4cos}\:^{\mathrm{2}} \mathrm{2}\theta=\mathrm{2}+\sqrt{\mathrm{3}} \\ $$$$\mathrm{2}\left(\mathrm{2cos}\:^{\mathrm{2}} \mathrm{2}\theta−\mathrm{1}\right)=\mathrm{2cos}\:\mathrm{4}\theta=\sqrt{\mathrm{3}} \\ $$$$\mathrm{co}{s}\:\mathrm{4}\theta=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\mathrm{4}\theta=\frac{\pi}{\mathrm{6}}+\mathrm{2}{k}\pi \\ $$$$\theta=\frac{\pi}{\mathrm{24}}+\frac{{k}\pi}{\mathrm{2}}\:\:\:\:\:\:\:\left({k}\in\mathbb{Z}\right) \\ $$
Answered by mr W last updated on 10/Mar/21
$$\mathrm{cos}^{\mathrm{2}} \:\theta=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$$$\mathrm{2cos}^{\mathrm{2}} \:\theta−\mathrm{1}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}}}=\frac{\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$\mathrm{cos}\:\mathrm{2}\theta=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\mathrm{cos}\:\mathrm{2}\theta=\mathrm{cos}\:\frac{\pi}{\mathrm{6}}×\mathrm{cos}\:\frac{\pi}{\mathrm{4}}+\mathrm{sin}\:\frac{\pi}{\mathrm{6}}×\mathrm{sin}\:\frac{\pi}{\mathrm{4}} \\ $$$$\mathrm{cos}\:\mathrm{2}\theta=\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}−\frac{\pi}{\mathrm{6}}\right) \\ $$$$\mathrm{2}\theta=\pm\left(\frac{\pi}{\mathrm{4}}−\frac{\pi}{\mathrm{6}}\right)=\pm\frac{\pi}{\mathrm{12}} \\ $$$$\Rightarrow\theta=\pm\frac{\pi}{\mathrm{24}}=\pm\mathrm{7}.\mathrm{5}°\left(+\mathrm{2}{k}\pi\right) \\ $$