Question Number 198558 by hardmath last updated on 22/Oct/23
Commented by mr W last updated on 22/Oct/23
$${hardmath}={shrinava}\:? \\ $$
Answered by mr W last updated on 22/Oct/23
$$\mathrm{cos}\:\frac{\pi}{\mathrm{6}}=\frac{\mathrm{1}}{\:\mathrm{2}}\sqrt{\mathrm{3}} \\ $$$$\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:\frac{\pi}{\mathrm{12}}−\mathrm{1}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{cos}\:\frac{\pi}{\mathrm{12}}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$$${similarly} \\ $$$$\Rightarrow\mathrm{cos}\:\frac{\pi}{\mathrm{24}}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}} \\ $$$$\Rightarrow\mathrm{cos}\:\frac{\pi}{\mathrm{48}}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}}} \\ $$$$\Rightarrow\mathrm{cos}\:\frac{\pi}{\mathrm{96}}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}}}} \\ $$$$\Rightarrow\mathrm{cos}\:\frac{\pi}{\mathrm{192}}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}}}}} \\ $$$${i}.{e}.\:\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}}}}}=\mathrm{2}\:\mathrm{cos}\:\frac{\pi}{\mathrm{192}} \\ $$
Commented by hardmath last updated on 28/Oct/23
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{professor} \\ $$