Question Number 111391 by Aina Samuel Temidayo last updated on 03/Sep/20
$$\mathrm{Compute}\:\mathrm{cos}\frac{\Pi}{\mathrm{12}} \\ $$
Answered by bemath last updated on 03/Sep/20
$$\frac{\pi}{\mathrm{6}}\:=\:\mathrm{2}×\frac{\pi}{\mathrm{12}}\:\Rightarrow\:\mathrm{cos}\:\frac{\pi}{\mathrm{6}}\:=\:\mathrm{cos}\:\left(\mathrm{2}.\frac{\pi}{\mathrm{12}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}}\:=\:\mathrm{2cos}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{12}}\right)−\mathrm{1} \\ $$$$\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\mathrm{1}\:=\mathrm{2}\:\mathrm{cos}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{12}}\right)\:\Rightarrow\:\mathrm{cos}\:\frac{\pi}{\mathrm{12}}=\sqrt{\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{4}}} \\ $$$$\mathrm{cos}\:\frac{\pi}{\mathrm{12}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$
Commented by bemath last updated on 03/Sep/20
$$\mathrm{haha}..\mathrm{why}\:\mathrm{you}\:\mathrm{think}\:\mathrm{it}\:\mathrm{not}\:\mathrm{equivalent}? \\ $$
Commented by bemath last updated on 03/Sep/20
Commented by bemath last updated on 03/Sep/20
$$\mathrm{look}\:\mathrm{it}\: \\ $$
Commented by bobhans last updated on 03/Sep/20
$$\mathrm{creative}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{like}\:\mathrm{it} \\ $$
Answered by ajfour last updated on 03/Sep/20
$$\mathrm{cos}\:\left(\frac{\pi}{\mathrm{12}}\right)=\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}−\frac{\pi}{\mathrm{4}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{cos}\:\frac{\pi}{\mathrm{3}}\mathrm{cos}\:\frac{\pi}{\mathrm{4}}+\mathrm{sin}\:\frac{\pi}{\mathrm{3}}\mathrm{sin}\:\frac{\pi}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\:\:\mathrm{cos}\:\frac{\pi}{\mathrm{12}}\:=\:\frac{\sqrt{\mathrm{2}}+\sqrt{\mathrm{6}}}{\mathrm{4}}\:. \\ $$
Commented by Aina Samuel Temidayo last updated on 03/Sep/20
$$\mathrm{Thanks}. \\ $$
Commented by bemath last updated on 03/Sep/20
$$\sqrt{\left(\frac{\sqrt{\mathrm{2}}+\sqrt{\mathrm{6}}}{\mathrm{4}}\right)^{\mathrm{2}} }\:=\:\sqrt{\frac{\mathrm{8}+\mathrm{2}\sqrt{\mathrm{12}}}{\mathrm{16}}}\:=\:\sqrt{\frac{\mathrm{8}+\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{16}}} \\ $$$$=\:\sqrt{\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{4}}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$$$\mathrm{clearr}??? \\ $$
Commented by Aina Samuel Temidayo last updated on 03/Sep/20
$$\mathrm{Yes}.\:\mathrm{Thanks}. \\ $$