Question Number 149408 by mathdanisur last updated on 05/Aug/21
$${cos}\left(\frac{\pi}{\mathrm{2}}\:-\:\frac{\mathrm{1}}{\mathrm{2}}\:{arccos}\:\frac{\mathrm{4}}{\mathrm{5}}\right)\:=\:? \\ $$
Commented by liberty last updated on 05/Aug/21
$$\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arccos}\:\frac{\mathrm{4}}{\mathrm{5}}\right)=\mathrm{sin}\:\mathrm{u} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arccos}\:\frac{\mathrm{4}}{\mathrm{5}}=\mathrm{u}\:\Rightarrow\mathrm{arccos}\:\frac{\mathrm{4}}{\mathrm{5}}=\mathrm{2u} \\ $$$$\mathrm{cos}\:\mathrm{2u}=\frac{\mathrm{4}}{\mathrm{5}}\Rightarrow\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \mathrm{u}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\mathrm{sin}\:\mathrm{u}=\sqrt{\frac{\mathrm{1}}{\mathrm{10}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{10}}} \\ $$$$ \\ $$
Commented by mathdanisur last updated on 05/Aug/21
$${Thank}\:{You}\:{Ser} \\ $$