Question Number 208704 by hardmath last updated on 21/Jun/24
$$\mathrm{Find}: \\ $$$$\mathrm{arccos}\:\left(\mathrm{arctg}\:\frac{\mathrm{5}}{\mathrm{12}}\:\:+\:\:\mathrm{arcsin}\:\frac{\mathrm{4}}{\mathrm{5}}\right)\:=\:? \\ $$
Commented by mr W last updated on 21/Jun/24
$${non}−{sense}. \\ $$$${arccos}\:\left({x}\right)\:{is}\:{only}\:{defined}\:{for}\: \\ $$$$−\mathrm{1}\leqslant{x}\leqslant\mathrm{1}. \\ $$$${do}\:{you}\:{mean} \\ $$$$\mathrm{cos}\:\left(\mathrm{arctg}\:\frac{\mathrm{5}}{\mathrm{12}}\:\:+\:\:\mathrm{arcsin}\:\frac{\mathrm{4}}{\mathrm{5}}\right)\:=\:? \\ $$
Commented by hardmath last updated on 28/Jun/24
$$\mathrm{sorry}\:\mathrm{dear}\:\mathrm{professor},\:\mathrm{yes}\:\:\:\mathrm{cos}… \\ $$
Answered by mr W last updated on 28/Jun/24
$$\alpha=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{5}}{\mathrm{12}}\:\Rightarrow\mathrm{sin}\:\alpha=\frac{\mathrm{5}}{\mathrm{13}},\:\mathrm{cos}\:\alpha=\frac{\mathrm{12}}{\mathrm{13}} \\ $$$$\beta=\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{5}}\:\Rightarrow\mathrm{sin}\:\beta=\frac{\mathrm{4}}{\mathrm{5}},\:\mathrm{cos}\:\beta=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\mathrm{cos}\:\left(\alpha+\beta\right)=\mathrm{cos}\:\alpha\:\mathrm{cos}\:\beta−\mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta \\ $$$$=\frac{\mathrm{12}}{\mathrm{13}}×\frac{\mathrm{3}}{\mathrm{5}}−\frac{\mathrm{5}}{\mathrm{13}}×\frac{\mathrm{4}}{\mathrm{5}}=\frac{\mathrm{16}}{\mathrm{65}}\:\checkmark \\ $$
Commented by hardmath last updated on 28/Jun/24
$$\mathrm{thankyou}\:\mathrm{dear}\:\mathrm{professor} \\ $$