Question Number 207638 by hardmath last updated on 21/May/24
Commented by TonyCWX08 last updated on 22/May/24
$${arcsin}\left({sin}\mathrm{5}\right)=\mathrm{5} \\ $$$${arccis}\left({cos}\mathrm{6}\right)=\mathrm{6} \\ $$$${arctan}\left({tan}\mathrm{2}\right)=\mathrm{2} \\ $$$$\mathrm{5}+\mathrm{6}+\mathrm{2}=\mathrm{13} \\ $$
Commented by mr W last updated on 22/May/24
$${arcsin}\left(\mathrm{sin}\:\mathrm{5}°\right)=\mathrm{5}°\:\left({in}\:{degree}\right) \\ $$$${arcsin}\left(\mathrm{sin}\:\mathrm{5}\right)\neq\mathrm{5}\:\:\left({in}\:{rad}\right) \\ $$$${therefore}\:{please}\:{make}\:{clear}\:{what} \\ $$$${you}\:{mean}\:{with}\:“\mathrm{5}''.\:\mathrm{5}°\:{or}\:\mathrm{5}\:{rad}? \\ $$$${usually}\:\mathrm{5}\:{means}\:\mathrm{5}\:{rad}. \\ $$
Commented by mr W last updated on 22/May/24
Commented by mr W last updated on 22/May/24
Answered by Simurdiera last updated on 21/May/24
$$=\mathrm{13}\: \\ $$
Commented by hardmath last updated on 21/May/24
$$\mathrm{How} \\ $$
Answered by mr W last updated on 22/May/24
$${assume}\:{the}\:{unit}\:{is}\:{rad}. \\ $$$$\mathrm{5}=\mathrm{2}\pi−\left(\mathrm{2}\pi−\mathrm{5}\right) \\ $$$$\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{sin}\:\mathrm{5}\right)=−\left(\mathrm{2}\pi−\mathrm{5}\right)=\mathrm{5}−\mathrm{2}\pi \\ $$$$\mathrm{6}=\mathrm{2}\pi−\left(\mathrm{2}\pi−\mathrm{6}\right) \\ $$$$\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{cos}\:\mathrm{6}\right)=\mathrm{2}\pi−\mathrm{6} \\ $$$$\mathrm{2}=\pi−\left(\pi−\mathrm{2}\right) \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{tan}\:\mathrm{2}\right)=−\left(\pi−\mathrm{2}\right)=\mathrm{2}−\pi \\ $$$$\Rightarrow\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{sin}\:\mathrm{5}\right)+\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{cos}\:\mathrm{6}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{tan}\:\mathrm{2}\right) \\ $$$$=\mathrm{5}−\mathrm{2}\pi+\mathrm{2}\pi−\mathrm{6}+\mathrm{2}−\pi=\mathrm{1}−\pi \\ $$$$\approx−\mathrm{2}.\mathrm{14159} \\ $$