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Question Number 208704 by hardmath last updated on 21/Jun/24

$$\mathrm{Find}:$$$$\arccos (\mathrm{arctg}\frac{5}{12}+\arcsin \frac{4}{5})=?$$

Commented by mr W last updated on 21/Jun/24

$$non-sense.$$$$arccos\left(x\right)isonlydefinedfor$$$$-1⩽x⩽1.$$$$doyoumean$$$$\cos (\mathrm{arctg}\frac{5}{12}+\arcsin \frac{4}{5})=?$$

Commented by hardmath last updated on 28/Jun/24

$$\mathrm{sorry}\mathrm{dear}\mathrm{professor},\mathrm{yes}\cos ...$$

Answered by mr W last updated on 28/Jun/24

$$\alpha ={\tan }^{-1}\frac{5}{12}\Rightarrow \sin \alpha =\frac{5}{13},\cos \alpha =\frac{12}{13}$$$$\beta ={\sin }^{-1}\frac{4}{5}\Rightarrow \sin \beta =\frac{4}{5},\cos \beta =\frac{3}{5}$$$$\cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta$$$$=\frac{12}{13}\times \frac{3}{5}-\frac{5}{13}\times \frac{4}{5}=\frac{16}{65}✓$$

Commented by hardmath last updated on 28/Jun/24

$$\mathrm{thankyou}\mathrm{dear}\mathrm{professor}$$

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