Find-arccos-arctg-5-12-arcsin-4-5-

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

Find :

\arccos (arctg \frac{5}{12} + \arcsin \frac{4}{5}) = ?

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

n o n - s e n s e .

a r c c o s (x) i s o n l y d e f i n e d f o r

- 1 ⩽ x ⩽ 1 .

d o y o u m e a n

\cos (arctg \frac{5}{12} + \arcsin \frac{4}{5}) = ?

Commented by hardmath last updated on 28/Jun/24

sorry dear professor, yes \cos \dots

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

α = \tan^{- 1} \frac{5}{12} \Rightarrow \sin α = \frac{5}{13}, \cos α = \frac{12}{13}

β = \sin^{- 1} \frac{4}{5} \Rightarrow \sin β = \frac{4}{5}, \cos β = \frac{3}{5}

\cos (α + β) = \cos α \cos β - \sin α \sin β

= \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

thankyou dear professor