Funkcje cyklometryczne

[lukasz037]

Witam. Mam problem z jednym zadaniem, mianowicie:

sin(arcsin\(\frac{2}{3}\) + arccos\(\frac{1}{3}\))

Prosze o pokaznie krok po kroku jak takie cudo rozwiazac. Z gory dziekuje

[artuditus]

\(arcsin \frac{2}{3} \Rightarrow sin \alpha = \frac{2}{3}\)
\(arccos \frac{1}{3} \Rightarrow cos \beta = \frac{1}{3}\)
\(sin(arcsin \frac{2}{3}+arccos \frac{1}{3} = sin( \alpha + \beta )=sin \alpha cos \beta +sin \beta cos \alpha\)
\(sin \beta = \sqrt{1-{cos \beta}^2 }= \frac{2 \sqrt{2} }{3}\)
\(cos \alpha = \sqrt{1-{sin \alpha}^2 }= \frac{ \sqrt{5} }{3}\)
\(sin(arcsin \frac{2}{3}+arccos \frac{1}{3})= \frac{2}{9}+ \frac{2 \sqrt{2}+ \sqrt{5} }{3}\)

[radagast]

\(arcsin \frac{2}{3} \Rightarrow sin \alpha = \frac{2}{3}\)
\(arccos \frac{1}{3} \Rightarrow cos \beta = \frac{1}{3}\)
\(sin(arcsin \frac{2}{3}+arccos \frac{1}{3} = sin( \alpha + \beta )=sin \alpha cos \beta +sin \beta cos \alpha\)
\(sin \beta = \sqrt{1-{cos \beta}^2 }= \frac{2 \sqrt{2} }{3}\)
\(cos \alpha = \sqrt{1-{sin \alpha}^2 }= \frac{ \sqrt{5} }{3}\)
\(sin(arcsin \frac{2}{3}+arccos \frac{1}{3})= \frac{2}{9}+ \frac{2 \sqrt{2}+ \sqrt{5} }{3}\)

Troszkę poprawię zapis:
\(arcsin \frac{2}{3}= \alpha \Rightarrow sin \alpha = \frac{2}{3}\)
\(arccos \frac{1}{3}= \beta \Rightarrow cos \beta = \frac{1}{3}\)
\(sin(arcsin \frac{2}{3}+arccos \frac{1}{3} )= sin( \alpha + \beta )=sin \alpha cos \beta +sin \beta cos \alpha\)
\(sin \beta = \sqrt{1-{cos^2 \beta} }= \frac{2 \sqrt{2} }{3}\)
\(cos \alpha = \sqrt{1-{sin ^2\alpha} }= \frac{ \sqrt{5} }{3}\)
\(sin(arcsin \frac{2}{3}+arccos \frac{1}{3})= \frac{2}{9}+ \frac{2 \sqrt{2}+ \sqrt{5} }{3}\)