Firstly, I have opened the brackets and solved using both compositions of trig and inverse trig functions and using right triangle (the results were the same):$$\arcsin(\frac{40}{41})=\gamma$$$$\frac{40}{41}=\sin\gamma.$$ Coming from the Pythagoras theorem, the adjacent side to $\gamma^\circ$ is 9, so $\cos(\gamma)=\frac{9}{41}.$ $$-\arcsin(\frac{4}{5})=\theta$$ $$-\frac{4}{5}=\sin\theta.$$ So, $\cos\theta=-\frac{3}{5};$ $$\cos(\arcsin(\frac{40}{41})-\arcsin(\frac{4}{5}))=\cos(\arcsin(\frac{40}{41}))-\cos(\arcsin(\frac{4}{5}))=\frac{9}{41}-\frac{3}{5}=-\frac{78}{205}.$$But it didn't turn out to be the right answer. Where did I make a mistake?
[Math] $\cos(\arcsin(\frac{40}{41})-\arcsin(\frac{4}{5}))=?$
trigonometry
Best Answer
You're on the right track you have that$$\cos(a-b)=\cos a\cos b+\sin a\sin b\\\cos(\arcsin\frac{40}{41})\cos(\arcsin(\frac{4}{5}))+\sin(\arcsin\frac{40}{41})\sin(\arcsin(\frac{4}{5}))=\frac{9}{41}\cdot\frac{3}{5}+\frac{40}{41}\cdot\frac{4}{5}$$