$\cos(\arcsin(\frac{40}{41})-\arcsin(\frac{4}{5}))=?$

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?

• $\cos(a-b)\neq \cos a-\cos b$ – kingW3 Jun 23 '17 at 14:43

You're on the right track you have that$$\cos(a-b)=\cos a\cos b+\sin a\sin b\\\cos(\arcsin\frac{40}{41})\cos(\arcsin(\frac{4}{5}))+\sin(\arcsin\frac{40}{41})\sin(\arcsin(\frac{4}{5}))=\frac{9}{41}\cdot\frac{3}{5}+\frac{40}{41}\cdot\frac{4}{5}$$
• Why all these $-$ signs in the last line? The argument of the cosine in the question is between $0$ and $\pi/2$. – Bernard Jun 23 '17 at 14:50
• @Bernard Yeah you're right, thanks. I've misread the question as $\cos(\arcsin(40/41)-\arcsin(-4/5))$ – kingW3 Jun 23 '17 at 14:56
There's your problem right there. You can't distribute $\cos$ across a sum or difference.