# arccos x + arccos y = ...

• January 29th 2013, 01:05 PM
darence
arccos x + arccos y = ...
Proove that for real numbers $x,y\in[-1,1]$ is $\arccos x+\arccos y =\begin{cases}\arccos(xy-\sqrt{1-y^2}\sqrt{1-x^2} & ,x +y \geq0\\2\pi-\arccos(xy-\sqrt{1-y^2}\sqrt{1-x^2} & ,x+y< 0\end{cases}$
• January 29th 2013, 04:01 PM
chiro
Re: arccos x + arccos y = ...
Hey darence.

Have you tried setting up two right hand triangles and using Pythagoras and trig identities?
• January 30th 2013, 04:19 AM
darence
Re: arccos x + arccos y = ...
I tried this :
$\alpha=\arccos x\\\beta=\arccos y\\\Rightarrow \cos\alpha=x, \cos\beta=y\\\cos(\alpha+\beta)=\cos\alpha\cos\bet a-\sin\alpha\sin\beta\\\sin\alpha=\sqrt{1-\cos^2\alpha}=\sqrt{1-x^2}\\ \sin\beta=\sqrt{1-\cos^2\beta}=\sqrt{1-y^2}\\ \Rightarrow\cos\gamma=xy-\sqrt{1-x^2}\sqrt{1-y^2}\\$
where
$\gamma=\arccos(xy-\sqrt{1-x^2}\sqrt{1-y^2})$

But, where to use given conditions ?