Prove geometrically: If $\cos^{-1}x + \cos^{-1}y = A$, then $x^2 - 2xy\cos A + y^2 = \sin^2A$ 
If $$\cos^{-1}x + \cos^{-1}y = A$$ 
  prove that $$x^2 - 2xy\cos A + y^2 = \sin^2A$$

It can be proved using simple inverse trig. formulas, but it seems to me that there is also some geometry inside, because of that cosine formula.
 A: In the unit circle, partition the full angle at the centre into $2A$, $2\sin^{-1}x$, and $2\sin^{-1}y$, thereby dermining three points on the circle. One angle of the resulting trinagle is $A$. Find the side lengths.
A: Consider the circle with unit diameter ($|AD|=|A'B|=1$):

The target relation is asserting the Law of Cosines in $\triangle ABC$:
$$z^2 = x^2 + y^2 - 2 x y \cos A$$
A: Denote
\begin{align}
\alpha&=\frac{\pi}{2}-\cos^{-1}x,\\
\beta&=\frac{\pi}{2}-\cos^{-1}y,
\end{align}
then $\alpha+\beta+A=\pi$. Assume all angles are positive here, hence, there is a triangle with angles $\alpha$, $\beta$ and $A$. Now
\begin{align}
x&=\cos\left(\frac{\pi}{2}-\alpha\right)=\sin\alpha,\\
y&=\cos\left(\frac{\pi}{2}-\beta\right)=\sin\beta,
\end{align}
which means (from the law of sines) that $x$, $y$ and $z=\sin A$ can be takes as the sides in the triangle (against the angles $\alpha$, $\beta$ and $A$, correspondingly). Applying the law of cosines, it gives the required identity
$$
z^2=x^2+y^2-2xy\cos A.
$$
