Suppose that real numbers $x, y, z$ satisfy:
$$\frac{\cos x + \cos y + \cos z}{\cos(x + y + z)}
=
\frac{\sin x + \sin y + \sin z}{\sin (x + y + z )}
= p$$
Then prove that:
$$\cos (x + y) + \cos (y + z ) + \cos (x + z) = p$$
I am not even getting where to start? Please help.
Above, $O$ is the centre of the sphere, $x$ is its radius. Writing the area of trapezoid $ABCD$ as a sum of three triangles, we can easily solve for $x$ to get:
Best Answer
note $$\cos{(x+y)}=\cos{[(x+y+z)-z]}=\cos{(x+y+z)}\cos{z}+\sin{(x+y+z)}\sin{z}$$ and $$\cos{(y+z)}=\cos{(x+y+z)}\cos{x}+\sin{(x+y+z)}\sin{x}$$ $$\cos{(z+x)}=\cos{(x+y+z)}\cos{y}+\sin{(x+y+z)}\sin{y}$$ add this three \begin{align*} &\cos{(x+y)}+\cos{(y+z)}+\cos{(x+z)}\\ &=(\cos{x}+\cos{y}+\cos{z})\cos{(x+y+z)}+(\sin{x}+\sin{y}+\sin{z})\sin{(x+y+z)}\\ &=p\cos^2{(x+y+z)}+p\sin^2{(x+y+z)}\\ &=p \end{align*}