If $\cos(x) = \sin(y)-\cos(z)$, prove that $xyz=\frac{\pi}{2}$

Question

I have no clue where to start from.

I tried to rewrite $\sin(y)$ as $\cos\left(\frac{\pi}{2}-y\right)$, but don't know what to do next.

Answer 1

$x=0,y=\pi /2,z=\pi /2$ is a counterexample.

Answer 2

Relationship

$$\cos(x)=\sin(y)-\cos(z) \tag{1}$$

is invariant separately by

$$x \leftrightarrow -x, \ \ \ \ \ z \leftrightarrow -z, \ \ \ \ \ y \leftrightarrow \pi-y, \cdots \tag{2}$$

Thus all relationships that are supposed to be consequences of (1) must be invariant by transformations (2), which is evidently not the case for $xyz=\pi/2$.

Generaly speaking, I don't know any trigonometric formula involving the product of some angles.