If $\cos(x+a)=\cos(x+y+z)$, then can we deduce that $a=y+z$?

Question

If I have the following equation:

$$\cos(x+a)=\cos(x+y+z)$$

Can I take the inverse cos of both sides?

$$\begin{align} \cos^{-1}(\cos(x + a))&=\cos^{-1}(\cos(x+y+z)) \\ x+a&=x+y+z \\ a&=y+z \end{align}$$

Is this correct?

Answer 1

No it'is wrong, let instead consider the trigonometric circle to derive that

$$\cos \theta = \cos \alpha \implies \theta = \alpha+2k\pi \quad \lor \quad \theta = -\alpha+2k\pi$$

Refer also to the related

• Solving $\cos(3x) = \cos(2x)$

Answer 2

Notice that due to $\cos\theta$ having a period of $2\pi$, for any $\lambda, \mu\in\Bbb Z, \theta\in \Bbb R$, we have:

$$\cos(\theta+2\lambda\pi)=\cos(\theta+2\mu\pi)$$

$\lambda=\mu$ is not a necessity to this end.

Effectively in your expression we have that $a=y+z+2\lambda\pi, \lambda \in \Bbb Z$.

$\lambda=0$ is a possibility, and there $a=y+z$, but there are infinitely many other possibilities as well.