Trigonometric equation $A\cos(ax+b_1)+B\cos(ax+b_2)=C\cos(ax+b_3)$

Question

I'm having a lot of trouble in the following task: consider the real numbers $A,B,C,a,b_1,b_2,b_3 \in \mathbb{R}$ such that the following trigonometric equation holds for every $x \in \mathbb{R}$:

$A\cos(ax+b_1)+B\cos(ax+b_2)=C\cos(ax+b_3)$

with $A,B,C,a>0$. Prove that $A+B=C$.

$ $

My incomplete attempt:

Let's differentiate both sides wrt x. Then we have:

$A\sin(ax+b_1)+B\sin(ax+b_2)=C\sin(ax+b_3)$

Taking the square of both sides in both the equations and then adding them together we obtain:

$\cos(b_1-b_2)=\frac{A_3^2-A_1^2-A_2^2}{2A_1A_2}$. Now if $b_1=b_2$ we have that $\cos(b_1-b_2)=1$ and then $A_1+A_2=A_3$. However, here I'm taking $b_1=b_2$ without any reason. Every effort I make to show that $b_1=b_2$ take me, after a lot of calculation, to an identity. Thank you for any help or hint!

Answer 1

Prove this holds for all $x$: $$ \cos(x + \pi/3) + \cos(x - \pi/3) = \cos(x) $$