How to prove $a\cos(\omega t + \phi_1) + b\cos(\omega t + \phi_2) = c\sin(d t + \phi_3)$

Question

One of my tutoring students has been given an equation representing the first half of the equality in the title, and has to find $c$, $d$ and $\phi_3$ for the second half of the inequality.

I know there's a way to do it for the LHS without the phis, but I can't figure out how to do it with the phis included.

It's quite possible that $d = \omega$, but I figured making the question more general might help the solution be more useful.

Thank you all in advance for any help you may be able to give me.

Answer 1

Expand $\cos$ and $\sin$ in the LHS to get $A\cos(\omega t) + B\sin(\omega t)$ in the LHS ($A$ and $B$ are the new coefficients of $\cos$ and $\sin$).

Then, we can write this as $$\sqrt{A^2+B^2} \sin(\omega t + c)$$ where $$c = \cos^{-1} \frac{B}{\sqrt{A^2+B^2}}$$

Equating with the RHS, we get $c = \sqrt{A^2+B^2}$, $d = \omega$, and $\phi_3 = c$