$A=B$ if $A\cos(\omega_1t+\phi_1)=B\cos(\omega_2t+\phi_2)$ for every $t∈ℝ_{≥0}$ I'm having a lot of trouble about an apparently simple task. I have the following trigonometric equation:
$A\cos(\omega_1t+\phi_1)=B\cos(\omega_2t+\phi_2)$
which holds for every $t \in [0,+\infty)$, where $\omega_1,\omega_2,\phi_1,\phi_2 \in \mathbb{R}$ are fixed, with $\omega_1 \neq 0$ and $\omega_2 \neq 0$.
With $A$ and $B$ positive, I need to show that $A=B$, but I'm really stuck. I tried to find a value of $t$ such that $\omega_1t+\phi_1=\pi$ and so on, but after a lot of calculation I can't conclude anything. Any help or hint would be really appreciated!
 A: The easiest way to prove this is not with a direct solution, but with a contrapositive proof. That is, instead of trying to prove that if
$$\forall t \in [0, \infty)\ [A \cos(\omega_1 t + \phi_1) = B \cos(\omega_2 + \phi_2)]$$
implies
$$A = B$$
try to prove that
$$A \ne B$$
implies
$$\exists t \in [0, \infty)\ [A \cos(\omega_1 t + \phi_1) \ne B \cos(\omega_2 t \phi_2)]$$
which is the logical negative of the statement you gave before. Note that this is an existential statement, so we only need to find one $t$ that works. And that's easy to do.
Cosine has a range of $[-1, 1]$. Hence $t \mapsto A \cos(\omega_1 t + \phi_1)$ has a range of $[-A, A]$. Similarly, the one involving $B$ has a range of $[-B, B]$.
Suppose $A > B$. Consider a point $t$ where $A \cos(\omega_1 t + \phi_1) = A$. At this same point, $B \cos(\omega_2 t + \phi_2)$ cannot equal $A$, because this latter expression can only get as large as $B$. Hence we have found the value $t$ for which the two sides are unequal.
The case for $A < B$ is pretty much the same way. Thus we have proven the contrapositive, and so too, proven the original statement.
A: You know that $\cos \theta = 0$ if and only if $\theta = (2 k + 1) \pi / 2$, use that to find a relation between $\omega_1 t + \phi_1$ and $\omega_2 t + \phi_2$. Then the value of $A \cos(\omega_1 t + \phi_1)$ and $B \cos(\omega_2 t + \phi_2)$ must agree when $\omega_1 t + \phi_1 = 0$. You'll find two solutions, $A = B$ and $A = -B$.
Alternatively, $\cos \theta$ has maxima exactly at $\theta = 2 k \pi$, use that to match up the functions like above.
