Show that a function of the form
$x(t) = K_1 \cos\beta t + K_2 \sin\beta t$
Can be written as
$x(t) = K\cos(Bt-\phi)$
Where $K = \sqrt {K_1^2 + K_2^2}$
I know that linear systems with complex coefficients are sometimes expressed in this form, however I'm not sure if/how that would be useful to solve this problem.
Any suggestions on how to approach this would be greatly appreciated.
$$x(t) = K\left( \frac{K_1}{K}\cos(\beta t) + \frac{K_2}{K}\sin(\beta t) \right) $$ Since $K \geq K_1, K_2$ we get $\frac{K_1}{K}, \frac{K_2}{K} \leq 1$, and since $\left(\frac{K_1}{K}\right)^2 +\left(\frac{K_2}{K}\right)^2 =1$ I can choose an angle $\phi$ such that $\cos(\phi) = \frac{K_1}{K}$ and $\sin(\phi)= \frac{K_2}{K}$. So then $$ x(t) = K (\cos(\phi)\cos(\beta t) + \sin(\phi)\sin(\beta t)) = K\cos(\beta t - \phi)$$ Where the last equality holds by the subtraction formula for cosine.