Rewrite $f(x) = 3 \sin (\pi x) + 3\sqrt{3} \cos (\pi x)$ in the form $f(x) = A \sin (Kx+D)$ I got a question like that said "Rewrite $f(x) = 3 \sin (\pi x) + 3\sqrt{3} \cos (\pi x)$ in the form $f(x) = A \sin (Kx+D)$".
I'm inclined to think that since the periods are the same ($2$), that the amplitudes will just add up. But, I'm not sure.
I also need to know the rules for combining sinusoids with different periods. I know that when you're multiplying them, the one with the longer period acts as a sort of envelope for the one with the smaller period. But what do you do with different periods when you add them?
Thanks!
evamvid

This is a standard thing:
$$
\begin{align}
& \phantom{{}={}}3\sin(\pi x) + 2\sqrt{3}\cos(\pi x) = \sqrt{\Big(3\Big)^2+\Big(3\sqrt{3}\Big)^2}\Big(\frac{3}{\bullet}\sin(\pi x)+\frac{3\sqrt{3}}{\bullet}\cos(\pi x)\Big) \\[10pt]
& = 6\Big(\frac{3}{6}\sin(\pi x)+\frac{3\sqrt{3}}{6}\cos(\pi x)\Big) = 6\Big( \frac 1 2 \sin(\pi x)+\frac{\sqrt{3}}{2}\cos(\pi x) \Big) \\[10pt]
& = 6\Big(\cos\alpha\sin(\pi x) + \sin\alpha\cos(\pi x)\Big) \\[10pt]
& = 6 \sin(\pi x + \alpha).
\end{align}
$$
I leave it to you to figure out what number $\alpha$ is.
