Finding max and min in $f(x) = 5\sin x + 12\sin(x+\frac{\pi}{3})$ For the function $f(x) = 5\sin x + 12\sin(x+\frac{\pi}{3})$, find the max and min value the function can be.
Own thoughts
I first noted that the function had no constant, and so the max = |amplitude|, that also means that min = -|amplitude|.
I tried what I could rewriting the function, because I know that a trig function's coefficient is its amplitude; alas I did not succeed.
$5\sin x+12\sin(x+\frac{\pi}{3}) = 5\sin x + 12(\sin x\cos\frac{\pi}{3}+\cos x\sin\frac{\pi}{3}) =\\= 5\sin x+ 6\sin x + 6\sqrt3\cos x = 11\sin x + 6\sqrt3\cos x$
Still a $\cos$ term. How can I solve this?

Hint:  Once you have an expression in the form $a\sin x+b\cos x$, think of it as being in the form $$\sqrt{a^2+b^2}\left({a\over\sqrt{a^2+b^2}}\sin x+{b\over\sqrt{a^2+b^2}}\cos x \right)$$
Then think of $a/\sqrt{a^2+b^2}$ and $b/\sqrt{a^2+b^2}$ as the sine and cosine (or vice versa) of some angle $\phi$.
