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?

  • $\begingroup$ how about deriving? $\endgroup$
    – Lolman
    Oct 13, 2014 at 19:48

1 Answer 1


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 $\phi$.


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