# Maximum of a trigonometric function without derivatives

I know that I can find the maximum of this function by using derivatives but is there an other way of finding the maximum that does not involve derivatives? Maybe use a well-known inequality or identity?

$f(x)=\sin(2x)+2\sin(x)$

The idea is to use $\sin^2 x + \cos^2x = 1$ to reduce to dealing with only 1 trigonometric function, and then proceed as a standard 1-variable inequality.

We wish to find the maximum of $f(x) = \sin 2x + 2 \sin x = 2 \sin x ( 1 + \cos x )$.

It is clear that we may assume $\sin x \geq 0, \cos x \geq 0$, to maximize this product.

Let's consider $$[f(x)]^2 = 4 \sin^2 x ( 1 + \cos x)^2 = 4 (1-\cos^2x ) ( 1 + \cos x)^2 = 4 ( 1 - \cos x ) ( 1 + \cos x)^3$$

By AM-GM, applied to $3(1-\cos x) , (1 + \cos x), (1 + \cos x), (1 + \cos x)$, we get that

$\sqrt{3 ( 1 - \cos x ) ( 1 + \cos x)^3} \leq \frac{ 6}{4}$, or that $( 1 - \cos x ) ( 1 + \cos x)^3 \leq \frac{27}{16}$.

Hence, $[f(x)]^2 \leq \frac{27}{4}$ so $f(x) \leq \frac{3 \sqrt{3} } {2}$.

It remains to verify that equality can occur, which it does at $3(1-\cos x) = 1 + \cos x$ of $\cos x = \frac{1}{2}$.

• You beat me to it! +1 – Namaste Jun 21 '13 at 23:49
• Very nice! Exactly what I was looking for. Thanks Calvin – EricAm Jun 21 '13 at 23:51
• @AdamYac The motivation is pretty clear, and can be routine, if you're familiar with 1 variable inequalities. E.g. try to find the max of $( 1 + \sin x) ( 1 + \cos x)$. – Calvin Lin Jun 21 '13 at 23:53
• Edit: Never mind, I carried out the steps by hand and I see why now. – EricAm Jun 22 '13 at 18:22

As Calvin Lin has observed $\sin x,\cos x>0 \implies 0\le x\le \frac\pi2$ $\implies 0\le 2x\le \pi\implies0\le \pi-2x\le \pi$

As sine function is concave in $[0,\pi]$ $$\sin2x+2\sin x$$

$$=\sin(\pi-2x)+\sin x+\sin x\le 3\sin\frac{(\pi-2x+x+x)}3=3\sin\frac{\pi}3=\frac{3\sqrt3}2$$ the equality occurs when $x=x=\pi-2x$ i.e., when $x=\frac\pi3$

• Doesn't showing that something is concave/convex use calculus? Of course, you can try and do it by definition, but i think showing $\lambda \sin x + (1-\lambda) \sin y \leq \sin ( \lambda x +(1-\lambda)y)$ directly would be tricky. – Calvin Lin Jun 23 '13 at 16:17