# Prove that $\sin x \cdot \sin (2x) \cdot \sin(3x) < \tfrac{9}{16}$ for all $x$

Prove that
$$\sin (x) \cdot \sin (2x) \cdot \sin(3x) < \dfrac{9}{16} \quad \forall \ x \in \mathbb{R}$$

I thought about using derivatives, but it would be too lengthy.

Any help will be appreciated.
Thanks.

• Derivatives sound like a good way to solve this. You should give it a try! – Lynn May 29 '16 at 11:25
• Wolfram says $max(sin(x).sin(3x))=\frac{9}{16}$ for a certain x at which sin(x) isn't 1. Doesn't seem easy to prove though – Evariste May 29 '16 at 11:27
• The period of that function is $2\pi$, so one way to go about it could be to graph the function in that interval and then check the value of the maximum. Another way to attack this would be to expand $sin(2x)$ and $sin(3x)$, and then work with the resulting expression. – user1790813 May 29 '16 at 12:03
• Since the maximum of $\sin(x)\cdot\sin(3x)$ is $\frac{9}{16}$ (still quite lengthy to calculate) and $\sin(2x)$ has a maximum value of $1$, you are left to proof, that there is no $x_0$ where $\sin(2x_0)=1$ and $\sin(x_0)\cdot\sin(3x_0)=\frac{9}{16}$, which would mean the expression is never $\geq \frac{9}{16}$. – qwertz May 29 '16 at 12:17
• What is amazing is that, playing with a few trigonometric identities, the derivative can write $2 \sin ^2(x) (4 \cos (2 x)+3 \cos (4 x)+2)$ from which we can easily get $\cos(2x)$ then $\sin(x)=\frac{1}{2} \sqrt{\frac{1}{3} \left(8-\sqrt{10}\right)}$ which is almagest's result. – Claude Leibovici May 29 '16 at 12:18

One has $$2\sin x\sin(3x)=\cos(2x)-\cos(4x)\ ,$$ and therefore $$2\sin x\sin(2x)\sin(3x)=(1+u-2u^2)\sqrt{1-u^2}=:f(u)\qquad(-1\leq u\leq1)\ ,$$ where we have put $\cos(2x)=:u$. One computes $$\sqrt{1-u^2}f'(u)=6u^3-2u^2-5u+1$$ with zeros at $$u\in\left\{1, -{\sqrt{10}+2\over6},{\sqrt{10}-2\over6}\right\}\ .$$ The third of these leads to the maximal value of $f(u)$, which we then have to divide by $2$. The result can be simplified to $$\sin x\sin(2x)\sin(3x)\leq{68+5\sqrt{10}\over108\sqrt{2}}\doteq0.548737<{9\over16}\ .$$

We have the following equality:

$$\sin(x) \sin(3x) = \frac{\cos(3x - x) - \cos(3x + x)}{2} = \frac{\cos(2x) - \cos(4x)}{2} = \frac{c - 2c^2 + 1}{2}$$

where $c = \cos(2x)$. The function $f(x) = -2x^2 + x + 1$ attains its global maximum at $x = 1/4$, which means that we have

$$\sin(x)\sin(2x)\sin(3x) \leq \sin(x)\sin(3x) = \frac{f(\cos(2x))}{2} \leq \frac{f(1/4)}{2} = \frac{9}{16}$$

if at most one of $\sin(x), \sin(2x), \sin(3x)$ are negative.

Note: To show that the inequality is strict, it suffices to observe that $\sin(2x) < 1$ when $\cos(2x) = 1/4$.

There should be some way to fix this solution so it applies more generally. I will revisit this answer sometime in the future if I find such an argument. If you find a way, feel free to suggest an edit.

• @K.Power This is not easily seen because it is not true in this generality: For $x=\frac54\pi$ we have $\sin x>0$, $\sin 2x<0$, $\sin 3x<0$ – Hagen von Eitzen May 29 '16 at 13:02
• @HagenvonEitzen thank you I see that is true. I was too hasty in assuming the truth of the inequality. – K.Power May 29 '16 at 13:10
• @Starfall Your edit is only correct if the minimum of $f(\cos(2x))$ is greater than $-f(1/4)$ – K.Power May 29 '16 at 13:12