When is $\sin\theta > \sin 3\theta$?

When is $\sin\theta > \sin 3\theta$?

Question

Determine for what range of values of $\theta$ between $0$ and $2\pi$ $\sin\theta > \sin 3\theta.$

Suggestion

Could we sketch the two curves $y = \sin \theta$ and $y = \sin(3\theta)$?

If $\sin A=\sin B$, what can we say about $A$ and $B$?

Alternatively, can we express $\sin 3\theta$ in terms of $\sin\theta$?

Solution

The best thing to do is sketch the two curves $y=\sin\theta$ and $y=\sin 3\theta$.

So we get intersections when $\sin \theta=\sin 3 \theta.$

Now $\sin A= \sin B \iff A = B + 2n\pi \quad \text{OR} \quad A= \pi - B + 2n\pi,$ where $n$ is a whole number.

So here we have $\theta = 3\theta + 2n\pi \implies \theta = -n\pi \quad \text{OR} \quad \theta = \pi - 3\theta + 2n\pi \implies \theta = \dfrac{(2n+1)\pi}{4}.$

So the intersection points for $y = \sin \theta$ and $y = \sin (3\theta)$ in the given range are $0, \dfrac{\pi}{4},\dfrac{3\pi}{4},\pi,\dfrac{5\pi}{4},\dfrac{7\pi}{4}, 2\pi.$

Now looking back at our sketch, we can pinpoint the ranges where

$\sin\theta > \sin 3 \theta$

as

$$\begin{align*} \frac{\pi}{4} &< \theta < \frac{3\pi}{4} \\ \pi &< \theta < \frac{5\pi}{4} \\ \frac{7\pi}{4} &< \theta < 2\pi. \end{align*}$$

Alternatively, we could say

$$\sin 3 \theta = \sin (\theta + 2\theta)=\sin\theta\cos (2\theta) +\sin (2 \theta) \cos \theta.$$

That is, we have the equation

$$\begin{align*} \sin \theta &= \sin\theta(1-2\sin^2 \theta) +2\sin \theta \cos^2 \theta \\ &= \sin \theta - 2\sin^3\theta + 2\sin\theta(1-\sin^2 \theta)\\ &= 3\sin\theta - 4\sin^3\theta. \end{align*}$$

Thus we want to solve $\sin \theta (2\sin^2 \theta -1) =0.$ The first part of the equation, $\sin\theta = 0$, gives the solutions $\theta = 0,\pi,2\pi$.

The second part of the equation, $2\sin^2 \theta -1 =0$, can be rearranged to $\sin\theta = \pm \dfrac{1}{\sqrt{2}}$, and so gives the solutions $\theta = \dfrac{\pi}{4},\,\dfrac{3\pi}{4},\,\dfrac{5\pi}{4},\,\dfrac{7\pi}{4}$ in this range.