# To find the roots of $3\cos(3\theta)\sin\theta+\sin(3\theta)\cos\theta$

I want to find the points in the polar plane where the tangent to the curve $r=\sin(3\theta)$ is parallel to the polar axis. I computed the slope of the tangent to be $3\cos(3\theta)\sin\theta+\sin(3\theta)\cos\theta$. Its pretty obvious that $n\pi$, $(2n+1)\frac{\pi}{2}$ are roots but a search on wolfram reveals that $θ = π n-tan^{-1}(\sqrt{3/5})$ are also roots but I cant seem to be able to show that.

I tried using the triple angle relation for sine an cosine but to no avail. Any kind of help would be great.

• One observation: The stated equation remains valid if one replaces $\theta\to -\theta$, meaning that roots come in $\pm$ pairs. The additional tan^{-1} solutions identified by WA, however, are not symmetric in this sense. So these aren't the only solutions. – Semiclassical Aug 8 '16 at 16:28

As noted in the OP, there are obvious roots at $\theta=n\pi$ corresponding to the zeros of $\sin \theta$. These are a bit of a distraction; to get rid of these, we assume $\theta\neq n\pi$ and divide both sides by $\sin\theta$ to get

$$3\cos 3\theta+\frac{\sin 3\theta}{\sin\theta}\cos\theta=0.$$

The advantage of this is that this equation can be expressed as a polynomial in the variable $x=\cos\theta$. To see this, we recall the triple angle formulas to write $$\cos 3\theta=4\cos^3\theta-3\cos\theta=4x^3-3x,$$ $$\dfrac{\sin 3\theta}{\sin\theta}=3-4\sin^2\theta=4\cos^2\theta-1=4x^2-1.$$

Therefore the above equation becomes $$3(4x^3-3x)+(4x^2-1)x=16x^3-10x=0\implies x=0\text{ or }x=\pm \sqrt{5/8}.$$ The solutions to $x=\cos\theta=0$ correspond to the $\theta=n\pi +\pi/2$ roots noted in the OP. But we also have the solution set $\cos\theta=\pm \sqrt{5/8},$ which can also be expressed as $\tan\theta=\pm \sqrt{3/5}$. This is the second solution set identified by Wolfram Alpha.

• I should note that, more generally, one finds that $\cos(n\theta)$ and $\sin(n\theta+\theta)/\sin\theta$ are polynomials of degree $n$ in the variable $\cos\theta$. (More precisely, these are the so-called Chebyshev polynomials of the first and second kinds.) So this strategy would also have worked had we been working, say, with the polar curve $r=\sin n\theta$ instead. – Semiclassical Aug 8 '16 at 16:38

Notice, for finding the roots:

$$3\cos\left(3\theta\right)\sin\left(\theta\right)+\sin\left(3\theta\right)\cos\left(\theta\right)=2\cos\left(\theta\right)\sin\left(\theta\right)\left(4\cos\left(2\theta\right)-1\right)$$

So, for the roots we split it into three equations:

• $$\cos\left(\theta\right)=0$$
• $$\sin\left(\theta\right)=0$$
• $$4\cos\left(2\theta\right)-1=0$$

Using $\cos3\theta=4\cos^3\theta-3\cos\theta$ and $\sin3\theta=3\sin\theta-4\sin^3\theta,$

$$3\cos\left(3\theta\right)\sin\left(\theta\right)+\sin\left(3\theta\right)\cos\left(\theta\right) =\cos\theta\sin\theta\{3(4\cos^2\theta-3)+3-4\sin^2\theta\}$$

Now use $\cos2\theta=2\cos^2\theta-1=1-2\sin^2\theta$

$x = r \cos \theta\\ y = r \sin \theta\\ \frac {dx}{d\theta} = \frac {dr}{d\theta} \cos \theta - r \sin \theta\\ \frac {dy}{d\theta} = \frac {dr}{d\theta} \sin \theta + r \cos \theta\\ r = \sin 3\theta \frac {dr}{d\theta} = 3 \cos 3\theta\\ \frac {dy}{dx} = \frac {3 \cos 3\theta sin\theta + sin 3\theta\cos\theta}{3\cos3\theta\cos\theta - \sin3\theta\sin\theta}=0\\ 3 \cos 3\theta \sin\theta + \sin 3\theta\cos\theta = 0$

This is where you left of...

$\sin (3\theta + \theta) = \sin 3\theta \cos\theta + \cos 3\theta \sin\theta\\ \sin (3\theta - \theta) = \sin 3\theta \cos\theta - \cos 3\theta \sin\theta\\ \sin 4\theta + \sin 2\theta = 2\sin 3\theta \cos\theta\\ \sin 4\theta - \sin 2\theta= 2\cos 3\theta \sin\theta$

$3 \cos 3\theta \sin\theta + \sin 3\theta\cos\theta = 0\\ \frac12 (3\sin 4\theta - 3 \sin 2\theta + \sin 4\theta + \sin 2\theta)= 0\\ 2\sin 4\theta - \sin 2\theta = 0 \\ 4\sin 2\theta \cos 2\theta - \sin 2\theta = 0 \\ (\sin 2\theta)(4\cos 2\theta - 1) = 0 \\ \sin 2\theta = 0, \theta = \frac {n\pi}{2}\\ 4\cos 2\theta = 1\\ \theta = \pm \frac 12 \cos^{-1} \frac 14 + n\pi$

I am going to leave it to you to determine if: $\frac 12 \cos^{-1} \frac 14 = \tan^{−1}\sqrt{\frac35}$
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