Question
A line $OA$ of length $r$ starts from its initial position $OX$ and traces an angle $\angle AOB=\alpha$ in the anticlockwise direction. It then traces back in the clockwise direction $\angle BOC=3\theta$. L is the foot of the perpendicular from C on OA, such that $\frac{\sin^3\theta}{CL}=\frac{\cos^3\theta}{OL}=1$.
Then what is the value of $\frac{1-r\cos\alpha}{r\sin\alpha}$?
My attempt:
Diagram: (I tried my best in GeoGebra :( and that $\angle AOB=\alpha\neq30^\circ$)
My approach is that $r^2=\cos^6\theta+\sin^6\theta=1-3\sin^2\theta\cos^2\theta$
Also, $r\sin(3\theta-\alpha)=CL=\sin^3\theta$. But now I am stuck in a dead end. I wish to expand this $\sin(3\theta-\alpha)$ (to get those $\cos\alpha$ and the $\sin\alpha$) but I am afraid doing so will be too complicated.
Is there any other approach? Thanks!
EDIT: I just realized that the diagram I gave above was not present in the original question, and that while I have drawn OC to be on the other side of OA, it can also be on the same side of OA as OB is. (as nothing is given about the position of OC and OB wrt OA in the orignial question) Though I don't think it should affect the final answer but I still noted just in case.
Edit :
If we assume that $\alpha\gt 3\theta$, then the result does equal $\tan(2\theta)$.
We have $$\sin^3\theta=-r\sin(3\theta-\alpha)=-r\sin(3\theta)\cos\alpha+r\cos(3\theta)\sin\alpha\tag3$$ and $$\cos^3\theta=r\cos(3\theta-\alpha)=r\cos(3\theta)\cos\alpha+r\sin(3\theta)\sin\alpha\tag4$$
$\sin(3\theta)\times (3)-\cos(3\theta)\times(4)$ gives $$\sin^3\theta\sin(3\theta)-\cos^3\theta\cos(3\theta)=-r\sin^2(3\theta)\cos\alpha-r\cos^2(3\theta)\cos\alpha=-r\cos\alpha$$
$\cos(3\theta)\times (3)+\sin(3\theta)\times(4)$ gives $$\sin^3\theta\cos(3\theta)+\cos^3\theta\sin(3\theta)=r\cos^2(3\theta)\sin\alpha+r\sin^2(3\theta)\sin\alpha=r\sin\alpha$$
It follows from these that, writing $s:=\sin\theta$ and $c:=\cos\theta$, $$\begin{align}\frac{1-r\cos\alpha}{r\sin\alpha}&=\frac{1+\sin^3\theta\sin(3\theta)-\cos^3\theta\cos(3\theta)}{\sin^3\theta\cos(3\theta)+\cos^3\theta\sin(3\theta)}\\\\&=\frac{1+s^3(3s-4s^3)-c^3(4c^3-3c)}{s^3(4c^3-3c)+c^3(3s-4s^3)}\\\\&=\frac{-4(s^6+c^6)+3(s^4+c^4)+1}{3cs(-s^2+c^2)}\\\\&=\frac{-4((s^2+c^2)^3-3s^4c^2-3s^2c^4)+3((s^2+c^2)^2-2s^2c^2)+1}{3cs(c^2-s^2)}\\\\&=\frac{-4+12c^2s^2+3-6s^2c^2+1}{3cs(c^2-s^2)}\\\\&=\frac{2cs}{c^2-s^2}\\\\&=\tan(2\theta)\end{align}$$
Too long for a comment.
In the following, we assume that $3\theta\gt\alpha$.
We have $$\sin^3\theta=r\sin(3\theta-\alpha)=r\sin(3\theta)\cos\alpha-r\cos(3\theta)\sin\alpha\tag1$$ and $$\cos^3\theta=r\cos(3\theta-\alpha)=r\cos(3\theta)\cos\alpha+r\sin(3\theta)\sin\alpha\tag2$$
$\sin(3\theta)\times (1)+\cos(3\theta)\times(2)$ gives $$\sin^3\theta\sin(3\theta)+\cos^3\theta\cos(3\theta)=r\sin^2(3\theta)\cos\alpha+r\cos^2(3\theta)\cos\alpha=r\cos\alpha$$
$-\cos(3\theta)\times (1)+\sin(3\theta)\times(2)$ gives $$-\sin^3\theta\cos(3\theta)+\cos^3\theta\sin(3\theta)=r\cos^2(3\theta)\sin\alpha+r\sin^2(3\theta)\sin\alpha=r\sin\alpha$$
It follows from these that, writing $s:=\sin\theta$ and $c:=\cos\theta$, $$\begin{align}\frac{1-r\cos\alpha}{r\sin\alpha}&=\frac{1-(\sin^3\theta\sin(3\theta)+\cos^3\theta\cos(3\theta))}{-\sin^3\theta\cos(3\theta)+\cos^3\theta\sin(3\theta)}\\\\&=\frac{1-s^3(3s-4s^3)-c^3(4c^3-3c)}{-s^3(4c^3-3c)+c^3(3s-4s^3)}\\\\&=\frac{4(s^6-c^6)-3(s^4-c^4)+1}{-8s^3c^3+3cs(s^2+c^2)}\\\\&=\frac{4(s-c)(s^2+sc+c^2)(s+c)(s^2-sc+c^2)-3(s^2+c^2)(s^2-c^2)+1}{-8s^3c^3+3cs}\\\\&=\frac{4(s-c)(1+sc)(s+c)(1-sc)-3(s^2-c^2)+1}{sc(3-8s^2c^2)}\\\\&=\frac{4(s^2-c^2)(1-s^2c^2)-3(s^2-c^2)+1}{sc(3-8s^2c^2)}\\\\&=\frac{(c^2-s^2)(4s^2c^2-1)+1}{sc(3-8s^2c^2)}\\\\&=\frac{-(\cos2\theta)(1-\sin2\theta)(1+\sin2\theta)+1}{sc(4(1-2s^2c^2)-1)}\\\\&=\frac{-\cos^32\theta+1}{sc(4(s-c)^2-1)}\\\\&=\frac{2-2\cos^3(2\theta)}{\sin(2\theta)(3-2\sin^2(2\theta))}\\\\&=\frac{\frac{2}{\cos^3(2\theta)}-2}{\frac{3\tan(2\theta)}{\cos^2(2\theta)}-2\tan^3(2\theta)}\\\\&=\frac{\frac{2}{\cos(2\theta)}(1+\tan^2(2\theta))-2}{3\tan(2\theta)(1+\tan^2(2\theta))-2\tan^3(2\theta)}\\\\&=\frac{-2+\frac{2}{\cos(2\theta)}+\frac{2}{\cos(2\theta)}\tan^2(2\theta)}{\tan^3(2\theta)+3\tan(2\theta)}\\\\&=\frac{-2\cos(2\theta)+2+2\tan^2(2\theta)}{\sin(2\theta)(\tan^2(2\theta)+3)}\end{align}$$ It seems that this is not equal to $\tan(2\theta)$.