# Eliminating $\theta$ from trigonometric system

Eliminate $$\theta$$ from the equation

$$x\cos(3\theta)+y\sin(3\theta)=a\cos(\theta)\\ x\sin(3\theta)+y\cos(3\theta)=a\cos(\theta+\frac{\pi}{6})$$

I tried squaring a adding but got nowhere. Also got $$x$$ and $$y$$ as linear equations in form of $$\theta$$ but cant see what to do next.

Not elegant at all!

Squaring & adding $$\dfrac{x^2+y^2+2xy\sin6\theta}{a^2}= \cos^2\theta+\cos^2\left(\theta+\dfrac\pi6\right) =1+\cos\dfrac\pi6\cos\left(2\theta+\dfrac\pi6\right)\ \ \ \ (1)$$

Adding & squaring $$\dfrac{(x+y)^2(1+\sin6\theta)}{a^2}= \left[\cos\theta+\cos\left(\theta+\dfrac\pi6\right)\right]^2=\left(1+\cos\dfrac\pi6\right)\left(1+\cos\left(2\theta+\dfrac\pi6\right)\right)\ \ \ \ (2)$$

using Prosthaphaeresis Formula $$\cos C+\cos D$$ and Double angle formula

Set $$2\theta+\dfrac\pi6=t,6\theta=3t-\dfrac\pi2$$

From $$(1),$$ $$\dfrac{x^2+y^2-2xy\cos3t}{a^2}=1+\cos\dfrac\pi6\cos t\ \ \ \ (3)$$

From $$(2),$$ $$\dfrac{(x+y)^2(1-\cos3t)}{a^2}=\left(1+\cos\dfrac\pi6\right)\left(1+\cos t\right)\ \ \ \ (4)$$

Solve $$(3),(4)$$ for $$\cos t,\cos3t$$

and use $$\cos3t=4\cos^3t-3\cos t$$ to eliminate $$t$$