If $x=\cos t,y=\cos(2t+\pi/3)$ find an analytical relation between $x$ and $y$.

Question

I'm having a bit of trouble figuring this out. At the moment this is the near solution I have:

$$y=\frac12(2\cos^2 t-1)-\sqrt{3}\sin t\cos t.$$

I should be just about to solve it but find myself stuck. I appreciate any hint on how to "eliminate" that sine, or any other way in which I could express $x$ as an expression that depends on $y$. Thanks!

Answer 1

You're almost there. You have correctly used that $$ \cos\left(2t+\frac{\pi}{3}\right)=\frac{1}{2}\cos2t-\frac{\sqrt{3}}{2}\sin2t $$ so $$ 2y=2\cos^2t-1-2\sqrt{3}\sin t\cos t=2x^2-1-2x\sqrt{3}\sin t. $$ Thus $$ 2x\sqrt{3}\sin t=2x^2-1-2y. $$ Hence $$ 12x^2(1-x^2)=(2x^2-1-2y)^2. $$

Answer 2

• To find y as a function of x, just replace t with $\arccos x$.

• To find x as a function of y, just take the arccos of the second expression, and try to express t in terms of $\arccos y$.