Using trig identities to change from parametric to Cartesian equation

Question

$$x=\sin t\\ y= 3\cos (3t)$$

Find $y$ in terms of $x$. I have graphed the function and it appears to follow $y(x)=-4x^2 +2$ from $-1\le x\le 1$ and $-2\le y\le 2$ .

Thanks

Answer 1

$$y=3\cos3t=3\cos t(4\cos^2t-3)=3\cos t(1-4\sin^2t)=3\cos t(1-4x^2)$$

and use $$\cos t=\pm\sqrt{1-x^2}$$ to eliminate $t$

Answer 2

By expressing $\cos (3t)$ as $\cos (2t + t)$, you can derive the following identity: $$\cos3t=4\cos^3 t-3\cos t$$ Let's start by first expressing $\cos t$ in terms of $x$: $$x=\sin t \implies x^2=\sin^2 t \implies x^2=1-\cos^2 t \\ \implies \cos t=\pm \sqrt{1-x^2}$$ Now we subsitute this into the expression for $y$: $$y=3 \cos (3t)=12\cos^3 t-9\cos t=12(\pm \sqrt{1-x^2})^3\mp 9\sqrt{1-x^2}$$

Note that this is not a function, but can be expressed in terms of two separate functions if you take the case for the positive and negative square roots separately.

Answer 3

We have $$\cos 3 t = \cos t ( 4 \cos^2 t - 3 ) = \cos t ( 1- 4 \sin^2 t)$$ and therefore $$ (3 \cos 3 t)^2 = 9 \cos ^2 t( 1- 4 \sin^2 t)^2$$ It follows that $(x,y) = (\sin t, 3 \cos (3 t))$ lies on the curve of equation $$ y^2 = 9 (1-x^2)(1- 4 x^2)^2$$

One can show that every point $(x,y)$ on this curve that is not $(\pm 1/2, 0)$ is of the form $(\sin t, 3 \cos (3 t))$ for a unique point in $[0,2\pi]$. The points $(\pm 1/2, 0)$ are both covered twice. Indeed, let $(x,y) $ satisfying $y^2 = 9 (1-x^2)(1- 4 x^2)^2$. Then $(1-x^2) \ge 0$ and so $x = \sin t= \sin(\pi-t)$. Now calculate $9 (1-x^2)(1- 4 x^2)^2 = 9\cos ^2 t( 1- 4 \sin^2 t)^2 = (3\cos(3t))^2$. But since $(x,y)$ is on the curve we have $9 (1-x^2)(1- 4 x^2)^2 = y^2$. Therefore $(3\cos(3t))^2= y^2$ and so $y = \pm 3 \cos 3 t$. Now choose between $t$ and $\pi -t$ to match the sign.