Post History

We know $$\sin^2\theta+\cos^2\theta=1$$
So, we can write
$$(\sin^3 \theta)^{\dfrac{2}{3}} +(\cos^3\theta)^{\dfrac{2}{3}}=1$$

Let, $$\sin^3\theta=\frac{x}{a}$$
$$\cos^3\theta=\frac{y}{b}$$

$$(\frac{x}{a})^{\dfrac{2}{3}} +(\frac{y}{b})^{\dfrac{2}{3}}=1$$

So, they took $$x=a\sin^3 \theta$$
and, $$y=a \cos^3 \theta$$

That's what they did