# How to convert parametric equation into implicit equation?

Q:

Given
$$x = \sin(t)$$
$$y = \cos(t)$$
What is the implicit form of this equation?

My attempt at solving:

$$x = \sin (t)$$
$$t = \sin^{-1} (x)$$
Substituting into y:
$$y = \cos(\sin^{-1}x)$$

And I am stuck here...

In an attempt to understand this question, I broke down the question and used desmos to plot out this question, desmos graph link.

I can see that as $$t$$ increases or decreases, it goes around like a circle. But my final equation $$y = \cos(\sin^{-1}x)$$ only plots out a semi-circle, so my answer is definitely wrong.

Could someone please explain where my mistakes were and show me how you would solve this question? Thanks!

• no need of that ${sin}^2t+{cos}^2t=1$ or $x^2+y^2=1$ Commented Aug 22, 2020 at 6:45
• I don't get it .. I get that if I increase t, it's a circle ... but what is the relation between $sin^2(t) + cos^2(t) = 1$ and the 2 parametric equations? I mean, I can just say write down the equation of the circle as the answer? Sorry could you explain more please? Commented Aug 22, 2020 at 6:55

Hint: Square both equations to get $$x^2 = \sin^2(t)$$ and $$y^2 = \cos^2(t)$$, then add to get $$x^2+y^2 = \sin^2(t)+\cos^2(t)$$. By the Pythagorean Theorem, $$\sin^2(t) + \cos^2(t) = ?$$
In general for 2 parametric equations of the form $$x=a\sin\theta+b, y=a\cos\theta+c$$ we have $$\sin\theta=\frac{x-b}{a},\cos\theta=\frac{y-c}{a}$$ Using the identity $$\sin^2\theta+\cos^2\theta=1$$, we have $$\big(\frac{x-b}{a})^2+\big(\frac{y-c}{a})^2=1$$ and $$(x-b)^2+(y-c)^2=a^2$$ which is a circle wit centre $$(b,c)$$ and radius $$a$$. Obviously we could apply the same method if it was $$y=a\sin\theta+b$$etc. So the thing to look out for is whether the coefficient of $$\sin\theta$$ and $$\cos\theta$$ is the same. If it is, the curve is a circle.
With trigonometric parametric equations, you need to be thinking of trigonometric identities. Using the trigonometric identity $$\sin^{2}(\theta) + \cos^{2}(\theta) = 1$$, you can square both of your equations of $$x$$ and $$y$$. Which you can then substitute into the identity I just stated.
$$\sin^{2}(t) + \cos^{2}(t) = x^2 + y^2 = 1$$
You can now see, the implicit form is $$x^2 + y^2 = 1$$.
Your relation $$t=\sin^{-1}(x)$$ is only valid for $$t\in[-\pi/2,\pi/2],$$ given the definition of $$\sin^{-1}$$. Next you have, given that $$\cos(t)\geq0$$ in that interval, $$y=\cos(t)=\cos(\sin^{-1}(x))=\sqrt{1-\sin^2(\sin^{-1}(x))}=\sqrt{1-x^2},$$ that is part of the half circle $$x^2+y^2=1,\ y\geq0\qquad\implies\qquad y=+\sqrt{1-x^2}.$$ If you set $$t=\pi-\sin^{-1}(x),$$ that is another solution of $$x=\sin(t),$$ valid for $$t\in[\pi/2,3\pi/2],$$ where $$\cos(t)\leq0,$$ $$y=\cos(t)=\cos(\pi-\sin^{-1}(x))=-\cos(\sin^{-1}(x))=-\sqrt{1-\sin^2(\sin^{-1}(x))}=-\sqrt{1-x^2},$$ that is part of the other half circle $$x^2+y^2=1,\ y\leq0\qquad\implies\qquad y=-\sqrt{1-x^2}.$$