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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!

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    $\begingroup$ no need of that ${sin}^2t+{cos}^2t=1$ or $x^2+y^2=1$ $\endgroup$ Commented Aug 22, 2020 at 6:45
  • $\begingroup$ 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? $\endgroup$ Commented Aug 22, 2020 at 6:55

4 Answers 4

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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, $\sin^2(t) + \cos^2(t) = ?$

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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.

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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 $x^2 + y^2 = 1$.

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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 $$ x^2+y^2=1,\ y\leq0\qquad\implies\qquad y=-\sqrt{1-x^2}. $$

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