Hey everyone,

I'm asked to graph $\displaystyle x = \cos(\sin(t)), \ \ y = \sin(\sin(t)), \ \ 0 \leq t \leq 2\pi$.

So I graphed it and everything. But now I need to find an equation of this line in terms of X and Y.

So here is what I did:

$\displaystyle t = \arcsin(\arccos(x))$

$\displaystyle y = \sin(\arccos(x))$

So I figured that this must be some kind of trig identity and after checking up on wikipedia sure enough this gives me:

$\displaystyle y = \pm\sqrt{1 - x^2}$

So given my domain of $\displaystyle 0 \leq t \leq 2\pi$ I get:

$\displaystyle \boxed{y = \pm\sqrt{1 - x^2} \ \ , \ \ \cos(1) \leq x \leq 1}$

Basically what I'm wondering is if there is a quick derivation to go from $\displaystyle \sin(\arccos(x))$ to $\displaystyle \pm\sqrt{1 - x^2}$, or if am alright in just saying it's an identity.

Thanks,

TrevorP