# Thread: Parametric Graphs and Trig. Identity

1. ## Parametric Graphs and Trig. Identity

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

2. Draw a triangle with an angle $\displaystyle \theta$ where $\displaystyle \cos \theta = x$ so you draw the hypotenuse of the triangle of having length 1 and the adjacent side of having length x. Use pythagorus to find the length of the opposite side which is $\displaystyle \pm\sqrt{1 - x^2}$ and now you just find the sine of the angle $\displaystyle \theta$ and your done.

Bobak

3. Let $\displaystyle \theta = \arccos(x) \Rightarrow \cos(\theta) = x$

Draw a right trangle with an angle equal to $\displaystyle \theta$. Then $\displaystyle x = \frac{\text{adj}}{\text{hyp}}$ (adjacent side to the angle over hypotenuse). Set the hypotenuse equal to 1 (consider the unit circle). Then $\displaystyle x = \text{adj}$ and the opposite side of the triangle can be given in terms of $\displaystyle x$,
namely
$\displaystyle \text{opp} = \pm\sqrt{1 - x^2}$. (This follows from the Pythagorean Theorem).

Observe now that $\displaystyle \sin(\arccos(x)) = \sin(\theta) = \frac{\text{opp}}{\text{hyp}} = \text{opp} = \pm\sqrt{1 - x^2}$

4. Ohh wow thanks.