Parametric Graphs and Trig. Identity

**TrevorP** · March 24th 2008, 03:25 PM

Hey everyone,

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

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:

$t = \arcsin(\arccos(x))$
$y = \sin(\arccos(x))$

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

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

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

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

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

Thanks,

TrevorP

**bobak** · March 24th 2008, 03:38 PM

Draw a triangle with an angle $\theta$ where $\cos \theta = x$ 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 $\pm\sqrt{1 - x^2}$ and now you just find the sine of the angle $\theta$ and your done.

Bobak

**iknowone** · March 24th 2008, 03:42 PM

Let $\theta = \arccos(x) \Rightarrow \cos(\theta) = x$

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

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

**TrevorP** · March 24th 2008, 03:42 PM

Ohh wow thanks.

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