Results 1 to 4 of 4

Thread: Parametric Graphs and Trig. Identity

  1. #1
    Member
    Joined
    Jan 2008
    Posts
    91

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Oct 2007
    From
    London / Cambridge
    Posts
    591
    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Mar 2008
    Posts
    148
    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 $\displaystyle \sin(\arccos(x)) = \sin(\theta) = \frac{\text{opp}}{\text{hyp}} = \text{opp} = \pm\sqrt{1 - x^2}$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jan 2008
    Posts
    91
    Ohh wow thanks.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Parametric Equations graphs
    Posted in the Calculus Forum
    Replies: 9
    Last Post: Dec 2nd 2009, 09:10 PM
  2. Parametric Equations of Graphs
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: Nov 3rd 2009, 06:53 AM
  3. trig graphs
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: Jan 27th 2009, 01:41 PM
  4. Trig. Graphs
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Dec 23rd 2008, 07:17 AM
  5. Trig graphs
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: Jul 19th 2008, 01:22 PM

Search Tags


/mathhelpforum @mathhelpforum