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Math Help - Parametric Graphs and Trig. Identity

  1. #1
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    Parametric Graphs and Trig. Identity

    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
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  2. #2
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    London / Cambridge
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    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
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  3. #3
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    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}
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  4. #4
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    Ohh wow thanks.
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