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Math Help - Parametric function involving trig

  1. #1
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    Parametric function involving trig

    Hi all
    I have this high school maths problem where i am given two equations which are parametric and am asked to find a cartesian equation from them. Normally i would be able to do this on my own, but i have never seen a question like this that involves two sin functions rather than a cos and sin

    here are the equations

    x=-4sin(Pi(t))/6) -(2pi)/3
    and y=-3sin((Pi(t))/3) -pi

    Any help would be greatly appreciated

    Many thanks
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  2. #2
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    Re: Parametric function involving trig

    =simoro1;735620Here are the equations

    x=-4sin(Pi(t))/6) -(2pi)/3
    and y=-3sin((Pi(t))/3) -pi
    Hint: Use \sin2\theta = 2\sin \theta \cos \theta
    Spoiler:

    \sin (\frac{\pi t}{6}) = -\frac{x+\frac{2\pi}{3}}{4}\\ \cos (\frac{\pi t}{6}) = \sqrt{1-\left( \frac{x+\frac{2\pi}{3}}{4} \right)^2}\\ \sin (\frac{\pi t}{3}) = 2.\sin (\frac{\pi t}{6}).\cos (\frac{\pi t}{6}) \implies -\frac{y+\pi}{3} =-2\frac{x+\frac{2\pi}{3}}{4}\sqrt{1-\left( \frac{x+\frac{2\pi}{3}}{4} \right)^2} \implies \left( \frac{y+\pi}{3} \right)^2 =  4 \left( \frac{x+\frac{2\pi}{3}}{4} \right)^2 \left( 1-\left( \frac{x+\frac{2\pi}{3}}{4} \right)^2 \right)
    Last edited by kalyanram; September 12th 2012 at 12:09 AM.
    Thanks from simoro1
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  3. #3
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    Re: Parametric function involving trig

    thanks man you've been a huge help
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