# Parametric function involving trig

• Sep 11th 2012, 11:34 PM
simoro1
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
• Sep 12th 2012, 12:02 AM
kalyanram
Re: Parametric function involving trig
Quote:

=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:

$\sin (\frac{\pi t}{6}) = -\frac{x+\frac{2\pi}{3}}{4}\\ \cos (\frac{\pi t}{6}) = \sqrt{1-\left( \frac{x+\frac{2\pi}{3}}{4} \right)^2}\\ \sin (\frac{\pi t}{3}) = 2.\sin (\frac{\pi t}{6}).\cos (\frac{\pi t}{6}) \implies -\frac{y+\pi}{3} =-2\frac{x+\frac{2\pi}{3}}{4}\sqrt{1-\left( \frac{x+\frac{2\pi}{3}}{4} \right)^2} \implies \left( \frac{y+\pi}{3} \right)^2 = 4 \left( \frac{x+\frac{2\pi}{3}}{4} \right)^2 \left( 1-\left( \frac{x+\frac{2\pi}{3}}{4} \right)^2 \right)$
• Sep 12th 2012, 12:16 AM
simoro1
Re: Parametric function involving trig
thanks man you've been a huge help