# Eliminating the parameter when sin and cos are involved.

• Jan 24th 2010, 12:58 PM
buddyp450
Eliminating the parameter when sin and cos are involved.
Hello! I've just started Calculus in college but my Trig class in high-school was a joke and I didn't get much out of it (mostly my fault) so I thought this question would belong in the pre-university trig forum. Please correct me if I'm wrong.

a) Eliminate the parameter to find a Cartesian equation of the curve.

x = sin1/2theta , y = cos1/2theta , -pi <= theta <= pi

(sorry I'm also a noob on writing math nicely on a computer, a link to a tutorial would be nice)

I thought about using the rule of sin^2theta + cos^2theta = 1 but I'm not sure how to get it into that form algebraically or even what the hints are that I should do that...?

Thanks
• Jan 24th 2010, 01:05 PM
skeeter
Quote:

Originally Posted by buddyp450
Hello! I've just started Calculus in college but my Trig class in high-school was a joke and I didn't get much out of it (mostly my fault) so I thought this question would belong in the pre-university trig forum. Please correct me if I'm wrong.

a) Eliminate the parameter to find a Cartesian equation of the curve.

x = sin1/2theta , y = sin1/2theta , -pi <= theta <= pi

(sorry I'm also a noob on writing math nicely on a computer, a link to a tutorial would be nice)

I thought about using the rule of sin^2theta + cos^2theta = 1 but I'm not sure how to get it into that form algebraically or even what the hints are that I should do that...?

Thanks

note that y = x
• Jan 24th 2010, 01:07 PM
buddyp450
fixed
• Jan 24th 2010, 01:21 PM
skeeter
you have the right idea ...

$\displaystyle x^2 = \sin^2\left(\frac{\theta}{2}\right)$

$\displaystyle y^2 = \cos^2\left(\frac{\theta}{2}\right)$

$\displaystyle x^2+y^2 = 1$

the equation of the unit circle ... however, since $\displaystyle y = \cos\left(\frac{\theta}{2}\right) > 0$ for $\displaystyle -\pi \le \theta \le \pi$

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

the upper semicircle of radius 1.
• Jan 24th 2010, 05:55 PM
buddyp450
Am I correct in saying that

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

is the same as

$\displaystyle y = \sqrt{1-\sin^2\left(\frac{\theta}{2}\right)}$

and that the latter would be the proper Cartesian equation of the curve?

• Jan 24th 2010, 06:45 PM
mr fantastic
Quote:

Originally Posted by buddyp450
Am I correct in saying that

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

is the same as

$\displaystyle y = \sqrt{1-\sin^2\left(\frac{\theta}{2}\right)}$

and that the latter would be the proper Cartesian equation of the curve?

Why would you do this? Especially since the point is to elminate the parameter!