Eliminating the parameter when sin and cos are involved.

**buddyp450** · January 24th 2010, 12:58 PM

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

**skeeter** · January 24th 2010, 01:05 PM

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

**buddyp450** · January 24th 2010, 01:07 PM

fixed

**skeeter** · January 24th 2010, 01:21 PM

you have the right idea ...

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

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

$x^2+y^2 = 1$

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

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

the upper semicircle of radius 1.

**buddyp450** · January 24th 2010, 05:55 PM

Am I correct in saying that

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

is the same as

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

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

**mr fantastic** · January 24th 2010, 06:45 PM

Originally Posted by buddyp450

Am I correct in saying that

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

is the same as

$ 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!

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