# Parametric to Cartesian - # 5

#### Jason76

Find the Cartesian equation from the parametric one.

$$\displaystyle $x = \cos (\pi - t), y = \sin (\pi - t), 0 \leq t \leq \pi$$$

Since we have some stuff on the inside, rather than coefficients can we just say it's

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

#### topsquark

Forum Staff
Find the Cartesian equation from the parametric one.

$$\displaystyle $x = \cos (\pi - t), y = \sin (\pi - t), 0 \leq t \leq \pi$$$

Since we have some stuff on the inside, rather than coefficients can we just say it's

$$\displaystyle x^{2} + y^{2} = 1$$
Yup!

See if you can figure out what the difference between the problems ( x = cos(t), y = sin(t) ) and ( x = cos(pi - t) , y = sin(pi - t) ) is. The y(x) equations are the same. So what's different?

-Dan

Sorry, beats me.

#### topsquark

Forum Staff
Sorry, beats me.
Make a list of a couple of t values. In particular, what are x and y for both problems at t = 0...

-Dan

#### thisuserhas

Sorry, beats me.
Employ the following Trig Identities -
$$\displaystyle \cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B)$$
$$\displaystyle \sin(A + B) = \sin(A)\cos(B) + \sin(B)\cos(A)$$
eg -
$$\displaystyle \cos(\pi - t) = \cos(\pi)\cos(t) - \sin(\pi)\sin(t)$$

1 person

#### Jason76

Employ the following Trig Identities -
$$\displaystyle \cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B)$$
$$\displaystyle \sin(A + B) = \sin(A)\cos(B) + \sin(B)\cos(A)$$
eg -
$$\displaystyle \cos(\pi - t) = \cos(\pi)\cos(t) - \sin(\pi)\sin(t)$$
$$\displaystyle y = \cos(\pi - x) = \cos(\pi)\cos(x) - \sin(\pi)\sin(t)$$ ??

#### Plato

MHF Helper
Find the Cartesian equation from the parametric one.
$$\displaystyle $x = \cos (\pi - t), y = \sin (\pi - t), 0 \leq t \leq \pi$$$
Since we have some stuff on the inside, rather than coefficients can we just say it's
$$\displaystyle x^{2} + y^{2} = 1$$
Given this contour is a semicircle. To see that $(\pi-t)$ goes $(-1,0)\to(1,0)$ as $0\le t\le \pi$.
$\Large y=\sqrt{1-x^2}$

1 person

#### Jason76

Given this contour is a semicircle. To see that $(\pi-t)$ goes $(-1,0)\to(1,0)$ as $0\le t\le \pi$.
$\Large y=\sqrt{1-x^2}$
How can you tell it's a semi-circle?

#### Plato

MHF Helper
How can you tell it's a semi-circle?
It is is very simple: I know the basics.
I have observed that you are trying to skip the grunt work.
You need to go back and take a solid pre-calculus course. It cannot be done online.

1 person

#### thisuserhas

$$\displaystyle y = \cos(\pi - x) = \cos(\pi)\cos(x) - \sin(\pi)\sin(t)$$ ??
$$\displaystyle \cos(\pi) = -1 \:,\: \sin(\pi) = 0$$
Sub in