# How is $\cos^2t + \sin^2t = 1$?

I have to find the parametric functions that represent the curve:

$$\left(\frac{x - x_0}a\right)^2 + \left(\frac{y - y_0}b\right)^2 = 1$$

The notes simplify this to

$$\frac{(x - x_0)^2}{a^2} + \frac{(y - y_0)^2}{b^2} = 1$$

and then jump to saying that since $$\cos^2t + \sin^2t = 1$$,

$$\frac{x - x_0}a = \cos t\text{ and }\frac{y - y_0}b = \sin t$$

Where did the $$t$$ come from? and how is $$\cos^2t + \sin^2t = 1$$? I know how the $$\cos$$ and $$\sin$$ functions look, but im not sure how they got this formula and where they got $$t$$ from.

• They do the change of variables and allow $t$ be the parameter for the angle. – Sean Roberson Nov 3 '18 at 17:33

A less insightful way to demonstrate this result is by showing, with the help of differential calculus, that the function $$cos(t)^2+sin(t)^2$$ doesn't change over different $$t$$ i.e. it's a constant function
$$\frac{d}{dt} \left( cos(t)^2+sin(t)^2\right) = 0$$
then by plugging in any value, e.g. $$t=0,\ cos(0)^2+sin(0)^2 = 1$$ we arrive at the desired result.