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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.

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  • $\begingroup$ They do the change of variables and allow $t$ be the parameter for the angle. $\endgroup$ – Sean Roberson Nov 3 '18 at 17:33
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This is arguably the most important trigonometric identity --- the Pythagorean trigonometric identity.

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  • $\begingroup$ So from my understanding, t considered theta here? $\endgroup$ – user2072374 Nov 3 '18 at 17:34
  • $\begingroup$ @user2072374 You could say that. Different variable names, but the same identity nevertheless. $\endgroup$ – user1337 Nov 3 '18 at 17:35
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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.

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