Find the equation of the line tangent to the curve at the point defined by t. $t=\frac{-1}{6}$, $x=\sin(2\pi t)$, $y=cos(2\pi t)$. How would I go about solving this problem. I have taken the derivative of both $x$ and $y$. Then, I have tried to use the quotient rule. I do not know what to do next.
This is what I have done so far:
$x'=2\pi\cos(2\pi t)$
$y'= -2\pi\sin(2\pi t)$
$\frac{dy}{dx}= (-2\pi\sin(2\pi t))/(2\pi\cos(2\pi t)$