I am trying to figure out how to find the eq to the tan to the curve x = cost + cos2t, y = sint + sin2t, (-1,1)
I'm lost at the point where you plug in t in order to find the slope. I have the derivative found for both x and y and have it set up as dy/dt / dx/dt cos(t) + 2cos(2t)/ -sin(t) - 2sin(2t)
I don't know a) what t is, the only examples I have is when the point are the same numbers (i.e (0,0)) and b) how to handle it if it is -1 (which I think it is). For the cases of 2cos(2t), how do you handle a negative number?
If the curve $x = \cos t + \cos 2t$, $y=\sin t + \sin 2t$ passes through the point $(-1,1)$, then there must be some value of $t$ such that $-1 = \cos t + \cos 2t$ and $1 = \sin t + \sin 2t$. Your first task is to find that value of $t$. Once you have that, you can evaluate the derivative (which you have correctly calculated) at that value of $t$, which gives you the slope of the tangent line at that value of $t$ (which, again, corresponds to the point $(-1,1)$). Finally, you now have a point together with the slope, so you can develop the equation of the tangent line.