Mathmatics

dy/dx = (dy/dt) / (dx/dt) = -(cost+2cos2t)/(sint+2sin2t)
so, at (-1,1) where t=π/2,
dy/dx = -(-1)/(1) = 1
so the line is y = x+2

Our parametric curve in terms of t is:
c(t) = (x(t), y(t)) = ([cos(t)+cos(2t)], [sin(t)+sin(2t)]).

We're given a point (-1, 1), but we can't find the slope of a line with a single point. We either need two points for that or an equation (slope function) that will give us our slope.

A tangent line to our curve, in terms of x and y, can be found using the point-slope form of y-y_1=m(x-x_1) where m = dy/dx and (x_1,y_1) = (-1,1).

But since c(t) = (x(t), and y(t)), our dy/dx needs to be found in terms of t. We must use the definition of dy/dx in terms of t: dy/dx = (dy/dt)/(dx/dt).

Our slope in terms of t is:
dy/dx = [cos(t)+2cos(2t)]/[-sin(t)-2sin(2t)]

However, we're given a point (-1, 1) in terms of x and y. Values for x and y in our tangent line depend on what t is at x(t) and y(t).

To evaluate the derivative and find the slope, we need to find a t where x(t) = cos(t) + cos(2t) = -1 (because our x-coordinate is -1). That occurs at t = pi/2. Verify this by evaluating x(t) = x(pi/2) = -1.

Similarly, we need a t where y(t) = sin(t) + sin(2t) = 1 (because our y-coordinate is 1). That also occurs at pi/2 within the equation's domain in terms of t. Verify this by evaluating y(t) = y(pi/2) = 1.

We now have our t values for both y and x, so we can now evaluate the first derivative, our slope function, dy/dx, which is in terms of t, at those values of t.

Revisiting dy/dx up above:
dy/dx = (dy/dt)/(dx/dt)
dy/dt evaluated at t = pi/2 = y'(pi/2) = cos(pi/2) + 2cos(2*pi/2) = -2
dx/dt evaluated at t = pi/2 = x'(pi/2) = -sin(pi/2) - 2sin(2*pi/2) = -1
So dy/dx = -2/-1 = 2
Our slope, m = dy/dx, is therefore 2.

We now have all we need to find the tangent line:
y-1 = 2(x -(-1))
y = 2x + 2 + 1 = 2x + 3

So an equation of a line tangent to the curve c(t) at the point (-1, 1) in terms of x and y, is: y = 2x + 3