find the equation of tangent and normal to the curve x=1-cos theta, y= theta-sin theta at theta=pi/4.
x = 1 - cosθ and y = θ - sinθ
find dy/dx
dy/dθ = 1 - cosθ
dy/dx = sinθ
therefore dy/dx = (dy/dθ) /(dx/dθ) = (1 - cosθ)/sinθ
slope dy/dx at θ = pi/4 = [1 - cos(π/4)] / sin(π/4) = (√2) - 1
We need to find the coordinates of the point θ = π/4 in cartesian coordinates.
y1 = π/4 - sin(π/4) = π/4 - 1/(√2)
x1 = 1 - cos(π/4) = 1 - 1/(√2)
Therfore the equation of tangent
Y - y1 = dy/dx(X - x1)
Y - (π/4 - 1/(√2)) = [(√2) - 1] [X - (1 - 1/(√2)]
You can rearrange it according to your aswer.
Y - (√2) - 1)X = π/4 + 2 - √2
For equation of normal. use negative reciprocal of slope -dx/dy = -1/[(√2) - 1]
-dx/dy = -[(√2) + 1] I have rationalized the denominator.
Using Y - y1 = -dx/dy(X - x1)
gives
Y + [√2) + 1]X = π/4