1. The problem statement, all variables and given/known data Find the equation of normal to curve x=cos b , y=sin b at the point where b=pi/4 . Find the coordinates of the point on the curve where the normal meets the curve again . 2. Relevant equations 3. The attempt at a solution Gradient of normal : dx/db=-sin b dy/db=cos b dy/dx=-cot b gradient of normal = tan b , when b=pi/4 , gradient of normal =1 The normal cuts the curve at point (1/sqrt(2) , 1/sqrt(2)) equation of normal is y-1/sqrt(2)=1(x-1/sqrt(2)) y=x --- 1 The cartesian equation of the curve is cos^(-1) x=sin^(-1) y sin y=cos x --- 2 combine 1 and 2 , sin x=cos x tan x=0 x=0 , pi , 2pi Since the domain is not mentioned , i take it as 0<=x<=2pi Therefore , the normal would cut the curve at points (0,pi/2) , (pi,3pi/2), (2pi, 5pi/2) Am i correct ?