Expert: Frederick Koh - 12/21/2011

Question
the normals to the curve y=cos2x at the points A(pi/4,0) and B(3pi/4,0) meet at the point C. find the coordinates of the point C, and the area of the triangle ABC

Answer
y=cos2x
Differentiating both sides wrt x gives

dy/dx=-2sin2x

Gradient function of the normal at any point on the curve = -1/ (-2sin2x) = 1/(2 sin2x)

At A (pi/4, 0), substituting x=pi/4, gradient of normal = 1/2 (sin pi/2) =1/2

Equation of normal at A is y-0=(1/2) (x-pi/4)
         y=(1/2) (x- pi/4)--------------(1)

At A (pi/4, 0), substituting x=3pi/4, gradient of normal = 1/2 (sin 3pi/2) =-1/2

Equation of normal at A is y-0=-(1/2) (x- 3pi/4)
         y=-(1/2) (x- 3pi/4)-------------------(2)

When the two normals meet, (1)=(2)
ie   (1/2) (x- pi/4)=-(1/2) (x- 3pi/4)

         x- pi/4 = - (x- 3pi/4) = -x + 3pi/4
         2x= pi
         x= pi/2 ,  y=(1/2) (pi/2- pi/4)= pi/8

Area of triangle ABC = 1/2 * length * height
         = 1/2 * (3pi/4 - pi/4 )* (pi /8)
         = 1/2 * (pi/2) *(pi/8)
         = (pi)^2/32  sq units (shown)
(You may wish to plot out the points in the Cartesian space to see things more clearly)

Hope this helps. Peace.