Student should use a combination of trigonometric identities, product rule and chain rule to find dy/dx.This can be done by applying product rule, obtainingdy/dx = sin(2x). d[cos(x)^2]/dx + cos(x)^2. d[sin(2x)]/dxthen using the chain rule to find d[cos(x)^2]/dx = -2cos(x)sin(x) andd[sin(2x]/dx = 2cos(2x).Alternatively, the student can rewrite cos(x)^2 as ½[cos(2x)-1] then differentiate in this form.Student should finddy/dx = 2cos(x)[cos(x)cos(2x) – sin(x)sin(2x)].
Substituting x= pi/3 into equation of curve returns corresponding y-value at that point, and substituting x = pi/3 into dy/dx returns gradient of tangent at that point. Diving -1 by this returns the gradient of the normal at that point. Using equation of a straight line y=mx+b, student should obtain b by rearranging at substituting x and y values found earlier. Students should find y = x + [(3sqrt(3) – 8pi)/24)]