Posted by Tina on Friday, November 26, 2010 at 5:51pm.
let's take f '(Ø)
f '(Ø) = -2sinØ + 2sinØcosØ
= 0 for max/mins
sinØ - sinØcosØ = 0
sinØ(1 + cosØ) = 0
sinØ = 0 or cosØ = -1
Ø = 0,π,2π or Ø = 3π/2
f(0) = 2+0 = 2
f(π) = -2+0 = -2
f(2π) = 2 + 0 = 2
f(3π/2) = 0 + 1 = 0
max value of the function is 2 when Ø = 0 or 2π
min value of the function is -2 when Ø = 3π/2
2sinθcosθ-1=0
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