back to Trigonometry lessons

Find maximum and minimum values of the function

Question

Fand the maximum and minimum values of the function f(x) = 3 cos 2x + 2 sin2 x - cos x

Solution

Using the formula, cos 2x = 2 cos2 x - 1 and sin2 x + cos2 x = 1.

f(x) = 3 (2 cos2 x - 1) + 2 ( 1 - cos2 x) - cos x
= 6 cos2 x - 3 + 2 - 2 cos2 x - cos x
= 4 cos2 x - cos x - 1
= 4[cos2 x - (1/4) cos x + (1/8)2 - (1/8)2] - 1
= 4[cos2 x - (1/4) cos x + (1/8)2] - 4 × (1/64) - 1
= 4[cos2 x - (1/4) cos x + (1/8)2] - 1/16 - 1
= 4[cos2 x - (1/4) cos x + (1/8)2] - 17/16
= 4 (cos x - 1/8)2 - 17/16

So, when cos x = -1, f(x) has maximum value and fmax = 4(-1 - 1/8)2 - 17/16 = 4(-9/8)2 - 17/16 = 4 × 81/64 - 17/16 = (81 - 17)/16 = 4

When cos x = 1/8, f(x) has minimum value and fmin = -17/16.