Calculus- Find the open interval

Question

Find the open interval(s) on which the function f(x) = cos^2(2x)  is increasing in the interval (0,2π). Round
numerical values in your answer to three decimal places.

Answer 1

Recall that cos(2a) = cos^2(a) - sin^2(a) = 2 cos^2(a) - 1

2 cos^2(a) - 1 = cos(2a)

2 cos^2(a) = 1 + cos(2a)

cos^2(a) = 1/2 + 1/2 cos(2a)

Thus, f(x) = cos^2(2x) = 1/2 + 1/2 cos(4x)

This function is increasing when f'(x) > 0

f'(x) = -2 sin(4x)

-2 sin(4x) > 0 on (0, 2 pi) iff sin(4x) < 0 on (0, 2 pi)

sin(x) is negative on (pi, 2 pi), (3 pi, 4 pi), (5 pi, 6 pi), (7 pi, 8 pi), so

sin(4x) is negative on (pi/4, pi/2), (3 pi/4, pi), (5 pi/4, 3 pi/2), (7 pi/4, 2 pi)