Question

sin(x+π/3)*sin(x+π/2)的最小正周期,

Answer 1

方法一: f(x) = cos [π / 2 - (x + π / 3)] cos [π / 2 - (x + π / 2)] = cos (π / 6 - x) cos (-x) = (√3 / 2 cos x + 1 / 2 sin x) cos x =√3 / 2 (cos x)^2 + 1 / 2 sin x cos x =√3 / 4 2 (cos x)^2 + 1 / 4 2 sin x cos x =√3 / 4 [1 + cos (2 x)] + 1 / 4 sin (2 x) =√3 / 4 + 1 / 2 [√3 / 2 cos (2 x) + 1 / 2 sin (2 x) =√3 / 4 + 1 / 2 sin (2 x + π / 3) 2π / 2 = π , 所以,f(x)的最小正周期是 π. 方法二: f(x) = sin (x + π / 3) cos [π / 2 - (x + π / 2)] = sin (x + π / 3) cos (-x) = sin (x + π / 3) cos x = sin [(x + π / 6) + π / 6] cos [(x + π / 6) - π / 6] = [√3 / 2 sin (x + π / 6) + 1 / 2 cos (x + π / 6)] [√3 / 2 cos (x + π / 6) + 1 / 2 sin (x + π / 6)] = 3 / 4 sin (x + π / 6) cos (x + π / 6) +√3 / 4 [cos (x + π / 6)]^2 + √3 / 4 [sin (x + π / 6)]^2 + 1 / 4 cos (x + π / 6) sin (x + π / 6) = (3 / 4 + 1 / 4 ) sin (x + π / 6) cos (x + π / 6) + √3 / 4 {[cos (x + π / 6)]^2 + [sin (x + π / 6)]^2} = 1 / 2 2 sin (x + π / 6) cos (x + π / 6) + √3 / 4 = 1 / 2 sin (2 x + π / 3) + √3 / 4 所以,f(x)的最小正周期是 π.