题目

化简f(x)=sin^2(45°+x)-sin^2(30°-x)-sin15°cos(15°+2x),并求出函数的最小正周期及最大最小

提问时间：2021-01-12

答案

sin²(45°+x) = [1- cos(90°+2x)] /2 = (1+sin2x)/2
sin²(30° - x) = [1 - cos(60° - 2x)] / 2
= 1/2 - 1/2 cos[90° - (2x+30°)]
= 1/2 - 1/2 sin(2x+30°)
sin15°cos(15°+2x)
= sin15°(cos15°cos2x - sin15°sin2x)
= sin15°cos15°cos2x - sin²15°sin2x
= 1/2 sin30° cos2x - (1-cos30°)/2 *sin2x
= 1/2 sin30° cos2x + 1/2 cos30°sin2x - 1/2 sin2x
= 1/2 sin(2x+30°) - 1/2 sin2x
∴f(x) = (1+sin2x)/2 - [1/2 - 1/2 sin(2x+30°)] - [1/2 sin(2x+30°) - 1/2 sin2x ] = sin2x
最小正周期T=π
最大值 1,最小值 -1