f(x)=sin(x-π)cos(π+x)+sin(x+π/2)cos(-x) ①f(x)最小正周期 ②[-π/6,π/2]f(x)最大值和最小值③f(x)递增区间
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/24 04:40:18
![f(x)=sin(x-π)cos(π+x)+sin(x+π/2)cos(-x) ①f(x)最小正周期 ②[-π/6,π/2]f(x)最大值和最小值③f(x)递增区间](/uploads/image/z/12529270-46-0.jpg?t=f%28x%29%3Dsin%28x-%CF%80%EF%BC%89cos%EF%BC%88%CF%80%2Bx%EF%BC%89%2Bsin%EF%BC%88x%2B%CF%80%2F2%EF%BC%89cos%28-x%29+%E2%91%A0f%28x%29%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F+%E2%91%A1%5B-%CF%80%2F6%2C%CF%80%2F2%5Df%28x%29%E6%9C%80%E5%A4%A7%E5%80%BC%E5%92%8C%E6%9C%80%E5%B0%8F%E5%80%BC%E2%91%A2f%28x%29%E9%80%92%E5%A2%9E%E5%8C%BA%E9%97%B4)
f(x)=sin(x-π)cos(π+x)+sin(x+π/2)cos(-x) ①f(x)最小正周期 ②[-π/6,π/2]f(x)最大值和最小值③f(x)递增区间
f(x)=sin(x-π)cos(π+x)+sin(x+π/2)cos(-x) ①f(x)最小正周期 ②[-π/6,π/2]f(x)最大值和最小值
③f(x)递增区间
f(x)=sin(x-π)cos(π+x)+sin(x+π/2)cos(-x) ①f(x)最小正周期 ②[-π/6,π/2]f(x)最大值和最小值③f(x)递增区间
(1)
注意
sin(x-π)= -sinx
cos(π+x)= -cosx
sin(x+π/2)= cosx
cos(-x)=cosx
所以
f(x)=(-sinx)*(-cosx) +cosx*cosx
=cos²x+sinx*cosx
由公式
cos²x=0.5cos2x+0.5
和2sinx*cosx=sin2x可以知道
f(x)=cos²x+sinx*cosx
=0.5cos2x+0.5sin2x+0.5
=√2 /2 *sin(2x+π/4) +0.5
所以
f(x)的最小正周期为2π/2=π
(2)
在区间[-π/6,π/2]上,
2x+π/4的取值范围为[-π/12,5π/4]
所以在区间[-π/6,π/2],
f(x)的最大值是f(π/8)=√2 /2 +0.5=(1+√2)/2
最小值是f(π/2)=√2 /2 *sin(5π/4) +0.5=0
(3)
f(x)=√2 /2 *sin(2x+π/4) +0.5
所以
f(x)的递增区间为
[-3π/8+kπ,π/8+kπ] ,k为整数