函数y=cos^2(x-π/12)+sin^2(x+π/12)-1的最小正周期为 A π/4 B π/2 C π D 2π

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/28 10:27:34
函数y=cos^2(x-π/12)+sin^2(x+π/12)-1的最小正周期为 A π/4 B π/2 C π D 2π
x){ھ qF 43@\mWgsnvӉ+͙d.G2RpR . Flid~ -]ML*hCm: #23X}#hC26LXԨ"$2ШH8|LM}#f(bkr-Ե/:m @!h

函数y=cos^2(x-π/12)+sin^2(x+π/12)-1的最小正周期为 A π/4 B π/2 C π D 2π
函数y=cos^2(x-π/12)+sin^2(x+π/12)-1的最小正周期为 A π/4 B π/2 C π D 2π

函数y=cos^2(x-π/12)+sin^2(x+π/12)-1的最小正周期为 A π/4 B π/2 C π D 2π
y=cos^2(x-π/12)+sin^2(x+π/12)-1
=[1+cos(2x-π/6)]/2+[1-vos(2x+π/6)]/2-1
=(cos2xcpsπ/6+sin2xsinπ/6-cos2xcosπ/6+sin2xsinπ/6)/2
=(sin2x)/2
T=2π/2=π
选C