三角函数题,函数y=cos^2(x-π/6)+sin^2(x+π/6)-1的最小正周期为要求有详细解答过程
问题描述:
三角函数题,函数y=cos^2(x-π/6)+sin^2(x+π/6)-1的最小正周期为
要求有详细解答过程
答
y=cos^2(x-π/6)+sin^2(x+π/6)-1 =1/2(cos(2x-π/3)+1)+1/2(1-sin(2x+π/3))-1 =1/2[1/2cos2x+根号3sin2x]-1/2[1/2sin2x+根号3/2cos2x] =(根号3-1)/4(sin2x-cos2x) =(根号6-根号2)/8sin(2x-π/4)所以最小正周期为:...