函数y=3sin^2x+cos2x的最小正周期为?y=3sin^2x+cos2x=3-3(cos2x+1)/2+cos2x这步是怎么做的,=-cos2x/2+3/2函数y=3sin^2x+cos2x的最小正周期为?重新做一次,写上全部步骤

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/06 04:51:21
函数y=3sin^2x+cos2x的最小正周期为?y=3sin^2x+cos2x=3-3(cos2x+1)/2+cos2x这步是怎么做的,=-cos2x/2+3/2函数y=3sin^2x+cos2x的最小正周期为?重新做一次,写上全部步骤

函数y=3sin^2x+cos2x的最小正周期为?y=3sin^2x+cos2x=3-3(cos2x+1)/2+cos2x这步是怎么做的,=-cos2x/2+3/2函数y=3sin^2x+cos2x的最小正周期为?重新做一次,写上全部步骤
函数y=3sin^2x+cos2x的最小正周期为?
y=3sin^2x+cos2x
=3-3(cos2x+1)/2+cos2x这步是怎么做的,
=-cos2x/2+3/2
函数y=3sin^2x+cos2x的最小正周期为?重新做一次,写上全部步骤

函数y=3sin^2x+cos2x的最小正周期为?y=3sin^2x+cos2x=3-3(cos2x+1)/2+cos2x这步是怎么做的,=-cos2x/2+3/2函数y=3sin^2x+cos2x的最小正周期为?重新做一次,写上全部步骤
1-0.5(cos2x+1)=1-0.5(cos^2 x- sin^2 x + sin^2 x+cos^2 x)
=1-cos^2 x=sin^2 x
3sin^2x=3-3(cos2x+1)/2
y=3sin^2x+cos2x
=3(1-cos^2 x)+cos2x
=3[1-0.5(cos^2 x- sin^2 x + sin^2 x+cos^2 x)]+cos2x
=3-3(cos2x+1)/2+cos2x
=-cos2x/2+3/2
周期为2π/2=π
PS:最小正周期为2π/n(n即为三角函数中x的系数)
再PS :同学~下次拜托人的时候记得说“谢谢”