早教吧作业答案频道 -->数学-->
=[-1+2(sinx)^2]/2+1/2-(sin2x)/2=(-1/2)(cos2x+sin2x)+1/2是怎么得来的(sinx-cosx)sinx=(sinx)^2-sinxcosx=[-1+2(sinx)^2]/2+1/2-(sin2x)/2=(-1/2)(cos2x+sin2x)+1/2=(-√2/2)sin(2x+φ)+1/2=(-√2/2)sin[2(x+φ/2)]+1/2∴最小正周期是(2π)/2=π
题目详情
=[-1+2(sinx)^2]/2+1/2-(sin2x)/2 =(-1/2)(cos2x+sin2x)+1/2 是怎么得来的
(sinx-cosx)sinx
=(sinx)^2-sinxcosx
=[-1+2(sinx)^2]/2+1/2-(sin2x)/2
=(-1/2)(cos2x+sin2x)+1/2
=(-√2/2)sin(2x+φ)+1/2
=(-√2/2)sin[2(x+φ/2)]+1/2
∴最小正周期是(2π)/2=π
(sinx-cosx)sinx
=(sinx)^2-sinxcosx
=[-1+2(sinx)^2]/2+1/2-(sin2x)/2
=(-1/2)(cos2x+sin2x)+1/2
=(-√2/2)sin(2x+φ)+1/2
=(-√2/2)sin[2(x+φ/2)]+1/2
∴最小正周期是(2π)/2=π
▼优质解答
答案和解析
因为(sinx)^2=[-1+2(sinx)^2]/2+1/2；sinxcosx=(sin2x)/2；
看了 =[-1+2(sinx)^2...的网友还看了以下：