sin^4-cos^4的最小正周期是我错在哪?f(x)=(sinx)^4+(cosx)^4=[(sinx)^2+(cosx)^2]^2 -2(sinx)^2(cosx)^2=1 -[sin(2x)]^2/2=1-[1-cos(4x)]/4=cos(4x)/4 +3/42π/4=π/2函数的最小正周期为π/2
问题描述：
sin^4-cos^4的最小正周期是
我错在哪?f(x)=(sinx)^4+(cosx)^4
=[(sinx)^2+(cosx)^2]^2 -2(sinx)^2(cosx)^2
=1 -[sin(2x)]^2/2
=1-[1-cos(4x)]/4
=cos(4x)/4 +3/4
2π/4=π/2
函数的最小正周期为π/2
答
是 f(x)=(sinx)^4+(cosx)^4还是f(x)=(sinx)^4-(cosx)^4如果是f(x)=(sinx)^4+(cosx)^4,你的答案没错如果是f(x)=(sinx)^4-(cosx)^4=[(sinx)^2+(cosx)^2][(sinx)^2-(cosx)^2]=-cos2x2π/2=π