Question

函数f(x)=sinx四次方+cosx平方的最小正周期

Answer 1

f(x)=(sinx)^4+(cosx)^2

=[(sinx)^2]^2+(cosx)^2

=[(1-cos2x)/2]^2+(1+cos2x)/2

=[1-2cos2x+(cos2x)^2+2+2cos2x]/4

=[3+(cos2x)^2]/4

=[3+(1+cos4x)/2]/4

=(1/8)cos4x+7/8

所以周期T=2π/4=π/2.最大值1最小值3/4