作业帮已知函数f(x）=sin^4x+cos^2x+1/4sin2xcos2x(x∈R）则f（x）化简的到式子是?
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已知函数f(x）=sin^4x+cos^2x+1/4sin2xcos2x(x∈R）则f（x）化简的到式子是?
已知函数f(x）=sin^4x+cos^2x+1/4sin2xcos2x(x∈R）则f（x）化简的到式子是?
已知函数f(x）=sin^4x+cos^2x+1/4sin2xcos2x(x∈R）则f（x）化简的到式子是?
f(x）=sin^4x+cos^2x+1/4sin2xcos2x
=(1/4)[1-cos2x]^2+1/2(cos2x+1)+(1/8)sin4x
=(1/4)-(1/2)cos2x+(1/4)(cos2x)^2+(1/2)cos2x+1/2+(1/8)sin4x
=3/4+(1/8)(1+cos4x)+(1/8)sin4x
=7/8+(1/8)(cos4x+sin4x)
=(√2/8)sin(4x+π/4)+7/8
其最小正周期T=2π/4=π/2
f(x）=sin^4x+cos^2x+1/4sin2xcos2x
=[(1-cos2x)/2]^2+(1+cos2x)/2+1/8sin4x
=1/4(cos2x)^2-(cos2x)/2+1/4+1/2+(cos2x)/2+1/8sin4x
=1/8(1+cos4x)+3/4+1/8sin4x
=1/8cos4x+1/8sin4x+7/8
=√2/8(sin45°cos4x+cos45°sin4x)+7/8
=√2/8sin(45°+4x)+7/8
1/4*[3+(sin2x+cos2x)cos2x)]
f(x)=[(sinx)^4+(cosx)^2+1]/(4sin2xcos2x)
=(1-cos2x)^2/4+cos2x/2]/(2sin4x)
=[1+(cos2x)^2]/(8sin4x)
=1/(8sin4x)+(1+cos4x)/(16sin4x)
=3/(16sin4x)+(1/16)cot(4x)