二倍角的三角函数题1.若已知sin(π/4+α)sin(π/4-α)=1/6,α∈(π/2,π),求(sin4α)/(1+cos^2α)2.求证:cos^8x - sin^8x + 1/4 *sin2xsom4x=cos2x第二题打错了...cos^8x - sin^8x + 1/4 *sin2xsin4x=cos2x

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二倍角的三角函数题1.若已知sin(π/4+α)sin(π/4-α)=1/6,α∈(π/2,π),求(sin4α)/(1+cos^2α)2.求证:cos^8x-sin^8x+1/4*sin2xsom4

二倍角的三角函数题1.若已知sin(π/4+α)sin(π/4-α)=1/6,α∈(π/2,π),求(sin4α)/(1+cos^2α)2.求证:cos^8x - sin^8x + 1/4 *sin2xsom4x=cos2x第二题打错了...cos^8x - sin^8x + 1/4 *sin2xsin4x=cos2x
二倍角的三角函数题
1.若已知sin(π/4+α)sin(π/4-α)=1/6,α∈(π/2,π),
求(sin4α)/(1+cos^2α)
2.求证:cos^8x - sin^8x + 1/4 *sin2xsom4x=cos2x
第二题打错了...
cos^8x - sin^8x + 1/4 *sin2xsin4x=cos2x

二倍角的三角函数题1.若已知sin(π/4+α)sin(π/4-α)=1/6,α∈(π/2,π),求(sin4α)/(1+cos^2α)2.求证:cos^8x - sin^8x + 1/4 *sin2xsom4x=cos2x第二题打错了...cos^8x - sin^8x + 1/4 *sin2xsin4x=cos2x
由sin(π/4+α)sin(π/4-α)
=cos(π/4-α)sin(π/4-α)
=sin(π/2-2α)/2
=cos2α/2
=1/6
cos2α=1/3
又α∈(π/2,π)
2α∈(π,2π)
sin2α