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sin^4θ+sin^4(π/3-θ)+sin^4(π/3+θ)化简求值
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sin^4(π/3-θ)+sin^4(π/3+θ)=[sin²(π/3-θ)+sin²(π/3+θ)]²-2sin²(π/3-θ)sin²(π/3+θ)
=[1/2-cos(2π/3-2θ)/2+1/2-cos(2π/3+2θ)/2]²-(1/2)[2sin(π/3-θ)sin(π/3+θ)]²
=[1-cos(2π/3)cos(2θ)]²-(1/2)[cos(2π/3)+cos(2θ)]²
=[1+cos(2θ)/2]²-(1/2)[-1/2+cos(2θ)]²=1+cos(2θ)+(1/4)cos²(2θ)-(1/8)+(1/2)cos(2θ)-(1/2)cos²(2θ)
=7/8+(3/2)cos(2θ)-(1/4)cos²(2θ)=7/8+(3/2)cos(2θ)-(1/8)[1+cos(4θ)]
=3/4+(3/2)cos(2θ)-(1/8)cos(4θ)
sin^4θ=[1/2-cos(2θ)/2]²=1/4-(1/2)cos(2θ)+(1/4)cos²(2θ)=1/4-(1/2)cos(2θ)+(1/8)+(1/8)cos(4θ)
所以sin^4θ+sin^4(π/3-θ)+sin^4(π/3+θ)
=9/8+cos(2θ)
=[1/2-cos(2π/3-2θ)/2+1/2-cos(2π/3+2θ)/2]²-(1/2)[2sin(π/3-θ)sin(π/3+θ)]²
=[1-cos(2π/3)cos(2θ)]²-(1/2)[cos(2π/3)+cos(2θ)]²
=[1+cos(2θ)/2]²-(1/2)[-1/2+cos(2θ)]²=1+cos(2θ)+(1/4)cos²(2θ)-(1/8)+(1/2)cos(2θ)-(1/2)cos²(2θ)
=7/8+(3/2)cos(2θ)-(1/4)cos²(2θ)=7/8+(3/2)cos(2θ)-(1/8)[1+cos(4θ)]
=3/4+(3/2)cos(2θ)-(1/8)cos(4θ)
sin^4θ=[1/2-cos(2θ)/2]²=1/4-(1/2)cos(2θ)+(1/4)cos²(2θ)=1/4-(1/2)cos(2θ)+(1/8)+(1/8)cos(4θ)
所以sin^4θ+sin^4(π/3-θ)+sin^4(π/3+θ)
=9/8+cos(2θ)
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