已知sin(α+β)sin(α-β)=1/3求证1/4sin^2(2α)+sin^2β+cos^4α为定值
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已知sin(α+β)sin(α-β)=1/3求证1/4sin^2(2α)+sin^2β+cos^4α为定值
1/4sin^2(2α)+sin^2β+cos^4α
=1/4(2sinacosa)^2+sin^2β+cos^4α
=sin^2acos^2a+sin^2β+cos^4α
=cos^2a(sin^a+cos^2a)+sin^2β
=sin^2β+cos^2a
=(1-cos2β)/2+(cos2a-1)/2
=1/2(cos2a-cos2β)
=1/2*(-2)(sin(2a+2β)/2sin(2a-2β)/2
=-sin(a+β)sin(a-β)=-1/3
所以本题得证
=1/4(2sinacosa)^2+sin^2β+cos^4α
=sin^2acos^2a+sin^2β+cos^4α
=cos^2a(sin^a+cos^2a)+sin^2β
=sin^2β+cos^2a
=(1-cos2β)/2+(cos2a-1)/2
=1/2(cos2a-cos2β)
=1/2*(-2)(sin(2a+2β)/2sin(2a-2β)/2
=-sin(a+β)sin(a-β)=-1/3
所以本题得证
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