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原式=[sin²(x+π/4)+sin²(π/4-x)][sin²(x+π/4)-sin²(π/4-x)]
={sin²(x+π/4)+cos²[π/2-(π/4-x)]}{sin²(x+π/4)-cos²[π/2-(π/4-x)]}
=[sin²(x+π/4)+cos²(π/4+x)][sin²(x+π/4)-cos²(π/4+x)]
=1*{-cos[2(x+π/4)]}
=-cos(2x+π/2)
=sin2x
={sin²(x+π/4)+cos²[π/2-(π/4-x)]}{sin²(x+π/4)-cos²[π/2-(π/4-x)]}
=[sin²(x+π/4)+cos²(π/4+x)][sin²(x+π/4)-cos²(π/4+x)]
=1*{-cos[2(x+π/4)]}
=-cos(2x+π/2)
=sin2x
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sin^4(x+π/4)-sin^4(x-π/4)
= [sin^2(x+π/4)+sin^2(x-π/4)][sin^2(x+π/4)-sin^2(x-π/4)]
= [cos^2(x-π/4)+sin^2(x-π/4)][cos^2(x-π/4)-sin^2(x-π/4)]
= 1×cos2(x-π/4)
= cos(2x-π/2)
= cos(π/2-2x)
= sin2x
= [sin^2(x+π/4)+sin^2(x-π/4)][sin^2(x+π/4)-sin^2(x-π/4)]
= [cos^2(x-π/4)+sin^2(x-π/4)][cos^2(x-π/4)-sin^2(x-π/4)]
= 1×cos2(x-π/4)
= cos(2x-π/2)
= cos(π/2-2x)
= sin2x
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