求sin^2 40°+cos^2 10°-sin40°sin80°的值.
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用到积化和差公式和倍角公式
sinαsinβ=-[cos(α+β)-cos(α-β)]/2
cos^2 α=1/2(1+cos2α)
解：
sin^2 40°+cos^2 10°-sin40°sin80°
=cos^2 50°+cos^2 10°+1/2[cos (40°+80°)-cos(40°-80°)
=1/2(1+cos100°)+1/2(1+cos20°)+1/2[cos (40°+80°)-cos(40°-80°)
=1+1/2(cos100°+cos20°)+1/2(cos120°-cos40°)
=1+1/2(cos100°+cos20°)-1/4-1/2cos40°
=3/4+cos60°cos40°-1/2cos40°
=3/4
sinαsinβ=-[cos(α+β)-cos(α-β)]/2
cos^2 α=1/2(1+cos2α)
解：
sin^2 40°+cos^2 10°-sin40°sin80°
=cos^2 50°+cos^2 10°+1/2[cos (40°+80°)-cos(40°-80°)
=1/2(1+cos100°)+1/2(1+cos20°)+1/2[cos (40°+80°)-cos(40°-80°)
=1+1/2(cos100°+cos20°)+1/2(cos120°-cos40°)
=1+1/2(cos100°+cos20°)-1/4-1/2cos40°
=3/4+cos60°cos40°-1/2cos40°
=3/4
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