sin^20°+sin80°sin40°=？ 且sin(a+b)sin(a-b)=sin^2a-sin^2b
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sin²20°+sin80°sin40°
=(1-cos40°)]/2 -[cos(80°+40°)-cos(80°-40°)]/2
=1/2 -cos40°/2-(cos120°-cos40°)/2
=1/2 -cos40°/2 -(-1/2 -cos40°/2)
=1/2 -cos40°/2 +1/2 +cos40°/2
=1
sin(a+b)sin(a-b)=-[cos(a+b+a-b) -cos(a+b-a+b)]/2
=-[cos(2a)-cos(2b)]/2
=-[(1-2sin²a)-(1-2sin²b)]/2
=-(2sin²b-2sin²a)/2
=sin²a-sin²b
=(1-cos40°)]/2 -[cos(80°+40°)-cos(80°-40°)]/2
=1/2 -cos40°/2-(cos120°-cos40°)/2
=1/2 -cos40°/2 -(-1/2 -cos40°/2)
=1/2 -cos40°/2 +1/2 +cos40°/2
=1
sin(a+b)sin(a-b)=-[cos(a+b+a-b) -cos(a+b-a+b)]/2
=-[cos(2a)-cos(2b)]/2
=-[(1-2sin²a)-(1-2sin²b)]/2
=-(2sin²b-2sin²a)/2
=sin²a-sin²b
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