题目内容
不查表求sin220°+cos280°+sin20°cos80°的值.
解法一: sin220°+cos280°+sin20°cos80°
= (1-cos40°)+ (1+cos160°)+ sin20°cos80°
=1-cos40°+cos160°+sin20°cos(60°+20°)
=1-cos40°+ (cos120°cos40°-sin120°sin40°)
+sin20°(cos60°cos20°-sin60°sin20°)
=1-cos40°-cos40°-sin40°+sin40°-sin220°
=1-cos40°-(1-cos40°)=
解法二: 设x=sin220°+cos280°+sin20°cos80°
y=cos220°+sin280°-cos20°sin80°,
则x+y=1+1-sin60°=,
x-y=-cos40°+cos160°+sin100°
=-2sin100°sin60°+sin100°=0
∴x=y=,即x=sin220°+cos280°+sin20°cos80°=.
= (1-cos40°)+ (1+cos160°)+ sin20°cos80°
=1-cos40°+cos160°+sin20°cos(60°+20°)
=1-cos40°+ (cos120°cos40°-sin120°sin40°)
+sin20°(cos60°cos20°-sin60°sin20°)
=1-cos40°-cos40°-sin40°+sin40°-sin220°
=1-cos40°-(1-cos40°)=
解法二: 设x=sin220°+cos280°+sin20°cos80°
y=cos220°+sin280°-cos20°sin80°,
则x+y=1+1-sin60°=,
x-y=-cos40°+cos160°+sin100°
=-2sin100°sin60°+sin100°=0
∴x=y=,即x=sin220°+cos280°+sin20°cos80°=.
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