不查表求值:cos40°•cos80°+cos80°•cos160°+cos160°•cos40°.
题型:不详难度:来源:
不查表求值:cos40°•cos80°+cos80°•cos160°+cos160°•cos40°. |
答案
原式=[cos120°+cos(-40°)+cos240°+cos(-80°)+cos200°+cos120°] =(-cos60°+cos40°-cos60°+cos80°-cos20°-cos60°) =[-+cos(60°-20°)+cos(60°+20°)-cos20°] =[-+cos60°cos20°+sin60°sin20°+cos60°cos20°-sin60°sin20°-cos20°] =[-+cos20°-cos20°] =- |
举一反三
cos54°-sin54°化为积的形式是( )A.cos9° | B.-cos9° | C.sin9° | D.-sin9° |
|
已知cos(α+β)cos(α-β)=,则cos2α-sin2β的值是( ) |
cos36°cos24°-sin36°sin24°=______. |
若α,β满足 | cos2(α-β)-cos2(α+β)= | (1+cos2α)(1+cos2β)= |
| | ,求tanαtanβ的值. |
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