求证:cosαcos(π/3+α)cos(π/3-α)=cos3α如题,求证:4cosαcos(π/3+α)cos(π/3-α)=cos3α不好意思 少打个4

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/16 06:45:50
求证:cosαcos(π/3+α)cos(π/3-α)=cos3α如题,求证:4cosαcos(π/3+α)cos(π/3-α)=cos3α不好意思 少打个4
xRN@~f3鴳|3?VVīe0BPo+ KĈ<;-^ L%Y5$s1s9ǍY9/(0{a>u~ػ=n2:y?P[Ywgtrr~\~2Fʀ넀&긧ޞ6 J,Kj^35ՂV+^׃0V9 壺V ,|džp-RA.zrFhPa̗v[Hr >!Jèľ4}ȮߋXb_i IŸmAloP j0)v$f4ݒd{//Ni~0Pov}U=ݲFyjJW@͛i6$G?R?x/Sex&r\Vԫvo/]2۔W

求证:cosαcos(π/3+α)cos(π/3-α)=cos3α如题,求证:4cosαcos(π/3+α)cos(π/3-α)=cos3α不好意思 少打个4
求证:cosαcos(π/3+α)cos(π/3-α)=cos3α
如题,
求证:4cosαcos(π/3+α)cos(π/3-α)=cos3α
不好意思 少打个4

求证:cosαcos(π/3+α)cos(π/3-α)=cos3α如题,求证:4cosαcos(π/3+α)cos(π/3-α)=cos3α不好意思 少打个4
点小图看大图

证明:由积化和差公式可得
cos(π/3+α)cos(π/3-α) =(cos2π/3+cos2a)/2=(cos2a-1/2)/2
所以4cosa(cos2a-1/2)/2=2cosa(cos2a-1/2)=2cosacos2a-cosa
=cos3a+cosa-cosa=cos3a
所以该等式成立