cos(X-B)的平方-Cos(X+B)的平方=1∕2,(1+cos2x)(1+cos2)=1/3求tanxtanb=急,最好今天就解决
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![cos(X-B)的平方-Cos(X+B)的平方=1∕2,(1+cos2x)(1+cos2)=1/3求tanxtanb=急,最好今天就解决](/uploads/image/z/13411776-48-6.jpg?t=cos%28X-B%29%E7%9A%84%E5%B9%B3%E6%96%B9-Cos%28X%2BB%29%E7%9A%84%E5%B9%B3%E6%96%B9%3D1%E2%88%952%2C%EF%BC%881%EF%BC%8Bcos2x%29%281%EF%BC%8Bcos2%29%3D1%2F3%E6%B1%82tanxtanb%3D%E6%80%A5%2C%E6%9C%80%E5%A5%BD%E4%BB%8A%E5%A4%A9%E5%B0%B1%E8%A7%A3%E5%86%B3)
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cos(X-B)的平方-Cos(X+B)的平方=1∕2,(1+cos2x)(1+cos2)=1/3求tanxtanb=急,最好今天就解决
cos(X-B)的平方-Cos(X+B)的平方=1∕2,(1+cos2x)(1+cos2)=1/3求tanxtanb=
急,最好今天就解决
cos(X-B)的平方-Cos(X+B)的平方=1∕2,(1+cos2x)(1+cos2)=1/3求tanxtanb=急,最好今天就解决
cos(X-B)²-cos(X+B)²=1∕2
【cos(X-B)+cos(X+B)】*【cos(X-B)-cos(X+B)】=1∕2
(cosXcosB+sinXsinB)+(cosXcosB-sinXsinB)
*(cosXcosB+sinXsinB)-(cosXcosB-sinXsinB)=1∕2
化简 得
4cosXsinXcosBsinB=1∕2……①
(1+cos2x)(1+cos2B)=1/3(你这里好像少打一个B吧)
(1+cosX²-sinX²)*(1+cosB²-sinB²)=1/3
∵ 1-sinX²=cosX²
∴ 2cosX²cosB²=1/3……②
①/② 得
tanxtanb= 3/2