(sina/2+cosa/2)^2+2sin^2(π/4-a/2)RT
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(sina/2+cosa/2)^2+2sin^2(π/4-a/2)RT
(sina/2+cosa/2)^2+2sin^2(π/4-a/2)
RT
(sina/2+cosa/2)^2+2sin^2(π/4-a/2)RT
1+sina+1-cos(π/2-a)=2
3sina=2cosa 则(cosa-sina/cosa+sina)+(cosa+sina/cosa-sina)=______
3sina=2cosa 则(cosa-sina/cosa+sina)+(cosa+sina/cosa-sina)=______
化简(sina+2cosa)/(sina-cosa)=
求证:1+sina+cosa/1+sina-cosa+1-cosa+sina/1+cosa+sina=2/sina
化简:(sina+cosa)^2
化简sina-2cosa
证明:cosa/(1+sina)-sina/(1+cosa)=2(cosa-sina)/(1+sina+cosa)
求证cosa/1+sina-sina/1+cosa=2(cosa-sina)/1+sina+cosa
求证:1+sina+cosa+2sina cosa/1+sina+cosa=sina+cosa
数学问题哦已知:sina+3cosa =2 ,s求(sina-cosa)/(sina +cosa)的值 求方法!谢谢!
Sina=2Cosa,则Sina^2+2Sina*Cosa
化简:sina^2/(sina-cosa)-(sina+cosa)/(tan^2-1)
sina*cosa+cosa*sina怎么会等于2sina*cosasina*cosa+cosa*sina怎么会等于2sina*cosa
证明:2(cosa-cosa)/(1+cosa+cosa)=cosa/(1+sina)-sina/(1+cosa).
证明 cosa/(1+sina0-sina/(1+cosa)=2(cosa-sina)/(1+sina+cos)cosa/(1+sina)-sina/(1+cosa)=2(cosa-sina)/(1+sina+cos)
若tan=2,求sina+cosa/sina-cosa +(cosa)^2的sina&cosa/sina-cosa 是个整体
化简,(1+sina-cosa)/(1+sina+cosa)+(1+sina+cosa)/(1+sina-cosa).答案是sina/2,为什么?
sinA=2-2cosA求sinA?