化简[sin(a+b)-2cosasinb]/[cos(a+b)+2sinasinb]
来源:学生作业帮 编辑:神马作文网作业帮 分类:数学作业 时间:2024/11/15 16:20:59
化简[sin(a+b)-2cosasinb]/[cos(a+b)+2sinasinb]
sin(a+b)-2cosasinb
=sinacosb+sinbcosa-2cos2sinb
=sinacosb-sinbcosa
=sin(a-b)
cos(a+b)+2sinasinb
=cosacosb-sinasinb+2sinasinb
=cosacosb+sinasinb
=cos(a-b)
所以:
[sin(a+b)-2cosasinb]/[cos(a+b)+2sinasinb]
=sin(a-b)/cos(a-b)
=tan(a-b)
=sinacosb+sinbcosa-2cos2sinb
=sinacosb-sinbcosa
=sin(a-b)
cos(a+b)+2sinasinb
=cosacosb-sinasinb+2sinasinb
=cosacosb+sinasinb
=cos(a-b)
所以:
[sin(a+b)-2cosasinb]/[cos(a+b)+2sinasinb]
=sin(a-b)/cos(a-b)
=tan(a-b)
sinAsinB+cosAcosB这个怎么变换!sinAcosB+cosAsinB为什么等于sin(A+B)
证明sin(a+b)=sinacosb+cosasinb
sin(A+B)=sinAcosB+cosAsinB
证明cos(A+B)=cosAcosB-sinAsinB
证cos(A-B)=cosAcosB+sinAsinB
cosacosb+sinasinb=cos(a-b)
一提三角函数转换题已知sinasinb=1,求cos((a+b)/2)
化简[sin(2A+B)]/sinA-2cos(A+B)
sin(A+B)=sinAcosB+cosAsinB//如何证明?
sin(A-B)=sinAcosB-cosAsinB怎么证明?
sin(A+B) = sinAcosB+cosAsinB公式的证明过程
sin(a-b)=sinacosb-cosasinb如何推导?