求证:tan(3x/2)-tan(x/2)=2sinx/(cosx+cos2x)
来源:学生作业帮助网 编辑:作业帮 时间:2024/12/01 07:13:58
x){F<
}#M]İ5*̫H/F6IEi/!L
`q[r>P5T H.bhl
' @7V?.Sf.i ;
M?tR`bz"ljӾO?|u~qAb( BW
求证:tan(3x/2)-tan(x/2)=2sinx/(cosx+cos2x)
求证:tan(3x/2)-tan(x/2)=2sinx/(cosx+cos2x)
求证:tan(3x/2)-tan(x/2)=2sinx/(cosx+cos2x)
tan(3x/2)-tan(x/2)
=sin(3x/2)/cos(3x/2)-sin(x/2)/cos(x/2)(通分)
=[sin(3x/2)cos(x/2)-cos(3x/2)sin(x/2)]/[cos(3x/2)cos(x/2)]
=sin(3x/2-x/2]/[(1/2)(cos2x+cosx)(积化和差)
=2sinx/(cosx+cos2x)
故原式成立.
求证tan(x+y)*tan(x-y)=tan^2x-tan^2y/1-tan^2xtan^2y
求证:tan(3x/2)-tan(x/2)=(2sinx)/(cosx+cos2x)
求证:tan(3x/2)-tan(x/2)=2sinx/(cosx+cos2x)
求证tan^2x+1/tan^2x=[2(3+cos4x)]/(1-cos4x)
求证:(tan的平方x)+(1/tan的平方x)=2*(3+cos4x)/(1-cos4x)
求证:tan的平方x+1/tan的平方x=2(3+cos4X)/1-COS4x
求证:tan平方x+1/tan平方x=2(3+cos4x)/1-cos4x
tan^2x+1/tan^2=2(3+cos4x)/1-cos4x求证
求证tan^x-sin^2x=tan^2x*sin^2x
2/tan(x)=3/tan(45-x)怎么解?
tan(x/2)+tan(x/3)的周期如何计算?
1 求证:tan(x-y)+tan(y-z)+tan(z-x)=tan(x-y)tan(y-z)tan(z-x)2 已知a+b+c=npai(n属于Z),求证:tan(a)+tan(b)+tan(c)=tan(a)tan(b)tan(c)(提示:在等式a+b=npai-b同时取正切)
求证:tan x/2=sin x/(1+cos x)
求证:1-sin(x)=cos(x)*tan(x/2)
1-tan*2x/1+tan*2x=cos*2x-sinx求证
已知sin(2x+y)=5siny,(y不等于K派),求证;3tan=2tan(x+y)
求证:tan(x/2+π/4)+tan(x/2-π/4)=2tanx
求证:tan(x/2+派/4)+tan(x/2-派/4)=2tanx