sin2x/(1+cos2x)*cosx/(1+cosx)*sinx/(1-cosx)化简出来是1
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sin2x/(1+cos2x)*cosx/(1+cosx)*sinx/(1-cosx)化简出来是1
sin2x/(1+cos2x)*cosx/(1+cosx)*sinx/(1-cosx)
化简出来是1
sin2x/(1+cos2x)*cosx/(1+cosx)*sinx/(1-cosx)化简出来是1
1 + cos 2x = 2 cos^2 x
sin 2x = 2sin x cos x
(1 + cos x)(1 - cos x ) = 1 - cos^2 x = sin^2 x
因此上式等于
(2*sinx*cosx*cosx*sinx) / 2 * cos^2 x * sin^2 x = 1
(1+sin2x-cos2x)/(1+sin2x+cosx)
化简 (sin2x)/(1+cos2x)*(cosx)/(1+cosx)
化简:(sinx+sin2x)/(1+cosx+cos2x)
sinx×cos2x-sin2x×cosx
(1+cos2x)/2cosx=sin2x/(1-cos2x)
求证:(1-2sinx×cosx)/cos2x-sin2x=(cos2x-sin2x)/(1+2sinx×cosx)
化简(1+cos2x/sin2x)乘以(1+cosx/cosx)
怎么证明cos2x/1-sin2x = cosx+sinx/cosx-sinx
已知1+sinx+cosx+sin2x+cos2x=0,求tanx
cos2x+sin2x=1 那么sinx-cosx等于多少
求证sinx+sin3x+sin2x=1+cos2x+cosx
解方程;SinX+Sin2X+Sin3X = 1+CosX+Cos2X.
Cosx+sinx分之1+sin2x+cos2x=2cos 证明
sin2x/(1+cos2x)*cosx/(1+cosx)*sinx/(1-cosx)化简出来是1
化简 (1+sin2x-cos2x)/(1+sin2x+cos2x)
若2sin2x+cos2x=1 [2(cosx)^2+sin2x]/(1+tanx)的值
已知(sin2x)^2+sin2x*cosx-cos2x=1,x属于(0,π/2),求x
cos2x+cosx=0,则sin2x+sinx