1sin2x

化简（1+sin2x-cos2x)/(1+sin2x+cos2x)（1+sin2x-cos2x)/(1+sin2x+cos2x)=(2sin^2x+2sinxcosx)/(2cos^2x+2sinxcosx)=2sinx(sinx+cosx

(1+sin2x-cos2x)/(1+sin2x+cosx)(1+sin2x-cos2x)/(1+sin2x+cosx)=(2sin²x+2sinxcosx)/(2cos²x+2sinxcosx)=sinx(sinx+c

(1+sin2x)/(1+cos2x+sin2x)化简原式=(sinx+cosx)^2/[2(cosx)^2+2sinxcosx]=(sinx+cosx)^2/2cosx(cosx+sinx)=(sinx+cosx)/(2cosx)=(1/

化简(1+sin2x)/(1+sin2x+cos2x)(1+sin2x)/(1+sin2x+cos2x)=(sin²x+cos²x+2sinxcosx)/(sin²x+cos²x+2sinxcosx+

∫sin2x/(1+e^sin2x)dx不定积分...这样做比较简单：令i=∫[(sinx)^2*cosx/(sinx+cosx)]dxj=∫[sinx*(∴i=-(1/8)(sin2x+cos2x)+(1/4)In|sinx+cosx|+

1-cos2x+sin2x=原式=sin2x-cos2x+1=√2(√2/2*sin2x-√2/2cos2x)+1=√2(sin2xcosπ/4-cos2xsinπ/4)+1=√2sin(2x-π/4)+11+根号2sin（2x-π/4）原

sin2x.sinx\1-cosxy=(sin2x·sinx)/(1-cosx)=2sinxcosxsinx/(1-cosx)=2sin^2xcosx/(1-cosx)=2(1-cos^2x)sinx/(1-cosx)y=(sin2x·si

求证1+sin2x-cos2x/1+sin2x+cos2x=tanx1+sin2x-cos2x=1+2sinxcosx-1+2sinx^2=2sinx(cosx+sinx)1+sin2x+cos2x=1+2sinxcosx+2cosx^2-

化简[1+SIN2X-COS2X]/[1+SIN2X+COS2X]的结果原式＝(-2cos2x/1+sin2x+cos2x)+1=(-2cos^2x+2sin^2x)/(1+2sinxcosx+cos^2x-sin^2x)+1=[2(sin

三角函数和角公式化简sin2x+1/1+cos2x+sin2x原式=（Sin2x+1）/(1+Cos2x+Sin2x)用1=(Sinx)^2+(Cosx)^2代换=（Sin2x+(Sinx)^2+(Cosx)^2）/((Sinx)^2+(C

(SIN2X)^2+SIN2X+COS2X=1求角Xsin²2x＋sin2x＋cos2x=1sn2x＋cos2x=1－sin²2x=cos²2xsin2x=cos²2x－cos2x----------

(SIN2X)^2+SIN2X+COS2X=1求角Xsin2x+cos2x=1-sin2x^2=cos2x^2设cos2x=a,则a+根号(1-a^2)=a^2-->1-a^2=(a^2-a)^2=a^2*(a-1)^2a=1时成立,若a不

2sin2x+2sinxcosx如何变换为1-cos2x+sin2x原式=2[（1-cos2x)/2]+sin2x=1-cos2x+sin2x,其实这就是几个基本公式应该是2sin²x+2sinxcosx=1-cos2x+sin2

函数y=1/2sin2x+sin2x值域设sinX=k,则-1sin2xin[-1，1]yin[-inf,-2]Union[2,Inf]

化简:cos2x/（1-sin2x）原式=（cos²x-sin²x)/(cos²x-2·sinx·cosx+sin²x)=[(cosx+sinx)(cosx-sinx)]/(cosx-sinx)

不定积分sin2x/(1+sinx^4)sin2xdx/(1+sinx^4)=d((sinx)^2)/(1+((sinx)^2)^2)=arctan((sinx)^2)+C

根号1-sin2x等于什么1－sin2x=1－2sinxcosx=sin²x－2sinxcosx＋cos²x=(sinx－cosx)²则：根号下(1－sin2x)=|sinx－cosx|∵1－sin2x=1－2

三角函数公式推导过程,急救啊(1-1/4*(sin2x)^2)/2-sin2x=(2+sin2x)/4原式=(1-0.5*2*sinxcosx*0.5*2*sinxcosx)/2-2sinxcosx=0.5-(sinxcosxsinxcos

sin2x/cos2x-sin2x怎么变成[sin2x·(1-cos2x)]/cos2x,是怎么变化的?原式=sin2x*(1/cos2x-1)=sin2x*(1-cos2x)/cos2x通分即sin2x/cos2x-sin2x*cos2x

为什么sin2xdx=1/2*sin2xdx打错了，是为什么sin2xdx=1/2*sin2xd2xdx=(1/2)*d2x