sin2x等于sinx

tanx/tan2x=tanx/[2tanx/1-(tanx)^2]=[1-(tanx)^2]/2X趋近于π=[1-(tanπ)^2]/2=1/2sinx/sin2x=sinx/2sinxcosx=1

(1)根据图信息可知sina=4/5cosa=3/5∴(sina^2+sin2a)/(cosa^2+cos2a)=(sina^2+2sinacosa)/(cosa^2+cosa^2-sina^2)=(

f(sinx)=cos2x=1-2sin^2xf(x)=1-2x^2f(cosx)=1-2cos^2x=-cos2x选C

如果是相乘的话,那么就是：sinx*sin2x*sin3x=sinX*2sinX*cosX*sin(2X+X)=2sinX*sinX*cosX*(sin2X*cosX+cos2X*sinX)=2sin

cos^2x+2SINXCOSX+SIN^2X=1(SINX+COSX)^2=1SINX+COSX=1或-1希望对你帮助

sin^2x+cos^2x=1(sinx+cosx)^2=1+2sinxcosx=1+sin2x=1-24/25=1/25sinx+cosx=1/25或sinx+cosx=-1/25又,x属于(-π/

sin^2x+cos^2x=1sin^2x+9sin^2x=1sin^2x=1/10Cos2X+Sin2X=cos^2x-sin^2x+2sinxcosx=9sin^2x-sin^2x+6sin^2x

题目条件不充分啊cos2x+sin2x=1cos2x=cos^2x-sin^2xsin2x=2sinxcosxcos2x+sin2x=cos^2x-sin^2x+2sinxcosx

楼主写错了吧,我只想告诉你lg（sinx+cosx）+lg（sinx+cosx）=lg【（sinx+cosx）*（sinx+cosx）】=lg（sinx平方+cosx平方+2sinx*cosx）=lg

d(f(sin²x))=f'(sin²x)d(sin²x)=f'(sin²x)2sinxcosxdx=sin(2x)f'(sin²x)dxy=x

sinx/cosx=tanx所以sin2x/cos2x=tan2x

等式左边=2sinxcosx/[sinx+(cosx-1)][sinx-(cosx-1)]=2sinxcosx/[sin^2x-(cosx-1)^2]=2sinxcosx/(sin^2x-cos^2x

设f(x)=sinx^2+sin2x-2cosx^2,此题实际上就是求f(x)的值域,具体解答如下:\x0df(x)=sinx^2+sin2x-2cosx^2\x0d=(sinx^2+cosx^2)+

sinx×cos2x-sin2x×cosx=sin(x-2x)=-sinx

答案如图所示,刚才有个错误,重传了一个答案.这里不考虑x使得分母为零的情况了,因为在分母为零处积分不存在

P=sin2xsinx-cosx还是P=sin2x（sinx-cosx）?

等式左边=2sinxcosx/[sinx+(cosx-1)][sinx-(cosx-1)]=2sinxcosx/[sin^2x-(cosx-1)^2]=2sinxcosx/(sin^2x-cos^2x

sin2xdx/(1+sinx^4)=d((sinx)^2)/(1+((sinx)^2)^2)=arctan((sinx)^2)+C

sin3x=sin(2x+x)=sin2xcosx+cos2xsinx=2sinxcosxcosx+cos2xsinx=2sinxcos²x+cos2xsinxsin3x-sin2x+sin

[（sinx）^2+(cosx)^2]^2=1(sinx)^4+2(sinx)^2(cosx)^2+(cosx)^4=12(sinx)^2(cosx)^2=4/9(sinx)^2(cosx)^2=2/