sinxcosx 根号1 sinx2的不定积分-搜答案-智慧作业帮

tanx=√2/2,所以sinx=√3/3,cosx=√6/3所以sinxcosx=√2/3sinx(sinx-3cosx)=1/3-√2

17/4根号3/2-321/64a/2带入后,平方,化简

f(x)=cos2x+√3sin2x+1=2(sin2x*√3/2+cos2x*1/2)+1=2(sin2xcosπ/6+cos2xsinπ/6)+1=2sin(2x+π/6)+1所以T=2π/2=π

∫sinxcosx/(1+sin^4x)dx=∫sinx/(1+sin^4x)d(sinx)=∫1/(1+sin^4x)d(1/2*sin²x)=(1/2)∫d(sin²x)/[1

f(x)=(1-cos2x)/2+(√3/2)sin2x+1/2=(√3/2)sin2x-(1/2)cos2x+1=sin(2x-π/6)+1（1）最小正周期T=2π/2=π（2）f(x)max=2,

1、sinx+cosx=根号2*sin(x+pai/4)

等式左边分母=(sinx+cosx-1)(sinx-cosx+1)=sin2x-(cosx-1)2=sin2x-cos2x+2cosx-1=2cosx-2cos2x与分子约去2cosx,左边得sinx

方法一：设tanx/2=t（后面写起来方便）原式=[2t/(1+t^2)]*[(1-t^2)/(1+t^2)]/[1+2t/(1+t^2)+(1-t^2)/(1+t^2)/=2t(1-t^2)/(2t

-根号3sinx^2+sinxcosx=-根号3*(1-cos2x/2)+1/2sin2x=根号3/2cos2x-根号3+1/2sin2x=sin(2x+60)-根号3/2f(a/2)=sin(a+6

①化简f（x）=1+sin(π/6-2x)∴T=2π/w=ππ/6-2x∈[-5/6π,π/6]f(x)min=1-1=0f(x)max=1+1/2=3/2②得sin(π/6-2x)=4/5π/6-2

(1)f(x)=(sinx-cosx)平方-1=(sinx)^2+(cosx)^2-2sinxcosx-1=-sin2x(2)f(x)=根号3sinxcosx-cosxsin(π/2+x)-1/2=根

(sinX+cosX)平方=2所以sinX平方+cosX平方+2sinXcosX=2因为sinX平方+cosX平方=1所以sinXcosX=0.5

f（x）=cosx^2-sinx^2+(2根号3)sinxcosx+1=cos2x+根号3sin2x+1=2(根号3/2sin2x+1/2cos2x)+1=2(sin2xcosπ/6+cos2xsin

sinx+cosx=1两边平方(sinx+cosx)²=1sin²x+cos²x+2sinxcosx=12sinxcosx=0∴sinx=0,cosx=1或cosx=0,

sinx/cosx

由于sinx*sinx+cosx*cox=1所以(sinx*sinx+cosx*cox+2sinxcosx)/(sinx+cosx)=(sinx+cosx)*(sinx+cosx)/(sinx+cos

二倍角公式

(1-2sinxcosx)(1+2sinxcosx)=(sin²x+cos²x-2sinxcosx)(sin²x+cos²x+2sinxcosx)=(sinx-

函数fx=2根号3sinxcosx+1-2sinX=根号3sin2x+cos2x=2sin（2x+30度）,fx的值域就是【-2,2】