若sinx cosx=根号下(1 sin2x) 求定义域x

3sinxcosx=3/2sin2x,cos平方x化为cos2x形式,再通过公式合并前面两项.

(1)f(x)=sinxcosx+√3cos²X-√3/2=sin2x/2+√3cos2x/2+√3/2-√3/2=sin(2x+π/3).(2)f(x)的最小正周期为π,值域是[-1,1]

y=√2/2*sin2x+(1+cos2x)/2-1/2=√2/2*sin2x+1/2*cos2x=√3/2*sin(2x+z)其中tanz=(1/2)/(√2/2)=√2/2所以T=2π/2=π

f(x)=1/2sin2x-√3/2(1-cos2x)=1/2sin2x+√3/2cos2x-√3/2=sin(2x+π/3)-√3/21、最小正周期T=2π/2=π2、0≤x≤π/2π/3≤2x+π

y=√3*sinxcosx+cos^2x=√3/2sin2x+(1+cos2x)/1=√3/2sin2x+1/2*cos2x+1/2=sin(2x+π/6)+1/2周期T=2π/2=π2Kπ-π/2≤

f(x)=(√3)sinxcosx+cos2x+1f(x)=(√3)(2sinxcosx)/2+cos2x+1f(x)=(√3/2)sin2x+cos2x+1f(x)=(√7/2)[(√3/2)(2/

f(x)=2cosx[1/2sinx+√3/2cosx]-√3sin^2(x)+sinxcosx=sinxcosx+√3cos^2(x)-√3sin^2(x)+sinxcosx=sin2x+√3cos

(sinx+cosx)^2=(sinx)^2+（cosx)^2+2sinxcosx=1+2*2/5=9/5∵cos²x=-cosx∴cosx0∴sinx

答：f(x)=(1/2)*(cosx)^2+(√3/2)sinxcosx+1=(1/2)*(cos2x+1)/2+(√3/4)sin2x+1=(1/2)[sin2xcosπ/6+cos2xsinπ/6

y=1/2*cos²x+√3/2*sinxcosx+1=1/4*(cos2x-1)+√3/4*sin2x+1=1/2*(1/2*cos2x+√3/2*sin2x)+3/4=1/2*(sin3

f(x)=5根号下3cos^2x+根号下3sin^2x-4sinxcosx=4根号下3cos^2x-4sinxcosx+根号下3=2根号下3(2cos^2x-1)-2sin2x+3根号下3=2根号3c

f(x)=5根号下3cos^2x+根号下3sin^2x-4sinxcosxx属于[π/4,7π/24]f(x)=根号3(5(1+cos2x)/2)+根号3(1-cos2x)/2-2sin2xf(x)=

(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

y=1/2cos²x+√3/2sinxcosx+1=1/2(1/2+1/2cos2x+√3/2sin2x)+1=1/2(sin2xcosπ/6+cos2xsinπ/6)+1+1/4=1/2s

-3

怎么感觉cosx应该是平方啊再问：嗯的，打错了再答：(cosx)^2=(1+cos2x)/2sinxcosx=1/2*sin2x所以原式=(1+cos2x)/2-根号3/2*sin2x+1=1/2*c

答：f(x)=√3sinxcosx+(cosx)^2=(√3/2)*2sinxcosx+(1/2)*[2(cosx)^2-1]+1/2=(√3/2)sin2x+(1/2)cos2x+1/2=sin(2

根号里的数据必须大于等于零,所以x>=0且-x>=0,得出x=0,则x+1=1

f(x)=√3sinxcosx+cos2x+1=(√3/2)sin2x+cos2x+1=[(√7)/2][(√3/√7)sin2x+(2/√7)cos2x]+1=[(√7)/2]sin(2x+α)+1

解:原式=√3sin2x＋cos2x＋1=2(√3/2sin2x＋1/2cos2x＋1=2cos(2x-pai/3)＋1.