y=2根三sinxcosx-2cos^2x 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

y=cos2x+sin2x=√2(√2/2*sin2x+√2/2cos2x)=√2(sin2xcosπ/4+cos2xsinπ/4)=√2sin(2x+π/4)所以T=2π/2=π值域是[-√2,√2

y=sinxcosx-cos^2x=1/2sin2x-1/2(1+cos2x)=1/2(sin2x-cos2x-1)=1/2[√2*sin(2x-派/4)-1]=√2/2*sin(2x-派/4)-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,

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

y=sinx+xcosxy'=cosx+cosx+x*(-sinx)=2cosx-xsinxy=2^x+log2xy'=2^xln2+1/(xln2)y=sinxcosxy'=cosx*cosx+si

y=(sinx+cosx)+(sinx+cosx)^2-(sinx^2+cosx^2)=(sinx+cosx)+(sinx+cosx)^2-1设sinx+cosx=t,t=√2sin(x+π/4)∈[

sinx+cosx=√2(sinxcosπ/4+cosxsinπ/4)=√2sin(x+π/4)y=sinx+cosx+2sinxcosx=sinx+cosx+2sinxcosxsin^2x+cos^

y=2(cosx)^2+2√3sinxcosx=cos2x+1+2√3sinxcosx=cos2x+√3sin2x+1=2[1/2cos2x+√3/2sin2x)+1=2sin(2x+π/6)+1

1．y=1/4(1+cos2x)+√3/4sin2x+1=1/2sin(2x+π/6)+5/4当2x+π/6＝2kπ+π/2,即x＝kπ＋π/6时,ymax＝7/4.自变量x的集合{x│x＝kπ＋π/

y=sin^x+2sinxcosx=1/2-cos2x/2+sin2x=根号下(5/4)*[2sin2x/根号5-cos2x/根号5]+1/2设cosa=2/根号5,sina=-1/根号5上式=根号下

解题思路：利用三角函数正弦的和公式sin（x+x）可得结果解题过程：解：因为sin2x=sin（x+x）=sinxcosx+cosxsinx=2sinxcosx,所以y=2sinxcosx=sin2x

hiy=1/2cos^2x+sinxcosx+3/2sin^2x=1/2*（（1+cos2x）/2）+1/2*sin2x+3/2*((1+cos2x)/2)=1/2*sin2x-1/2*cos2x+1

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

y=(1/2)sin2x＋(√3/2)[1＋cos2x]－√3=(1/2)sin2x＋(√3/2)cos2x－(√3/2)=sin(2x＋π/3)－(√3/2)最小正周期是π

cos(A-C)-cos(A+C),cosAcosC+sinAsinC-(cosAcosC-sinAsinC)=2sinAsinC=2sinB所以sinAsinC=sinBf(C)=2sin(2C+π

y=sinx+cosx+2sinxcosx+2=sinx+cosx+2sinxcosx+1+(sinx)^2+(cosx)^2=(sinx+cosx)^2+sinx+cosx+1=(sinx+cosx

∵y=sin2x+2sinxcosx=1−cos2x2+sin2x=sin2x-12cos2x+12=52sin（2x+φ）+12，（tanφ=-12）∴其周期T=2π2=π．故答案为：π．

y=1+2sinxcosx+sinx+cosx=sin²x+cos²x+2sinxcosx+sinx+cosx=(sinx+cosx)²+sinx+cosx=(sinx+

令u=sinx+cosx=√2sin(x+π/4)∈[-√2,√2]u²=sin²x+cos²x+2sinxcosx=1+2sinxcosx∴4sinxcosx=2(u&