已知函数y=(sinx cosx)平方 2cos^2x

（1）y=sin平方X+SinxCosx=(1-2Cos2x)/2+(Sin2x)/2根据辅助角公式得:y=根号（2∕4）sin（2x-π∕4）+1∕2所有周期为2π/2=π(2)因为sinx最大为1

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=2√3sinxcosx+2cos²x=√3sin2x+cos2x+1=2(√3/2sin2x+1/2*cos2x)+1=2(sin2xcosπ/6+cos2xsinπ/6)+1=2sin

这个题稍稍有些难度,我给你写前半部分的过程,相信后面的你就会了.我们知道（sinx+cosx）^2=sinx^2+cosx^2+2sinxcosx=1+2sinxcosx则2sinxcosx=(sin

y=sinxcosx+sinx+cosx=1/2(2sinxcosx+1-1)+sinx+cosx=1/2(sinx+cosx)^2-1/2+(sinx+cosx)=1/2[(sinx+cosx)^2

y=(1-cos2x)/2+(sin2x)/2+2=1/2(sin2x-cos2x)+5/2=√2/2sin(2x-π/4)+5/2因此y的最大值为：√2/2+5/2,最小值为-√2/2+5/2值域即

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

分析：由(sinx)^2+(cosx)^2=1进行解析如下,1.y=g(x)=a（sinx+cosx）+2sinxcosx,当a=1时,g(x)=sinx+cosx+2sinxcosx=sinx+co

令sinx-cosx=a则a^2=(sinx)^2+(cosx)^2-2sinxcosx=1-2sinxcosx所以sinxcosx=(1-a^2)/2所以y=1+a+(1-a^2)/2=-a^2/2

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=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

y=2cosxsin(x+π/3)-根号3*(sin^2)x+sinxcosx,后两项先提出一个sinx,然后括号内部分用叠加原理,得到y=2cosxsin(x+π/3)+2sinxcos(x+π/3

y=2根号3sinxcosx-2cos^2x+1＝根号3sin2x-cos2x=2sin(2x-π/6)T=2π/2=π2x-π/6=2kπ+π/2当x=kπ+π/3时,函数的最大值=22x-π/6=

-3

y=(sinx)^2+sinxcosx+2=(1-cos2x)/2+sin2x/2+2=5/2+(√2/2)[(√2/2)sin2x-(√2/2)cos2x]=5/2+(√2/2)sin(2x-π/4

y=sin²x+sinxcosx+2=(1-cos2x)/2+(sin2x)/2+2=(1/2)(sin2x-cos2x)+5/2=(1/2)*√2(sin2xcosπ/4-cos2xsin

y=cos²x-sin²x+2sinxcosx=cos2x+sin2x=√2sin(2x+π/4)所以值域为【-√2,√2】