y=sinx cosx=2(x属于(0,2 π))

解题思路：考查三角恒等变换解题过程：varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/include/read

y=sin²x+cos8x+2sinxcosx+2cos²x=1+sin2x+(1+cos2x)=sin2x+cos2x+2=√2sin(2x+π/4)+2-1

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/

y=cos²x-sin²x+2sinxcosx=cos2x+sin2x=√2(√2/2cos2x+√2/2sin2x)=√2sin(2x+π/4)所以可得此函数的值域为：[-√2,

y=1+sin(2x)+2cos^2(x)=1+sin(2x)+1+cos(2x)=2+sin(2x)+cos(2x)=2+√2sin(2x+π/4)所以周期为π2.-π/2+2kπ

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

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

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=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

y=sin²x+2sinxcosx-3cos²x=(sin²x+cos²x)+2sinxcosx-4cos²x=1+sin(2x)-2[1+cos(2

y=-(cos²x-sin²x）-√3sinxcosx=-cos2x-(√3/2)sin2x=-sin(2x+π/6)或者y=-(cos²x-sin²x）-√3

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

最小正周期是π

y=cos²x+2√3sinxcosx-sin²x=cos²x-sin²x+2√3sinxcosx=cos2x+2√3sinxcosx=cos2x+√3sin2

y=cos2x+sin2x=根号2*sin(2x+pai/4)故y(min)=-根号2

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】

y=7-4sin2x+4cos2x=7+4√2cos(2x+π/4）,y的最小值=7-4√2.