y=3sinxcosx-4cos2x

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

y=7-4sinxcosx+4cos2x-4cos4x=7-2sin2x+4cos2x（1-cos2x）=7-2sin2x+4cos2xsin2x=7-2sin2x+sin22x=（1-sin2x）2

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

求函数y=sin²x+3sinxcosx+5cos²x的值域y=1-cos²+3sinxcosx+5cos²x=4cos²x+3sinxcosx+1=

sinx/cosx=3sinx=3cosx(sinx)^2+(cosx)^2=110(cosx)^2=1(cosx)^2=1/10sinxcosx=3(cosx)^2=3/10再问：10(cosx)^

y=cos2x-sin2x+2sinxcosx=cos2x-2sinxcosx+2sinxcosx=cos2xx∈(0,4/π)2x∈(0,2/π)所以值域是(0,1)

sinxcosx=（sinx+cosx）的平方减1再除以2,然后把sinx+cosx看成整体,再根据平方法化简可得最后结果是(sinx+cosx+1)^2/2-1.

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

y=1-4(sinx)^2-2倍根号3sin（2x）=1-2（1-cos2x）-2倍根号3sin（2x）=4sin（π/6-2x）-1当x∈[0,π/2],π/6-2x属于[-5π/6,π/6]所以最

如果说化简应该不对结果应该是常数+sinT或者cosT你的结果还能继续化下去

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

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

-3

Y=4(cos⁡x)^2+4√3sin⁡xcos⁡x-2=2cos⁡2x+4√3sin⁡xcos⁡x=4sin⁡〖(

y=sin²x+√3sinxcosx=(1-cos2x)/2+√3/2sin2x=√3/2sin2x-1/2cos2x+1/2=cosπ/6sin2x-sinπ/6cos2x+1/2=sin

题目是不是错了,应该是：y=sin^4x+2√3sinxcosx-cos^4x吧y=sin^4x+2√3sinxcosx-cos^4x=sin^4x-cos^4x+√3sin2x=(sin²

y=(sinx)^4+2√3sinxcosx-(cosx)^4=(sin^2x+cos^2x)(sin^2x-cos^2x)+√3sin2x=(sin^2x-cos^2x)+√3sin2x=√3sin

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

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

y=sin2x+3sinxcosx+4cos2x=sin2x+(3/2)sin2x+4cos2x=(5/2)sin2x+4cos2x最大值为√[(5/2)^2+4^2]=√89/2最小值为-√89/2