y=sinxcosx,则dy dx

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=sinx+cosx+sinxcosx令sinx+cosx=T,(1)由同角三角函数关系sinxcosx=[（sinx+cosx)^2-(sinx^2+cosx^2)]/2把(1)式代入,得sinx

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

方法一：设tanx/2=t（后面写起来方便）原式=[2t/(1+t^2)]*[(1-t^2)/(1+t^2)]/[1+2t/(1+t^2)+(1-t^2)/(1+t^2)/=2t(1-t^2)/(2t

如图所示：

（1）y=sinxcosx+sinx+cosx令t=sinx+cosx=√2sin(x+∏/4),∴t∈[-√2,√2]则,t^2=(sinx+cosx)^2=1+2sinxcosx则,sinxcos

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

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

令sinx+cosx=√2sin(x+a)=t.t∈[-√2,√2]t²=(sinx+cosx)²=1+2sinxcosxsinxcosx=(t²-1)/2y=[(t&#

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=sinxcosx-1=1/2+sinxcosx-3/2=(1+2sinxcosx)/2-3/2=(sinx+cosx)^2/2-3/2=sin^2(x+π/4)-3/2所以最大值是1-3/2=-1

sinx+cosx=t√2sin(x+∏/4)=t-√2≤t≤√21+2sinxcosx=t²sinxcosx=(t²-1)/2y=1+sinx+cosx+sinxcosx=1+t

设t=sinx+cosx=2sin(x+π4)，则t∈[-2，2]．由（sinx+cosx）2=t2⇒sinxcosx=t2-12．∴y=1+t+t2-12=12（t+1）2．∴ymax=12（2+1

令sinx+cosx=T,1式由同角三角函数关系sinxcosx=[（sinx+cosx)^2-(sinx^2+cosx^2)]/2把1式代入,得sinxcosx=（T^2-1)/2所以y=T+（T^

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+