函数y=5sinXcosX-5根3(cosx)--2

你开玩笑啊?哪里的a?是b吗?由原式得：（1-根号3）*sin2x+b=y（y-b）/（1-根号3）=sin2x因为x属于[0,90度]所以sin2x属于[0,1]由其函数图像可知,当2x=0或=18

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

求函数y=sin²x+3sinxcosx+5cos²x的值域y=1-cos²+3sinxcosx+5cos²x=4cos²x+3sinxcosx+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=sinxcosx+sinx+cosx=1/2(2sinxcosx+1-1)+sinx+cosx=1/2(sinx+cosx)^2-1/2+(sinx+cosx)=1/2[(sinx+cosx)^2

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)+(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=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

解y=5cos²x+3sin²x-2sinxcosx=3cos²x+3sin²x+2cos²x-2sinxcosx=3+2cos²x-2si

f(x)=-1+√3sin2x+2cos2x=-1+√7sin(2x+t)想想,你题目cos2x前面的系数2是不是多余的?

y=3sin²x-2√3sinxcosx+3cos²x+2cos²x=1-√3sin2x+cos2x-1+1=cos2x-√3sin2x+2=2(cosπ/3*cos2x

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

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=(5/2)sin2x-(5√3/2)cos2x=(5/2)sin(π-2x)+(5√3/2)cos(π-2x)=5sin(π-2x+a)tana=(5√3/2)/(5/2)=√3a=π/3y=5s

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