sinx cosx=sin(x 派 4)

f(x)=sin²x+sinxcosx=[1-cos(2x)]/2+sin(2x)/2=sin(2x)/2-cos(2x)/2+1/2=(√2/2)sin(2x-π/4)+1/2最小正周期T

f(x)=√3sin²x+sinxcosx=√3[(1-cos2x)/2]+1/2sin2x=1/2sin2x-√3/2cos2x+√3/2=sin(2x-π/3)+√3/2∵x∈[π/2,

（2sinx-cosx）（sinx+cosx）-3（2sinx-cosx）=0（2sinx-cosx）（sinx+cosx-3）=0因为（sinx+cosx-3）

y=sin(2x-π/3)x∈[π/6,π/2]所以2x-π/3∈[0,2π/3]当2x-π/3=0即x=π/6时函数y有最小值为0当2x-π/3=π/2即5π/12时函数y有最大值为1所以值域为【0

sin²x-sinxcosx+2cos²x=1-sinxcosx+cos²x=1-½sin2x+(1+cos2x)/2=3/2+½cos2x-

f(x)=5√3cos^x+√3sin^x-4sinxcosx=√3+2√3（1+cos2x)-2sin2x=3√3+4cos(2x+π/6),π/4

sinx=-5/13x∈(П,3П/2)cosx=√1-(-5/13)^2=-12/13sin(x-П/4)=sinxcosП/4-cosxsinпП/4=√2/2(-5/13+12/13)=7√2/

再帮你一次吧~f(x)=(sinx)^2+sinxcosx=(1-cos2x)/2+sin2x/2=1/2+(sin2x-cos2x)/2=3/4所以sin2x-cos2x=1/2所以-根号2*cos

sinπ(x-1)=sin(πx-π)=sinπxcosπ-cosπxsinπ=-sinπx-0=-sinπx再问：那个。。。从第二步怎么化成第三步再答：不是有公式sin(a-b)=sinacosb-

f(Pai/4)=sinPai/4cosPai/4+(sinPai/4)^2=根号2/2*根号2/2+（根号2/2）^2=1/2+1/2=1

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

fx=2cosxsin（x+π/3）-√3sin^2x+sinxcosx+1=2cosx(√3/2cosx+1/2sinx)-√3sin^2x+sinxcosx+1=√3cos^2x-√3sin^2x

f(x)=(sinx)平方+sinxcosx=1/2(1-cos2x)+1/2sin2x=1/2-1/2cos2x+1/2sin2x=1/2+√2/2(sin2xcosπ/4-cos2xsinπ/4)

答：f(x)=2sin(x-π/3)cosx+sinxcosx+√3(sinx)^2=sin(x-π/3+x)+sin(x-π/3-x)+sinxcosx+(√3/2)(1-cos2x)=sin(2x

sin²x=(1-cos2x)/2sinxcosx=1/2*sin2xsin²X-sinXcosX=1/2-(cos2x+sin2x)/2=1/2-√2/2*sin(2x+45°)

sin(x+π/3)sin(x+11π/6)=1/2则有：sin(x+π/3)sin[(x-π/6)+2π]=1/2sin(x+π/3)sin(x-π/6)=1/2-----(1)由于：sin(x+π

f(x)=sin(π-x)cos(2π-x)/sin(π/2+x)tan(π+x)=sinxcosx/cosxtanx=cosxf（31π/3）=cos(10π+π/3)=cosπ/3=1/2再问：它

sqr(3)是根号3的意思f(x)=[(sqr(3)/2)*(1-cos2x)]+(1/2)*sin2x=(1/2)*sin2x-(sqr(3)/2)*cos2x+sqr(3)/2=sin(2x-π/

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

原式=sinxcosx+cos²x+(sin²x+cos²x)=1/2*sin2x+(1+cos2x)/2+1=1/2(sin2x+cos2x)+3/2=√2/2*sin