函数sinx cosx在[0,π 2 ]上的定积分等于

f（x）=cos^2x-2sinxcosx-sin^2x=-sin2x+cos2x=-√2*sin(2x-π/4)-π/2

y=√3sinxcosx+cos^2x-1/2=（√3/2）sin2x+（1/2）（cos2x+1）-1/2=（√3/2）sin2x+（1/2）cos2x=sin（2x+π/6）因为x属于[0,π/2

f(x)=根号3sin2x+cos2x=2（cosπ/6sin2x+sinπ/6cos2x）=2sin（2x+π/6）因为函数在区间[0,π/2]上所以π/6≤2x+π/6≤7π/6当2x+π/6=π

y=sinx^4-cosx^4+2*√3*sinxcosx=(sinx^2+cosx^2)(sinx^2-cosx^2)+√3sin2x=√3sin2x+(sinx^2-cosx^2)=√3sin2x

1.求函数y=sinx-cosx+sinxcosxx∈（0,π）的最大值最小值设t=sinx–cosx所以t²=1–2sinxcosx,则sinxcosx=1-t²/2因为t=si

f(x)=-√3sin^2x+sinxcosx=-√3(1-cos2x)/2+sin2x/2=sin2x/2+√3cos2x/2-√3/2=sin(2x+π/3)-√3/2因为X∈[0,π/2],所以

f(x)=1+sin2x+2cos^2x=1+sin2x+1+cos2x=√2sin(2x+π/4)+22kπ-π/2≤2x+π/4≤2kπ+π/2kπ-3π/8≤x≤kπ+π/8[0,π/8]

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

分析:将函数变成只含一个三角函数f(x)=-√3sin²x+sinxcosx=-√3×(1-cos(2x))/2+(sin(2x))/2(利用二倍角公式)=-√3/2+(√3/2)cos(2

令sinx+cosx=t，则sinxcosx=t2−12，∴y=sinxcosx+sinx+cosx=t+t2−12=12t2+t-12=12（t+1）2-1．∵x∈[0，π3]，t=sinx+cos

设sinx-cosx=t,则sinxcosx=(1-t*t)/20

设π/3+2x为t,第二个式子就变成π/3

f(x)=sinxcosx-(sinx+cosx)=0.5[1-(sinx+cosx)]²-1sinx+cosx=根号2sin(x+π/4)函数f(x)在(0,π/2）上的单调递增区间是(π

f(x)=√3sinxcosx-(sinx)^2-3/2=(√3/2)*sin2x-(1-cos2x)/2-3/2=(√3/2)*sin2x+(1/2)*cos2x-2=sin(2x+π/6)-2x∈

F[x]=sinxcosx+cos^2x-1/2=1/2sin2x+1/2(cos2x+1)-1/2=1/2(sin2x+cos2x)=√2/2sin(2x+π/4)最小正周期T=2π/W=π2x+π

设sinx-cosx=t，则（sinx-cosx）2=t2⇒sinxcosx=1−t22，∵x∈[0，π]，∴（x-π4）∈[-π4，3π4]，sin（x-π4）∈[-22，1]，∴t=sinx-co

f(x)=sinxcosx-m(sinx+cosx)=1/2sin2x-m(sinx+cosx)求导得到f’（x）=cos2x+m（sinx-cosx）函数f(x)在区间(0,π/2)是单调减函数所以

f(x)=√3/2*sin2x-(1-cos2x)/2-3/2=√3/2sin2x+1/2*cos2x-2=sin(2x+π/6)-2在[-π/2,0]上,-5π/6=再问：对不起看不清再答：哪步看不

f(x)=√2sin(2x+π/4)=√2sin2(x+π/8),x在（-3π/8,π/8）是增函数,在（π/8,5π/8）时减函数；f(x)在[0,π/2]上的最大值为x=π/8时,此时f(x)=√

f(x)=2√3sinxcosx+2cos²x+m=√3sin2x+1+cos2x+m=2sin(2x+π/6)+m+1.0再问：在三角形ABC中角ABC所对的边长abc若F（A）=1，si