设f(x)=sinXcosX-cos²(X π 4) 求f(x)的单调区间

解（1）f(x)＝32sin2x+1+cos2x2+a＝sin(2x+π6)+a+12，（2分）∴T=π．（4分）由π2+2kπ≤2x+π6≤3π2+2kπ，得π6+kx≤x≤2π3+kπ．故函数f（

f(x)=sin的平方x+根号3sinxcosx=(1-cos2x)/2+√3/2sin2x=1/2+sin2xcosπ/6-cos2xsinπ/6=1/2+sin(2x-π/6)最小正周期=2π/2

利用诱导公式和三角恒定公式来解f(x)=1/2*sin2x+√3cos^2f(x)=1/2*sin2x+√3*(1+cos2x)/2f(x)=1/2*sin2x+√3/2*cos2x+√3/2f(x)

解f(x)=√3cos²x+sinxcosx-√3/2=√3*(1+cos2x)/2+(1/2)sin2x-√3/2=(1/2)sin2x+(√3/2)cos2x=sin(2x+π/3)∴T

f(x)=-根号3cos2X-sin2x=-2(根号3/2cos2x+1/2sin2x)=-2sin(2x+π/3)(1).T=2π/w=π（2）.由X属于[-π/3,π/3]得2x+π/3∈【-π/

利用诱导公式和三角恒定公式来解f(x)=1/2sin2x+√3cos^2=1/2sin2x+√3(1+cos2x)/2=1/2sin2x+√3/2*cos2x+√3/2=sin(2x+π/3)+√3/

（Ⅰ）f(x)＝53sinxcosx+6cos2x+sin2x+32=53sinxcosx+5cos2x+52＝523sin2x+5•1+cos2x2+52=5sin(2x+π6)+5，由π6≤x≤π

（1）f（x）=32sin2x+1+cos2x2+a=sin（2x+π6）+a+12∴T=π由π2+2kπ≤2x+π6≤3π2+2kπ，得π6+kπ≤x≤2π3+kπ故函数f（x）的单调递减区间是[π

f(x)=(12)sin2x＋√3cos²x=(1/2)sin2x＋(√3/2)[1＋cos2x]=(1/2)sin2x＋(√3/2)cos2x＋(√3/2)=sin(2x＋π/3)＋(√3

原式=sin2x/2+cosx^2根号3=1/2sin2x+根号3/2+根号3/2*cos2x=sin（2x+π/3)+根号3/2因为T=2π/ww=2所以T=π所以最小正周期为π

令sinx+cosx=2sin(x+π/4)=t∵0≤x≤π/2,π/4≤x+π/4≤3π/4,∴-√2/2≤sin(x+π/4)≤1即-√2≤t≤2(sinx+cosx)^2=1+2sinxcosx

f(x)=sinxcosx-√3cos(π+x)cosx(x∈R)可化为f(x)=(sin2x)/2+√3((cos2x)/2+1))=(sin2x)/2+(√3cos2x)/2+√3=cosπ/3s

f(x)=sinxcosx-√3COS(π+x)cosx=sinxcosx+√3cos²x=1/2sin2x+√3/2cos2x+√3/2=sin(2x+π/3)+√3/2按向量b=﹙π／4

f(x)=sinxcosx-√3cos(x+π)cosx=(1/2)*sin2x+√3*(cosx)^2=(1/2)*sin2x+(√3/2)*cos2x+√3/2=sin(2x+π/3)+√3/2f

1.f(x)=1+cos2x-√3sin2x-1=2(1/2cos2x-√3/2sin2x)=2sin(π/6-2x).当sin(π/6-2x)=-1,即π/6-2x=2kπ-π/2,x=kπ-5π/

f(x)=2cosxsin(x+π/3)-√3sin²x+sinxcosx=2cosx[sinxcos(π/3)+cosxsin(π/3)]-√3sin²x+sinxcosx=2c

f(x)=-根号3sin^2x+sinxcosx=-√3/2(1-cos2x)+1/2sin2x=√3/2cos2x+1/2sin2x-√3/2=sin(2x+π/3)-√3/2(1)f((23π)/

f(x)=2sinxcosx+cos2x=sin2x+cos2xf(x/2)=sinx+cosx=1/5x∈（0,3π/4）,所以sinx=4/5cosx=-3/5cos2x=1-2(sinx)^2=

答：f(x)=6cos²x-2√3sinxcosx=3*(cos2x+1)-√3sin2x=3cos2x-√3sin2x+3=2√3*[(√3/2)cos2x-(1/2)sin2x]+3=2