已知sin(π-x)-cos(π x)=1-√3 2x是第二象限的脚

f(x)=cos^2x+sinxcosx=(1+cos2x)/2+1/2*sin2x=1/2+1/2(cos2x+sin2x)=√2/2*(√2/2*cos2x+√2/2sin2x)+1/2=√2/2

=1/2sin2x-sin^2x=12/sin2x-(1-2cos2x)/2=1/2sin2x+1/2cos2x-1/2=√2/2sin(2x+π/4)-1/2(1)T=2π/2=π(2)X属于(-π

1f(x)=√3sinπx+cosπx=2（（√3/2）sinπx+（1/2）cosπx）=2sin（πx+π/3）∴最小正周期T=2π/w=2π/π=2值域f（x）∈[-2,2]2-π/2+2kπ＜

-cosxtanx再问：根据上题求f（-31π/3）的值再答：上面的答案可化简为-sinxf（-31π/3）=f（-π/3）=-1/2

2sin(x-π/4)sin(x+π/4)=cos(x-π/4-x-π/4)-cos(x-π/4+x+π/4)=-cos2xf(x)=cos(2x-π/3)-cos2x=cos(2x-π/6-π/6)

f(x)=√2sin（2 x-π/4）+2f(x)max=√2+2；此时X=3π/8令（2 x-π/4）∈【-π/2,π/2】,解得x∈【-π/8,3π/8】,因为x∈(0&nbs

∵f（x）=[-sinx(-sinx)cos(π+x)]／[-2cosxsin(π-x)]=[sin²x(-cosx)]／(-2cosxsinx)=1/2sinx∴最小正周期T=2π∴函数图

因为cos(π/2+x)=-sinx,sin(x-π/2)=sin[π-(x-π/2)]=sin(π/2-x)=cosx,由cos(π/2+x)=sin(x-π/2),得：-sinx=cosx.所以[

f(sin(pai/2-x))=cos[3(pai/2-x)]f(cosx)=cos(3pai/2-3x)f(cospai/9)=cos(3pai/2-pai/3)=-sinpai/3=-根号3/2f

（1）∵f(x)＝3sinπx+cosπx=2(32sinπx+12cosπx)=2sin(πx+π6)，∴函数f（x）的最小正周期T＝2ππ＝2，又∵x∈R，∴−1≤sin(πx+π6)≤1，∴−2

f(x)=cos(-x/2)+sin(π-x/2)=cosx/2+sinx/2f(a)=cos(a/2)+sin(a/2)=（2√10）/5cos(a/2)+sin(a/2)=（2√10）/5平方1+

f(x)=cos(2x-π/3)-(cos^2x-sin^2x)=cos(2x-π/3)-cos2x=2sin(2x-π/6)sinπ/6=sin(2x-π/6)因为y=sinx的单减区间为[π/2+

(^2)x这是什么啊完全看不懂诶.再问：就是（sinx）^2再答：啊啊懂啦再答：

f(x)=cos(2x-π\3)+sin²x-cos²x=1/2cos2x+√3/2sin2x-cos2x=√3/2sin2x-1/2cos2x=-cos(2x+π\3)-1

根据题意可知：sinx>0cosx

原式=(-sinx*sinx)/(cos(1.5π-x)*sin(4.5π+x))=sinx/cosx=tanx=-3/4

sinx=2cosx,sin^2x=4cos^2xsin^2x=4-4sin^2x,sin^2x=4/5(cosx+sinx)/(cosx-sinx)+sin^2x=(1+tanx)/(1-tanx)

依题有2sin2x=sinθ+cosθsinx的平方=sinθ*cosθ又2sin2x=4sinx*cosxsinθ*cosθ=[(sinθ+cosθ)的平方-1]/2所以有sinx的平方=[(4si

f(x)=2cos(x+π/3)[sin(x+π/3)-√3cos(x+π/3)]=4cos(x+π/3)[1/2sin(x+π/3)-√3/2cos(x+π/3)]=4cos(x+π/3)[sin(

解:⑴f(x)=-1/2+sin(π/6-2x)+cos(2x-π/3)+(cosx)^2=-1/2+sinπ/6cos2x-sin2xcosπ/6+cos2xcosπ/3+sin2xsinπ/3+(