已知sin(x-2π)-cos(π-x)=1-根号3÷2,想为第二象限角,求

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

-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

先用tanx=sinx/cosx、倍角公式、诱导公式化简原函数：f（x）=sin²x+sinxcosx-sin[2（x+π/4）]=（1-cos2x）/2+1/2sin2x-sin（2x+π

(1)f(x)＝sin(2x＋π/6)＋sin(2x－π/6)＋2cos²x＝sin2xcosπ/6＋cos2xsinπ/6＋sin2xcosπ/6－cos2xsinπ/6＋2cos&sup

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)

1.f(x)=sin(2x+6/π)+2sin^2x=(√3/2)sin2x+(1/2)cos2x+1-cos2x=(√3/2)sin2x-(1/2)cos2x+1=sin(2x-π/6）+1所以f(

多一个sin(2x+π/6)吧!f(x)=sin(2x+π/6)+2cos²x=sin2xcosπ/6+cos2xsinπ/6+1+cos2x=√3/2sin2x+3/2cos2x+1=√3

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

f(x)=sin(2x+π/3)+sin(2x-π/3)+2cos^2x-1=sin2xcosπ/3+cos2xsinπ/3+sin2xcosπ/3-cos2xsinπ/3+cos2x=2sin2xc

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(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

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+(