已知cos(x 四分之pAI=五分之三)

f(x)=2sin(x+π/6)-2cosx=2sinxcosπ/6+2cosxsinπ/6-2cosx=√3sinx+cosx-2cosx=√3sinx-cosx,若sinx=4/5,x∈[π/2,

1）因为π

f（x）可化为=1/2+根号3*sin(2x+pai/2)(2x+pai/2)=π/2时有最大值1/2+根号3=1/2+根号3*sin(2x+pai/3)然后(2x+pai/3)=pai/2然后sin

①cos(x-Π/4)=√2/10→√2/2(cosx+sinx)=√2/10→cosx+sinx=1/5又因为sinx^2+cosx^2=1又x的范围（Π/2,3Π/4）得sinx的范围（√2/2,

先化简要求的式子,已知的等式左边将tan化成sin/cos,在化简试试,根据要求的式子来算,不要求tan,肯定有简便算法,我没算,看了看,你自己试试吧

f(x)=sin2x+2sin（π/4-x）cos（π/4-x）=sin2x+cos2x=√2sin（2x+π/4）所以T=2π/2=πx属于[-π/12,π/2]得到2x+π/4属于[π/12,5π

f(x)=cos(2pai-x)cos(pai/2-x)-sin^2x=cosxsinx-sin²x=(1/2)sin2x-(1-cos2x)/2=(1/2)(sin2x+cos2x)-1/

f(x)=sin(π-x)cos(3π/2+x)+sin(π+x)sin(3π/2-x)=(sinx)(sinx)+(-sinx)(-cosx)=sinx(sinx+cosx)f'(x)=cosx(s

cos(2(x-pai/4))=2(cos(x-pai/4))^2-1=2×(√2/10)^2-1=-24/25cos(2(x-/4))=cos(2x-pai/2)=sin2xsin2

f(x)=（1/根号2）sin(2w+pi/4)+1+2所以w=1,最小值是1,x=0时

f（x）=sin（πx/4-π/6）-2cos²πx/8+1={sinπx/4*cosπ/6-cosπx/4*sinπ/6}-{2cos²πx/8-1}={sinπx/4*cosπ

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

(1)f(x)=[cos(x-π/6)]^2-(sinx)^2f(π/12)=(cos(π/12))^2-(sin(π/12))^2=cos(π/6)=√3/2(2)f(x)=[cos(x-π/6)]

cos(x-π/6)+sinx=4√3/5cosxcosπ/6+sinxsinπ/6+sinx=4√3/5(√3/2)cosx+(3/2)sinx=4√3/5(1/2)cosx+(√3/2)sinx=

因为sin(pai/4+x)=sin[pai/2-(pai/4-x)](诱导公式)=cos(pai/4-x)(cosx是偶函数)=cos(x-pai/4)=根号2/10由x属于[pai/2,3pai/

f（x）=sin(2x+pai/6)+cos(2x-pai/3)=sin(2x+pai/6)+sin(2x-pai/3+pi/2)=2sin(2x+pai/6)当2x+pai/6=pi/2+2kpi时

f(x)=(√3/2)sin2x－cos²x.=(√3/2)sin2x－(1/2)cos2x－(1/2).=sin(2x－π/6)－(1/2)则函数f(x)的最大值是1/2,最小值是－3/2

f(x)=sin(2x+π/6)+sin(2x-π/6)+2cos²x=sin2xcosπ/6+sinπ/6cos2x+sin2xcosπ/6-sinπ/6cos2x+2cos²x

∵cosx=-4/5,x∈(π/2,π)∴sinx>0sinx=√[1-(cosx)^2]=3/5∴tanx=sinx/cosx=-3/4∴tan(π/4-x)=(tanπ/4-tanx)/(1+ta