f(x)=2cos(x π 4)cos(x-π 4)

f(x)=cos(-2x+a)(-π

f(x)=[1+cos2x)/2]/[1+(1-cos2x)/2]=(1+cos2x)/(3-cos2x)=-1+4/(3-cos2x)f'(x)=-4/(2-cos2x)^2*(2-cos2x)'=

由诱导公式f(x)=[6cos(π+x)+5sin^2(π-x)-4]/cos(2π-x)=(-6cosx+5(sinx)^2-4)/cosxf(-x)=[-6cos(-x)+5(sin(-x))^2

积分值=（变量替换x＝pi/2-t)积分（0到pi/2）f(cosx)/(f(sinx)+f(cosx)),两者相加（就是两倍的积分值）,被积函数是1,故积分值是pi/2,因此原积分值是pi/4

1)f(x)=1+cos(2x+π/3)-(1+cos2x)/2=1/2-sin2x根号3/2最小值1/2-根号3/2最小正周期π2）c带入得sinC=根号3/2C=π/3A=π-B-C=2π/3-a

根据公式：COS^2a=(1+COS2a)/2a=(X+π/12)说句不好听的,你还是基本知识没掌握好,不会活用知识.希望你多背多看,看清题,把公式活用.

f(x)=-√2sin(2x+π/4)+6sinxcosx-2cos²x+1=-√2(sin2xcosπ/4+cos2xsinπ/4)+3sin2x-2×(1+cos2x)/2+1=-√2(

不知道,sorry

化简f(x)=cosx(asinx-cosx)+cos（π/2-x）cos(π/2-x).=acosxsinx-cos^2x+sin^2x=a/2sin2x-cos2x2.化简f(x)=4cosxsi

(1)、f(x)=cos2xcos4π/3+sin2xsin4π/3+cos2x+1=-1/2cos2x-根号3/2sin2x+cos2x+1=1/2cos2x-根号3/2sin2x+1=cos(2x

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+

f(x)=cos(π/3+x)cos(π/3-x),=(cosxcosπ/3-sinxsinπ/3)(cosxcosπ/3+sinxsinπ/3)=(1/2cosx-√3/2sinx)(1/2cosx

f(x)=(1+√2*cos(2x-π/4))/sin(π/2-x)=[1+√2*cos(2x-π/4)]/cosx=[1+√2*(cos2xcosπ/4+sin2xsinπ/4)]/cosx=[1+

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

1）利用2倍角公式f(x)=sinx+sin[2(π/4+x/2)]=sinx+sin(π/2+x)=sinx+cosx=√2(√2/2sinx+√2/2cosx)=√2(cosπ/4sinx+sin

f(x)=sin(2x)+√2cos(x-π/4)=sin(2x)+√2[cosxcos(π/4)+sinxsin(π/4)]=sin(2x)+cosx+sinx=sin(2x)+√2sin(x+π/

f(x)=2√3sinxcosx-2cos(x+π/4)cos(x-π/4)=√3sin2x+2sin(x+π/4-π/2)cos(x-π/4)=√3sin2x+sin(2x-π/2)=√3sin2x

∵函数f（x）=2cos（π4-ωx）=cos（ωx-π4）（ω＞0）的最小正周期为π2，∴2πω=π2，ω=2，∴f（x）=cos（2x-π4）．令2kπ≤2x-π4≤2kπ+π，k∈z，求得kπ+

解原题应为f(x)=cosx(asinx-cosx)+cos^2(π/2-x)=acosxsinx-cos^2（x）+cos^2(π/2-x)=(a/2)sin2x-（1+cos2x）/2+（1+co

f（x）=2*（1-cos2（x+π\4））/2-（根号3/2cos2x-1/2sin2x）=1-sin2x-（根号3/2cos2x-1/2sin2x）=1-（1/2sin2x+根号3/2cos2x）