F(X)=4COSX*SIN(WX

题中显然是d(cosx)=(cosx)′dx=-sinxdx,不要遗忘前面的那一个负号!

f(x+π/2)=|sin(x+π/2)|+|cos(x+π/2)|+sin^4(2x+π)=|cosx|+|sinx|+sin^4(2x)=f(x)所以f(x)是以π/2为周期的周期函数所以在[0,

f(x)=2cos2x+sin^x-4cosx=2(2cos^x-1)+(1-cos^x)-4cosx=3cos^x-4cosx-1f(π/3)=3cos^(π/3)-4cosπ/3-1=3x(1/2

令a=sinx,b=cosx,则a^2+b^2=1分子=a^4+b^4+a^2b^2=a^4+2a^2b^2+b^4-a^2b^2=(a^2+b^2)^2-a^2b^2=1-a^2b^2=1-(sin

∵f(x)=2sin(π-x)cosx=2sinxcosx=sin2x1、最小正周期T=2π/2=π.2、∵-π/6≤x≤π/2∴-π/3≤2x≤π,∴-√3/2≤f(x)≤1,∴最大值1,最小值-√

1.原式=（cos^x+sin^x)(cos^x-sin^x)-2sinxcosx=(cos^x-sin^x)-sin2x=cos2x-sin2x=根号2sin(π/4-2x)T=2π/(π/4)=8

f(X)=2cos2x+sin²x-4cosx=4cos²x-2+1-cos²x-4cosx=3cos²x-4cosx-1

f(x)=cosx-sin²x-cos2x+7/4=cosx-(1-cos²x)-(2cos²x-1)+7/4=-cos²x+cosx+7/4令cosx=t,t

cosx=sinx是取得最大值,m=1再问：能给我详细过程吗？再答：f(x)=2(sin^4x+cos^4x)+m(sinx+cosx)^4=2(sin^2x+cos^2x)^2-4sin^2xcos

f(x)=|sinx+cosx|=|√2（√2/2*sinx+√2/2*cosx|=|√2（sinx*cosπ/4+cosx*sinπ/4）|=|√2sin(x+π/4)|.

f(x)=4cos^2(x)-2+1-cos^2(x)-4cos(x)Cos2a=2Cos^2(a)-1sin^2(α)+cos^2(α)=1=3cos^2(x)-4cos(x)-1

f(x)=4sinx*sin^2（(π+2x）/4)+（cosx+sinx）（cosx-sinx）=4sinx*[1-cos(π/2+x)]/2+cos²x-sin²x=sinx(

f(x)=2cos2x+sin^2x-4cosx=2(cosx)^2+(sinx)^2-4cosx=1+(cosx)^2-4cosx=(cosx-2)^2-3=(cosx-2+√3)(cosx-2-√

f(x)=2sin(派-x)cosx=2sinxcosx=sin2x最小正周期=2pi/2=pi(pi就是“派”）f(-pi/6)=sin(-pi/3)=-(根号3)/2f(pi/2)=sin(pi)

1).cosx≠0.x≠kπ+π/2.k∈Z.∴f(x)的定义域为{x│x≠kπ+π/2.k∈Z}.2)α是第四象限的角,切tanα=-4/3.sinα=-4/5,cosα=3/5.sin2α=2si

2-x2再问：写下过程吧？？！！

(1)cosx≠0x≠Kπ+π/2K∈Z(2)由tanα=-4/3可知:sinα=4/5,cosα=-3/5得：sin2α=2sinα*cosα=-24/25cos2α=1-2(sinα)^2=-7/

和差化积公式.sin(x+y)=sinx*cosy+cosx*siny

f(x)=sinx+cosx=√2（√2/2sinx+√2/2sinxcosx）-------(提一个√2出来)=√2（sinxcosπ/4+cosxsinπ/4）-（把（√2/2分别变为cosπ/4