f(x)=4sin(x-3分之π)cosx 根号3求最小

f(x)=2cos2x+sin平方x－4cosx=3cos^2x-4cosx-1=3(cosx-2/3)^2-7/3(1)f(π/3)=3*(cosπ/3)^2-4cosπ/3-1=3*(1/2)^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)

(1)首先利用降幂公式：f(x)=1-cos(π/2+2x)-根号3cos2x再利用诱导公式：f(x)=1+sin2x-√3cos2x最后是辅助角公式:f(x)=2*(1/2*sin2x-√3/2*c

f(x)的定义域为全体实数另g(x)=sin(2x+3π/2)f(x)与g(x)奇偶性一致（f(x)/g(x)=根号2>0）g(x)=sin(2x+3π/2)=-cos(2x)g(-x)=-cos(-

已知函数f(x)=2倍根号3sin(x-4分之π)cos(x-4分之π)-sin(2x-π)（1）求函数f（x）的单调递减区间2）试试画出函数f（x）在0到π上的图像(1)解析：∵函数f(x)=2倍根

f(x)=1/2*sin2x+√3/2*cos2x-√3/2=√(1/4+3/4)*sin[2x+arctan√3]-√3/2=2sin(2x+π/3)-√3/2所以T=2π/2=π0

f（x）=sin（4x+π/4）+cos（4x-π/4）=√2/2sin4x+√2/2cos4x+√2/2cos4x+√2/2sin4x=√2sin4x+√2cos4x=2sin（4x+π/4）这题没

解；f(x)=sinx+sinxcosπ/3+cosxsinπ/3=sinx+1/2sinx+√3/2cosx=3/2sinx+√3/2cosx=√3sin(x+π/6)当x+π/6=-π/2+2kπ

f(x)=cos(3x)*cos(2x)+sin(3x)*sin(2x)=cos(3x-2x)=cosxf'(x)=-sinx

f(x)=sin(x+3π/2)sin(x-2π)=-cosxsinx=-1/2sin2x最大值1/2最小值-1/2最小正周期2π/2=πf(π/6)=-1/2sinπ/3=-√3/4f(π/12)=

经济将积极

hello!^-^令u=2x-π/6,则f（x）=sin(2x-π/6)=sinu=fu).因为-π/12≤x≤π/2,所以-π/3≤2x-π/6≤5π/6,即-π/3≤u≤5π/6.所以根据函数图像

最大值:f(π/8)=(√2)sin(π/4+π/4)=(√2)sin(π/2)=√2最小值:f(-π/4)=(√2)sin(-π/2+π/4)=(√2)sin(-π/4)=√2*(-√2/2)=-1

f(x)=sin^2x+2√3sinxcosx+sin(x+π/4)sin(x-π/4)=(1-cos2x)/2+√3sin2x+(1/2)2sin(x-π/4)cos(x-π/4)=2-2cos2x

f(x)=sin(4x+π/3）+cos(4x-π/6)=sin(4x+π-π/6）+cos(4x-π/6)=sin(π+4x-π/6）+cos(4x-π/6)=-sin(4x-π/6）+cos(4x

f(x)=1/4(-cos2x-√3/2)+√3/2*（√2/2*sinx-√2/2*cosx)²=-1/4*cos2x-√3/8+√3/2*1/2（1-2sinxcosx）=-1/2*si

解:f(x)=1+√2sin(2x-π/4).1.函数f(x)的最小正周期为T=2π/2=π.当思念(2x-π/4)=1时,函数f(x)具有最大值,且f(x)min=1+√2.2.函数的增区间:∵si

答：0

一.求函数的定义域：1)、f(x)=√[lg(4-x)]；2)、f(x)=1/lg(2-x).1)、由4-x>0得x再问：我已经完全忘光了，lg这个要怎么算?比如lg(4-x)≧0这些内容是什么阶段的