2√3sin(π-x)sinx-(sinx-cosx)²

由(sinx+cosx)^2=1/25得2sinxcosx＝－24/25,(sinx-cosx)^2=48/25得sinx-cosx=-4√3/5,故sin^3x-cos^3x＝（sinx-cosx）

f(x)=[2sin(x+π/3)+sinx]cosx-根3sin^2x=[sinx+(√3）cosx+sinx]cosx-(√3）（sinx)^2=2sinxcosx+(√3）[(cosx)^2-(

1.f（x）=2cosx*sin(x+π/3)-√3sin^2x+sinx*cosx=2cosx*sin(x+π/3)-2sinx*[(√3/2)sinx-(1/2)cosx]=2cosx*sin(x

(1).f(x)=2cosx*sin(x+π/3)-√3sin^2x+sinx*cosx=2cosx(1/2sinx+√3/2cosx)-√3sin^2x+sinxcosx=2sinxcosx+√3c

5.（1）f(x)=(√3)cos2x+2sinxsin(x+π/2).=√3cos2x+2sinxcosx=sin2x+√3cis2x=2sin(2x+π/3).∴最小正周期T=2π/2=π,f(x

f（x）=2cosx*sin（x+π/3)-√3sinx^2+sinx*cosx=2cosx*(sinxcosπ/3+cosxsinπ/3))-√3sinx^2+sinx*cosx=sinxcosx+

由f(x)＝3sinx+cosx＝2sin(x+π6)⇒f(x)max＝2．故答案为：2

f(x)=2sinx*sin(π/2+x)-2sin^2x+1＝2sinxcosx+cos2x=sin2x+cos2x=√2sin(2x+π/4)因为f(x0/2)=根2/3所以sin(x0+π/4)

f(x)=2cos*sin(x+π/3)-^3sin^2x+sinx*cosx=2cosx(1/2sinx+√3/2cosx)-^3sin^2x+sinx*cosx=sin2x+√3cos2x=2si

(sinx+cosx)/(sinx-cosx)=3sinx+cosx=3sinx-3cosxsinx=2cosxtanx=sinx/cosx=2sinx=2cosx带入恒等式sin²x+co

sinx+cosx=√2(cos45°sinx+sin45°cosx)=√2sin(x+45°)==√2sin(x+π/4)

这类题目的一般解法是先化成asinx+bcosx=0,再化成√(a^2+b^2)sin(x+φ)=0,即可求出解集.

1.f(x)=2cosx*sin(x+π/3)-√3﹙sinx﹚^2+sinx*cosx=2cosx*﹙sinxcosπ/3+cosxsinπ/3﹚-√3﹙sinx﹚^2+sinx*cosx=cosx

这个简单：f(x)=2cosx(sinxcos(pi/3)+cosxsin(pi/3))-根号33sin^2x+sinx*cosx=2sinxcosx+根号3cos2x=2sin(x+pi/3)所以：

y=2sinx*cos(3π/2+x)+√3cosx*sin(π+x)+sin(π/2+x)*cosx=2sinx*sinx-√3cosx*sinx+cosx*cosx=1+(sinx)^2-√3co

f(x)=√3sinx+sin(π/2+x)=√3sinx+cosx=2sin(x+π/6)∴最大值是2

f(x)=sinx+sin(x+2π/3)=simx+simxcos2π/3+cosxsin2π/3=sinx-1/2sinx+√3/2cosx=1/2sinx+√3/2cosx=sin(x+π/3)

f(x)=[2(sinx*1/2+cosx*√3/2)+sinx]cosx-√3sin²x=(2sinx+√3cosx)cosx-√3sin²x=2sinxcosx+√3(cos&

先化简原式,得到f(x)=2sin(2x+π/6)你的那个式子应该错了,应该是f(A)=2吧这样得到角A=π/6向量AB*向量AC=边AB*边AC*cosA这样得到AB*AC=2再利用不等式[(AB)

(1)当x∈（0,π/2）时：y=|sin(x+π/2)|-|sinx|=sin（x+π/2）-sinxy′=cos（x+π/2）-cosx（2）当x∈（π/2,π）时：y=-sin（x+π/2）-s