2√3sin(π-x)sinx-(sinx-cosx)²
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/27 19:26:06
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