已知a=9(3sinx,1
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原式=(sin²x+sinxcosx)/(2sin²x+cos²x)1=sin²x+cos²x=(tan²x+tanx)/(2tan
tanx=sinx/cosx=2sinx=2cosx1(2cosx-3sinx)/(sinx+cosx)=(sinx-3sinx)/(sinx+sinx/2)=-2/(3/2)=-4/32sinx+c
f(x)=ab=2sinx(1+sinx)+(cosx+sinx)(cosx-sinx)=2sinx+2sin^2x+cos^2x-sin^2x=2sinx+sin^2x+cos^2x=2sinx+1
a和c的夹角有一个公式.cos()=(a点乘c)比上(a的模xc的模).题中即:(2分之根三,1/2)x(-1,0)比上1(根号下sinx的平方+cosx的平方+根号下(-1)的平方+0,的平方).即
1、先把第一题答案给你.a·b=(sinx)^2+sinxcosx=(1-cox2x)/2+(sin2x)/2=(sin2x-cos2x)/2+1/2=(√2/2)[(√2/2)sin2x-(√2/2
a.c=(sinπ/3,cosπ/3).(-1,0)cosz=-sinπ/3z=5π/6f(x)=αsinx(sinx+cosx)f'(x)=α[cosx(sinx+cosx)+sinx(cosx-s
1、f(x)=2sin²x+2√3sinxcosx=1-cos2x+√3sin2x=2sin(2x-π/6)+1.当x∈[0,π/2]时,f(x)∈[2,3];若f(x)关于直线x=a对称,
分析,f(x)=a*b-√3=2sinx*cosx+2√3sin²x-√3=sin(2x)-√3cos(2x)=2sin(2x-π/3)当,2kπ+π/2≦2x-π/3≦2kπ+3π/2∴k
a*b=2sinxcosx+(2根号3)sin^x=sin2x+根号3-(根号3)cos2x=2sin(2x-π/3)+根号3(|b-c|)^=b^+c^+2bc=cos^x+4sin^x+4cos^
(1)f(x)=a·b=2sin²x+2√3sinxcosx=1-cos2x+√3sin2x=2(√3/2*sin2x-1/2*cos2x)+1=2sin(2x-π/6)+1由2kπ-π/2
sinX+cosX=1/3sin^2x+cos^2x=1(sinx+cosx)^2=1+2sinxcosx=1/92sinx*cosx=sin2x=-8/9
1.f(x)=2(√3sinxcosx+(cosx)^2)+2m-1=√3sin2x+cos2x+2m=2sin(2x+pi/6)+2m最小正周期=pi2.x属于[0,pi/2]f(x)最小值=2si
f(x)=向量a.向量b.=(1+sin2x)*1+(sinx-cosx)*(sinx+cosx).=1+sin2x-(cos^2x-sin^2x).=1+sin2x-cos2x.=1+√2sin(2
已知公式:sinx*sinx+cosx*cosx=1(1);由原式知cosx=3sinx-1,两边平方得:cosx*cosx=9sinx*sinx-6sinx+1,代入(1)中,得10sinx*sin
[1].cosx+1/3=1,cosx=2/3,x=arxcos2/3.[2].cos^2x=4/9,sin^2x=5/9,sinx=√5/3.cos(2x+π/3)=cos2xcosπ/3-sin2
1.当a‖b时,-sinx-1.5cosx=0,即tanx=-1.5所以sin2x=2tanx/1+tan^2x=12/13cos2x=1-tan^2x/1+tan^2x=-5/13所以2cos^2x
f(x)=a·b=sin²x-√3sinxcosx²=1/2-(cos2x+√3sin2x)/2=1/2-sin(2x+π/6)单调递增区间2x+π/6∈[(2n+1/2)π,(2
向量a=(√3cosx,cosx),b=(0,sinx),c=(sinx,cosx),d=(sinx,sinx)(1)c•d=sin²x+sinxcosx=(1-cos2x)/2
sinx=2sinx/2cosx/2=1/3sinx/2cosx/2=1/6(sinx/2+cosx/2)^2=1+2sinx/2cosx/2=4/32π