大师作文网a=(sinx,1)b=(cosx,1),当X=4分之派时,求a b
来源：学生作业帮助网 编辑：作业帮 时间：2024/09/30 01:43:05
a=(1,-2sinx)b=(cos2x,cosx)f(x)=cos2x-2sinxcosx+a=cos2x-sin2x+a=√2cos(2x+π/4)+a(1).T=2π/2=π(2)0≤X≤π/2
f(x)=a^2+2aba^2=1,a*b=(sinx)^2+√3sinx*cosx=1/2-1/2cos2x+√3/2sin(2x)=1/2+sin(2x-π/6)所以f(x)=1+2(1/2+si
f(x)=ab=2sinx(1+sinx)+(cosx+sinx)(cosx-sinx)=2sinx+2sin^2x+cos^2x-sin^2x=2sinx+sin^2x+cos^2x=2sinx+1
f(x)=a(a+2b)=1+2(sin²x+√3sinxcosx)=1+(1-cos2x)+√3sin2x=2+√3sin2x-cos2x=2+2sin(2x+π/6)∴由2kπ-π/2≤
f(x)=2sinxcosx+2√3sinx^2-√3=sin2x+√3(1-cos2x)-√3=sin2x-√3cos2x+√3-√3f(x)=2sin(2x-∏/3)T=∏2x-∏/3=∏/2+k
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
画图：∴x∈[0,π/2]时 值域为[-2,2]
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
(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
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
(1)a*b=0sin2x-cos2x=0sqr(2)sin(2x-π/4)=0x=π/8+kπ/2,k∈Z(2)f(x)=sqr(2)sin(2x-π/4)x∈(3π/8+kπ,7π/8+kπ),k
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=sinx(6sinx+cosx)+cosx(7sinx-2cosx)=8sinxcosx+6(sinx)^2-2(cosx)^2=4sin2x+6-8(cosx)^2=4sin2x+
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
a*b=sin²x＋sinxcosx=sinx(sinx＋cosx)=(1/2)(sinx＋cosx),所以,sinx＋cosx=0或sinx=1/2.1、若sinx＋cosx=0,则tan