已知复数z1=sin2x+ti,z2=m+(m-根号3cos2x)i (i为虚数单位,t,m,x∈R),且z1=z2.⑴若t=0且0〈X〈π ,求 X的值 (2)设t=f(x),已知当x=a时,t=1/2,试求cos(4a+π/3)的值.(a—角)
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![已知复数z1=sin2x+ti,z2=m+(m-根号3cos2x)i (i为虚数单位,t,m,x∈R),且z1=z2.⑴若t=0且0〈X〈π ,求 X的值 (2)设t=f(x),已知当x=a时,t=1/2,试求cos(4a+π/3)的值.(a—角)](/uploads/image/z/5274713-65-3.jpg?t=%E5%B7%B2%E7%9F%A5%E5%A4%8D%E6%95%B0z1%3Dsin2x%2Bti%2Cz2%3Dm%2B%28m-%E6%A0%B9%E5%8F%B73cos2x%29i+%28i%E4%B8%BA%E8%99%9A%E6%95%B0%E5%8D%95%E4%BD%8D%2Ct%2Cm%2Cx%E2%88%88R%EF%BC%89%2C%E4%B8%94z1%3Dz2.%E2%91%B4%E8%8B%A5t%3D0%E4%B8%940%E3%80%88X%E3%80%88%CF%80+%2C%E6%B1%82+X%E7%9A%84%E5%80%BC+%282%29%E8%AE%BEt%3Df%28x%29%2C%E5%B7%B2%E7%9F%A5%E5%BD%93x%3Da%E6%97%B6%2Ct%3D1%2F2%2C%E8%AF%95%E6%B1%82cos%284a%2B%CF%80%2F3%EF%BC%89%E7%9A%84%E5%80%BC.%EF%BC%88a%E2%80%94%E8%A7%92%EF%BC%89)
已知复数z1=sin2x+ti,z2=m+(m-根号3cos2x)i (i为虚数单位,t,m,x∈R),且z1=z2.⑴若t=0且0〈X〈π ,求 X的值 (2)设t=f(x),已知当x=a时,t=1/2,试求cos(4a+π/3)的值.(a—角)
已知复数z1=sin2x+ti,z2=m+(m-根号3cos2x)i (i为虚数单位,t,m,x∈R),且z1=z2.
⑴若t=0且0〈X〈π ,求 X的值
(2)设t=f(x),已知当x=a时,t=1/2,试求cos(4a+π/3)的值.(a—角)
已知复数z1=sin2x+ti,z2=m+(m-根号3cos2x)i (i为虚数单位,t,m,x∈R),且z1=z2.⑴若t=0且0〈X〈π ,求 X的值 (2)设t=f(x),已知当x=a时,t=1/2,试求cos(4a+π/3)的值.(a—角)
第一个问题:
∵z1=z2,
∴m=sin2x,m-√3cos2x=t.联立两式消去m,得:sin2x-√3cos2x=t,而t=0,
∴2[(1/2)sin2x-(√3/2)cos2x]=0,∴sin2xcos(π/3)-cos2xsin(π/3)=0,
得:sin(2x-π/3)=0.
∵0<x<π,∴0<2x<2π,∴-π/3<2x-π/3<2π/3,∴2x-π/3=0,得:x=π/6.
第二个问题:
f(x)=t=m-√3cos2x=sin2x-√3cos2x.
∵f(a)=1/2,∴sin2a-√3cos2a=1/2,∴2[(1/2)sin2a-(√3/2)cos2a]=1/2,
∴sin(2a-π/3)=1/4,
而sin(2a-π/3)=-cos[π/2+(2a-π/3)]=-cos(2a+π/6),
∴cos(2a+π/6)=-1/4,
得:cos(4a+π/3)=2[cos(2a+π/6)]^2-1=2×(-1/4)^2-1=-7/8.
∵z1=z2
∴sin2x=m,t=m-√3cos2x
√3cos2x=m-t
(1)
当t=0时
sin2x=m,√3cos2x=m
想减得
sin2x-√3cos2x=0
2sin(2x-π/3)=0
2x-π/3=kπ(k∈Z)
x=π/6+kπ/2
∵0<x<π
∴x=π/6或2π/3
...
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∵z1=z2
∴sin2x=m,t=m-√3cos2x
√3cos2x=m-t
(1)
当t=0时
sin2x=m,√3cos2x=m
想减得
sin2x-√3cos2x=0
2sin(2x-π/3)=0
2x-π/3=kπ(k∈Z)
x=π/6+kπ/2
∵0<x<π
∴x=π/6或2π/3
(2)
当t=f(x)时
即sin2x-√3cos2x=f(x)
2sin(2x-π/3)=1/2
sin(2x-π/3)=1/4
sin(2x-π/3+π/2-π/2)=1/4
sin(2x+π/6-π/2)=1/4
cos(2x+π/6)=-1/4
cos(4a+π/3)=cos2(2a+π/6)=2cos^2(2a+π/6)-1=-7/8
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