1.已知数列{an}的通项为an=b^n(b>0),推导出Sn
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1.已知数列{an}的通项为an=b^n(b>0),推导出Sn
2.在三角形ABC中,tanA=1/4,tanB=3/5
(1)求角C的大小(2)若三角形ABC的最大边长为根号17,求最小边长.
帮帮忙啊~
怎样求sinA啊?
2.在三角形ABC中,tanA=1/4,tanB=3/5
(1)求角C的大小(2)若三角形ABC的最大边长为根号17,求最小边长.
帮帮忙啊~
怎样求sinA啊?
1.
b=1时,
Sn=1+1+1+……+1=n;
b≠1时,
Sn=a1+a2+a3+……+a(n-1)+an
=b+b^2……+b^(n-1)+b^n
b*Sn=b^2+……+b^(n-1)+b^n+b^(n+1)
两式相减得:
(b-1)*Sn=b^(n+1)-b
Sn=[b^(n+1)-b]/(b-1).
2.
(1)
tanC=tan(π-A-B)
=-tan(A+B)
=-(tanA+tanB)/(1-tanA*tanB)
=-(1/4+3/5)/[1-(1/4)*(3/5)]
=-(5+12)/[20-3]
=-1
C=3π/4
(2)
A+B=π-C=π/4<C,A<π/4
C为最大角,c=√17,
tanA+tanB=1/4-3/5=-7/20<0
tanA<tanB
A<B,
A为最小,a为最小边,
Sin2A=2tanA/[1+(tanA)^2]
=2*(1/4)/[1+(1/4)^2]
=8/17
Cos2A=15/17,(因A<π/4,2A<π/2,所以取+)
2(sinA)^2=1-cos2A=2/17
sinA=1/√17,
SinC=Sin3π/4=√2/2
a/sinA=c/SinC
a=c*sinA/SinC
=√17*(1/√17)/(√2/2)
=√2.
b=1时,
Sn=1+1+1+……+1=n;
b≠1时,
Sn=a1+a2+a3+……+a(n-1)+an
=b+b^2……+b^(n-1)+b^n
b*Sn=b^2+……+b^(n-1)+b^n+b^(n+1)
两式相减得:
(b-1)*Sn=b^(n+1)-b
Sn=[b^(n+1)-b]/(b-1).
2.
(1)
tanC=tan(π-A-B)
=-tan(A+B)
=-(tanA+tanB)/(1-tanA*tanB)
=-(1/4+3/5)/[1-(1/4)*(3/5)]
=-(5+12)/[20-3]
=-1
C=3π/4
(2)
A+B=π-C=π/4<C,A<π/4
C为最大角,c=√17,
tanA+tanB=1/4-3/5=-7/20<0
tanA<tanB
A<B,
A为最小,a为最小边,
Sin2A=2tanA/[1+(tanA)^2]
=2*(1/4)/[1+(1/4)^2]
=8/17
Cos2A=15/17,(因A<π/4,2A<π/2,所以取+)
2(sinA)^2=1-cos2A=2/17
sinA=1/√17,
SinC=Sin3π/4=√2/2
a/sinA=c/SinC
a=c*sinA/SinC
=√17*(1/√17)/(√2/2)
=√2.
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