1.△ABC中A>B>C,A=2C,b=4,a+c=8,求a,c2.{an}a1=1,a(n+1)=2an+3(n≥1）则an=

Question

1.△ABC中 A>B>C,A=2C,b=4,a+c=8,求a ,c
2.{an} a1=1,a(n+1)=2an+3(n≥1）则an=

Answer 1

1.在△ABC中,已知∠A>∠B>∠C且∠A=2∠C,b=4,a+c=8,求a,c的长.
A=2C
sinB=sin(180-B)=sin(A+C)=sin3C
sinA=sin2C
由正弦定理得
b/sinB=(a+c)/(sinA+sinC)
4/sin3C=8/(sinA+sinC)
2sin3C=sin2C+sinC
之后用3倍2倍角及恒等式可得出8cos^2C-2cosC-3=0
所以cosC=3/4 (cosC=-1/2舍去)
sinC=√7/4
sinA=sin2C=3√7/8
a/c=sinA/sinc=3:2
所以a=24/5 c=16/5
2.{an} a1=1,a(n+1)=2an+3(n≥1）求an
A(n+1)=2An+3^n (两边同除以3^(n+1) )
A(n+1)/3^(n+1)=(2/3)*An/3^n+1/3
A(n+1)/3^(n+1)-1=(2/3)*(An/3^n-1)
[A(n+1)/3^(n+1)-1]/(An/3^n-1)=2/3
所以{An/3^n-1}是以A1/3-1=-2/3为首相q=2/3为公比的等比数列
An/3^n-1=(-2/3)(2/3)^(n-1)=-(2/3)^n
所以An=3^n-2^n