Question

已知△ABC中,角A,B,C所对的边分别是a,b,c,且a+c=根号2b

1.求证tanA/2.tanC/2的值
2.求证2/tanB/2=1/tanA+1/tanc

Answer 1

1.由正弦定理知：
a/sinA=b/sinB=c/sinC=2R
a=sinA·2R
b=sinB·2R
c=sinC·2R
而a+c=√2b
即sinA·2R+sinC·2R=√2sinB·2R
∴sinA+sinC=√2sinB
∵π-B=A+c
∴sinB=sin(π-B)=sin(A+C)
根据和差化积公式：sinA+sinC=2sin(A/2+C/2)cos(A/2-C/2)
倍角公式：sin(A+C)=2sin(A/2+C/2)cos(A/2+C/2)
则2sin(A/2+C/2)cos(A/2-C/2)=2√2sin(A/2+C/2)cos(A/2+C/2)
即cos(A/2-C/2)=√2cos(A/2+C/2)
cos(A/2)cos(C/2)+sin(A/2)sin(C/2)=√2[cos(A/2)cos(C/2)-sin(A/2)sin(C/2)]
两边同时除以cos(A/2)cos(C/2)，得：
1+tan(A/2)tan(C/2)=√2[1-tan(A/2)tan(C/2)]
令tan(A/2)tan(C/2)=x
1+x=√2(1-x)
1+x=√2-√2x
√2x+x=√2-1
(√2+1)x=√2-1
x=(√2-1)/(√2+1)
x=3-2√2
即tan(A/2)tan(C/2)=3-2√2

2.∵tanA=tan(A/2+A/2)=2tan(A/2)/[1-tan²(A/2)]
1/tanA=[1-tan²(A/2)]/2tan(A/2)
同理：tanC=tan(C/2+C/2)=2tan(C/2)/[1-tan²(C/2)]
1/tanC=[1-tan²(C/2)]/2tan(C/2)
则1/tanA+1/tanC
=[1-tan²(A/2)]/2tan(A/2)+[1-tan²(C/2)]/2tan(C/2)
=[tan(C/2)-tan(C/2)tan²(A/2)+tan(A/2)-tan(A/2)tan²(C/2)]/[2tan(A/2)tan(C/2)]
=[tan(A/2)+tan(C/2)][1-tan(A/2)tan(C/2)]/[2tan(A/2)tan(C/2)]
=[tan(A/2)+tan(C/2)]×[1-(3-2√2)]/[2×(3-2√2)]
=(√2+1)[tan(A/2)+tan(C/2)]
而B=π-(A+C)
B/2=π/2-(A+C)/2
tan(B/2)=tan[π/2-(A+C)/2]=cot(A/2+C/2)=1/tan(A/2+C/2)
则2/tan(B/2)
=2tan(A/2+C/2)
=2[tan(A/2)+tan(C/2)]/[1-tan(A/2)tan(C/2)]
=2[tan(A/2)+tan(C/2)]/[1-(3-2√2)]
=(√2+1)[tan(A/2)+tan(C/2)]
∴2/tan(B/2)=1/tanA+1/tanC