Question

在△ABC中,内角A,B,C所对的边分别为a,b,c,若B=π/4,0<A<π/2,且a^2,b^2,c^2成等差数列，则tanA=

Answer 1

由正弦定理得
a/sinA=b/sinB=c/sinC，又a²、b²、c²成等差数列，因此
2sin²B=sin²A+sin²C
1-cos(2B) =[1-cos(2A)]/2 +[1-cos(2C)]/2
cos(2A)+cos(2C)=2cos(2B)=2cos(π/2)=0
2cos(A+C)cos(A-C)=0
-2cosBcos(A-C)=0
cos(A-C)=0
A-C=π/2(C>0，0<A<π/2，舍去)或A-C=-π/2
C=A+π/2
A=π-B-C=π-π/4-(A+π/2)
2A=π/4
A=π/8
tan(2A)=tan(π/4)=1=2tanA/(1-tan²A)
整理，得
tan²A+2tanA=1
(tanA+1)²=2
tanA=-√2-1(0<A<π/2，tanA>0，舍去)或tanA=√2-1

tanA=√2-1

Answer 2

解：a^2,b^2,c^2成等差数列，设△ABC外接圆半径为R，差值为t，则
a^2=(2R*sinB)^2-t
b^2=(2R*sinB)^2
c^2=(2R*sinB)^2+t
而B=π/4 ∴a^2=2*R^2-t, b^2=2*R^2, c^2=2*R^2+t
cosB=(a^2+c^2-b^2)/(2ac)
∴sqrt(2)/2=2*R^2/(2*sqrt(2*R^2-t)*sqrt(2*R^2+t))
∴t=sqrt(2)*R^2
∴a^2=(2-sqrt(2))*R^2, b^2=2*R^2, c^2=(2+sqrt(2))*R^2
∴cosA=(b^2+c^2-a^2)/(2bc)
=(2+2*sqrt(2))*R^2/(2*R^2*sqrt(2)*sqrt(2+sqrt(2)))
=(1+sqrt(2))/(sqrt(2)*sqrt(2+sqrt(2)))
∴tanA=sqrt(1/(cosA)^2-1)
=sqrt(2*(2+sqrt(2))/(1+sqrt(2))^2-1)
=sqrt(2*(2+sqrt(2))-(1+sqrt(2))^2)/(1+sqrt(2))
=sqrt(4+2*sqrt(2)-1-2*sqrt(2)-2)/(1+sqrt(2))
=1/(1+sqrt(2))
=sqrt(2)-1
此即为所求。