Question

已知△ABC的三个内角A，B，C的对边分别为a,b,c,若a,b,c成等差数列，且2cos2B-8cosB+5=0,求角B的大小并判断△ABC的形状.

Answer 1

解法一：∵2cos2B-8cosB+5=0,?

∴2(2cos²B-1)-8cosB+5=0.                                                                                      ?

∴4cos²B-8cosB+3=0,即(2cosB-1)(2cosB-3)=0.??

解得cosB=或cosB=（舍去），∴cosB=.                                                      ?

∵0＜B＜π,∴B=.                                                                                                ?

∵a,B,c成等差数列，∴a+c=2B.                                                                       ?

∴cosB===,                                                      ?

化简得a²+c²-2AC=0,解得a=c.                                                                                 ?

∴△ABC是等边三角形.                                                                                          ?

解法二：∵2cos2B-8cosB+5=0,?

∴2(2cos²B-1)-8cosB+5=0.                                                                                      ?

∴4cos²B-8cosB+3=0,即(2cosB-1)(2cosB-3)=0.

解得cosB=或cosB=(舍去),∴cosB=.                                                           ?

∵0＜B＜π,∴B=.                                                                                                ?

∵a,B,c成等差数列,∴a+c=2B.                                                                                ?

由正弦定理得sinA+sinC=2sinB=2sin=,                                                    ?

∴sinA+sin(-A)=.?

∴sinA+sincosA-cossinA=.?

化简得sinA+cosA=,∴sin(A+)=1.?

∵0＜A＜π,∴A+=.?

∴A=,C=.                                                                                                        ?

∴△ABC是等边三角形.