Question

设△ABC的三内角A,B,C所对的长分别为a,b,c,向量m=(a，b).向量n=(b,c),

若向量m·向量n=2b^2，，且A-C=π/3，求cosB的值

Answer 1

解：
由题意，向量
m*n
=
(a,b)*(b,c)
=
ab+bc
=
2b²
==>
b(a+c-2b)
=0
三角形ABC中，显然边长
b≠0；
因此：a+c
=
2b
由正弦定理a/sinA
=
b/sinB
=c/sinC
=2R,可得
sinA
+
sinC
=2sinB；
==>
2sin[(A+C)/2]cos[(A-C)/2]
=
4sin(B/2)cos(B/2);
==>
2sin(π/2
-B/2)
cos(π/6)
=4sin(B/2)cos(B/2);
==>
cos(B/2)[4sin(B/2)
-
√3]=0;
三角形ABC
==>
0<
B
<π
==>
cos(B/2)
>0
∴
4sin(B/2)
-
√3
=
0
==>
sin(B/2)
=
√3/4
∴
cosB
=
1
-
2sin²(B/2)
=
5/8；