Question

在△ABC中，a.b.c分别是角A.B.C的对边，设a+c=2b,A-c=π/3求sinB的值?

Answer 1

因为
a
+
c
=
2b
由正弦定理，知：
sinA
+sinC
=
2sinB
2sin[(A+C)/2]
*
cos[(A-C)/2]
=
2sinB
sin[(A+C)/2]
*
cos(pi/6)
=
sinB
因为A
+
B
+
C
=
180
所以：(A+C)/2
=
pi/2
-
B/2
所以：cos(B/2)
*
√3/2
=
2sin(B/2)cos(B/2)
显然B/2不等于pi/2，cos(B/2)不等于0
所以：
sin(B/2)
=
√3/4
cos(B/2)
=
√13/4
sinB
=
2sin(B/2)cos(B/2)
=
√39/8