Question

数学三角题

三角形ABC中，设a+c=2b,∠A-∠C=π／3 求sinB的值

Answer 1

sinA+sinC=2sinB
2sin(A+C/2)cos(A-C/2)=4sinB/2cosB/2
cos(B/2)*根号3/2=2sinB/2cosB/2
sinB/2=根号3/4
cosB/2=根号13/4
sinB=2sinB/2cosB/2=根号39/8

Answer 2

好难啊。。。

Answer 3

由正弦定理,sinA+sinC=2sinB，左边和差化积得2sin((A+C)/2)cos((A-C)/2)=2sinB。cos((A-C)/2)=cosπ／6=（根号3）/2，A+C=π-B,得到（根号3）/2*cos(B/2)=2sin(B/2)cos(B/2),从而,sin(B/2)==（根号3）/4,进而求得sinB

Answer 4

用正弦定理得sinA+sinC=2sinB
计算可知A=120°-B/2,C=60°-B/2
sin(120-B/2)+sin(60-B/2)=2sinB
展开化简
√3cos(B/2)=4sin(B/2)cos(B/2)
cosB/2不可能=0
∴sin(B/2)=√3/4
cos(B/2)=√13/4
sinB=2sin(B/2)*cos(B/2)=√39/8