Question

在三角形ABC中，a+c=2b，A-C=60度，则sinB=_____

我要过程！

Answer 1

则由正弦定理，得：a=2RsinA,b=2RsinB,c=2RsinC,代入a+c=2b，得： 2RsinA+2RsinC=4RsinB,即sinA+sinC=2sinB
2sin[(A+C)/2] * cos[(A-C)/2] = 2 * 2 * sin(B/2) * cos(B/2)
sin[(п-B)/2] * cos30 = 2 * sin(B/2) * cos(B/2)
cos[B/2] * cos30 = 2 * sin(B/2) * cos(B/2)
因为0度<B/2<90度，所以cos(B/2)非零
cos30 = 2 * sin(B/2)
sin(B/2) = 根3 / 4
则cos(B/2)=√（1-sin^2(B/2)=√13/4
则sinB=2sin(B/2)cos(B/2)=√39/8

Answer 2

解：设三角形ABC的外接圆为R，则由正弦定理，得：a=2RsinA,b=2RsinB,c=2RsinC,将其代入a+c=2b，得：
2R(sinA+sinC)=2RsinB,即sinA+sinC=sinB,由和化积公式，得：
sinA+sinC=2sin[(A+C)/2]*cos[(A-C)/2]=2sinB
=2sin[(pi-B)/2]*cos[(A-C)/2]=2sinB
=2[cos(B/2)]*(cos30度）=sinB=2[sin(B/2)][cos(B/2)]…（1）
因为0度<B/2<90度，所以cos(B/2)非零
所以 2cos30度=2sin(B/2),所以sin(B/2)=根号3/2
所以B/2=arcsin(根号3/2）=60度，所以B=120度.

Answer 3

解：∵a+c=2b，A-C=60°
由正弦定理：
a:b:c=sinA:sinB:sinC
=>
sinA + sinC
=2sin((A-C)/2)*cos((A+C)/2)
=cos((π-B)/2)
=sin(B/2)
=sinB
=2sin(B/2)cos(B/2)
=>
cos(B/2)=1/2
B/2=π/3
B=2π/3
=>
即 sinB=根号3/2