Question

在△ABC中，a、b、c分别是角A、B、C的对边，且A=π/3，求[a/(b-c)]*sin(π/3 -C)的值

求解

Answer 1

解：∵A=π/3
∴B+C=2π/3
由正弦定理，可知
a/(b-c)=sinA/(sinB-sinC)
则原式=sinA*sin(π/3-C)/(sinB-sinC)
=sinA*sin(π/3-C)/{2[cos(B+C)/2]*[sin(B-C)/2]}
=sin(π/3)*sin(π/3-C)/{2cos(π/3)*sin[(B-C)/2]}
=(√3/2)*sin(π/3-C)/sin[(B-C)/2]
又B+C=2π/3
∴B-C=2π/3-2C
∴(B-C)/2=π/3-C
∴sin[(B-C)/2]=sin(π/3-C)
∴原式=√3/2