Question

若sina+根号3cosa=2sin(a+b)，且0＜b＜π，求b的值

Answer 1

答案为b=π/3原式可化为：2(sinacosb+cosasinb)=sina+√3cosa将等号两边各除以2可得：sinacosb+cosasinb=1/2sina+√3/2cosa等式两边比较可得：cosb=1/2;sinb=√3/2;又0<b<π,可得：b=π/3

Answer 2

sinA+√3cosA=2sin(A+B)
2[sinA.cos(π/3)+cosA.sin(π/3)]=2sin(A+B)
2sin(A+π/3) =2sin(A+B)
A+π/3=A+B
B=π/3