Question

高中数学三角函数化简

Answer 1

consider
sin(A+B)= sinA.cosB + cosA.sinB (1)
sin(A-B)= sinA.cosB - cosA.sinB (2)
(1)+(2)
2sinA.cosB = sin(A+B)+ sin(A-B)
A+B=2π/3 -C (3)
A-B=C (4)
(3)+(4)
A= π/3
(3)-(4)
B=π/3 -C

sin(2π/3 -C ) + sinC
=2sin(π/3).cos(π/3 -C)
=√3.cos(π/3 -C)

Answer 2

见图

Answer 3

原式=sin2π/3cosC-cos2π/3sinC+sin2π/3
=√3/2*cosC+1/2*sinC+sin2π/3
=√3/2*cosC+3/2*sinC
=√3*(√3/2*sinC+1/2*cosC)
=√3*(cosπ/6sinC+sinπ/6cosC)
=√3sin(C+π/6)