Question

高中三角函数变换问题

（2√3）sinB+（2√3）sin(2π/3 -B)如何化为3cosB+(3√3)sinB

Answer 1

（2√3）sinB+（2√3）sin(2π/3 -B)
=2√3(sinB+sin(2π/3 -B))
=2√3(sinB-sin(-π/3-B))
=2√3(sinB+sin(π/3+B))
=2√3(sinB+sinπ/3cosB+cosπ/3sinB)
=2√3(sinB+√3/2cosB+1/2sinB)
=2√3(3/2sinB+√3/2cosB)
=3cosB+(3√3)sinB

Answer 2

因为 sin(2π/3 -B)=sin(π/3 +B)
故（2√3）sinB+（2√3）sin(2π/3 -B)=（2√3）sinB+（2√3）sin(B+π/3 )
=2√3sinB+√3sinB+3cosB
=3cosB+(3√3)sinB

Answer 3

（2√3）sinB+（2√3）sin(2π/3 -B)=（2√3）sinB+（2√3）(sin(2π/3)cosB-cos(2π/3)sinB)=（2√3）sinB+（2√3）((√3/2)cosB+(1/2)sinB)=3cosB+(3√3)sinB