Question

求第二问函数化简详细过程

Answer 1

y=1 - cos2B - 2cosBcos(2π/3 - B)
=1 - cos2B - 2cosB[cos(2π/3)cosB + sin(2π/3)sinB]
=1 - cos2B - 2cosB[(-1/2)cosB + (√3/2)sinB]
=1 - cos2B + cos²B - √3sinBcosB
=1 - cos2B + (1 + cos2B)/2 - (√3/2)sin2B
=1 - cos2B + 1/2 + (1/2)cos2B - (√3/2)sin2B
=3/2 - (1/2)cos2B - (√3/2)sin2B
=3/2 - [(√3/2)sin2B + (1/2)cos2B]
=3/2 - [sin2Bcos(π/6) + cos2Bsin(π/6)]
=3/2 - sin(2B + π/6)