wangmiao_12
幼苗
共回答了227个问题 举报
√3sin2x-cos2x + 1 + m
= 2[(√3)*(1/2)sin2x - (1/2)cos2x] + 1 + m [辅助角公式]
= 2[sin(π/3)sin2x - cos(π/3)cos2x] + 1 + m
= -2cos(2x + π/3) + 1 + m
= -2sin[π/2 - (2x + π/3)] +1 + m
= -2sin[π/6 - 2x] + 1 + m
= 2sin[2x - π/6] + 1 + m
证明完毕。
1年前
2