求助，第一题怎么做

求助，第一题怎么做

cos(α +π/6) - sinα
=cosα * cos(π/6) - sinα * sin(π/6) - sinα
= (√3/2) * cosα - (1/2) * sinα - sinα
= (√3/2) * cosα - (3/2) * sinα
= (√3) * [(1/2) * cosα - (√3/2) * sinα]
= (√3) * [cos(π/3) * cosα - sin(π/3) * sinα]
= (√3) * cos(α + π/3)
= 4√3 /5
可以得到：
cos(α + π/3) = 4/5
α + π/3 = arccos(4/5)
α = arccos(4/5) - π/3
α + π/12 = arccos(4/5) - π/3 + π/12 = arccos(4/5) - π/4
所以，
sin(α + π/12)
=sin[arccos(4/5) - π/4]
=sin[arccos(4/5)] * cos(π/4) - cos[arccos(4/5)] * sin(π/4)
= 3/5 * √2/2 - 4/5 * √2/2
= (-1/5) * √2/2
= - √2/10
所以，正确的答案应该是 B