Question

已知+α∈[0,π]+,且+cosα=-1/3+,则+sin(α-π/(6))=？

Answer 1

已知+α∈[0,π]+,且+cosα=-1/3+,则+sin(α-π/(6))=？
根据三角函数的和差公式，有： sin(α-π/6) = sinαcos(π/6) - cosαsin(π/6) 因为cosα=-1/3，sin(π/6)=1/2，cos(π/6)=√3/2，代入上式得： sin(α-π/6) = sinα*(√3/2) + (1/3)(1/2) 由于α∈[0,π]，而cosα=-1/3<0，因此α在第二象限，sinα>0，所以： sin(α-π/6) = sinα(√3/2) + (1/3)*(1/2) = (√3/2)*√(1-cos^2α) + 1/6 = (√3/2)*√(1-1/9) + 1/6 = (√3/2)*√8/9 + 1/6 = (√6+√2)/6

Answer 2

sinα=根号（1-1/9）=2/3*根号2
sin(α-π/6)=2/3*根号2*根号3/2-(-1/3)*1/2
=根号6/3+1/6

Answer 3

α∈[0,π]
cosα=-1/3 =>sinα =2√2/3
sin(α-π/6)
=sinα.cos(π/6)-cosα.sin(π/6)
=(2√2/3)(√3/2)- (-1/3)(1/2)
=√6/3 + 1/6