设α为钝角,若sin(α+π/6)=1/3 则cos(2α+π/12)=?

设α为钝角,若sin(α+π/6)=1/3 则cos(2α+π/12)=?
设α为钝角,若sin(α+π/6)=1/3 则cos(2α+π/12)=?

设α为钝角,若sin(α+π/6)=1/3 则cos(2α+π/12)=?
cos(2α+π/12)
=cos(α+π/6+α+π/6-π/4)
=cos(α+π/6)cos(α+π/6-π/4)-sin(α+π/6)sin(α+π/6-π/4)
=(-2/3)√2cos(α+π/6-π/4)-1/3sin(α+π/6-π/4)
=(-2/3)√2[cos(α+π/6)cosπ/4+sin(α+π/6)sinπ/4]-1/3[sin(α+π/6)cosπ/4-cos(α+π/6)sin(π/4)]
=(-2/3)√2[(-2/3)√2*(√2/2)+1/3*(√2/2)]-1/3[(1/3)*(√2/2)-(-2/3)√2*(√2/2)]
=(7√2-8)/18