Ответ Автор - hlopushinairina sinα=8/17;π/2<α<π;cos(π/6-α)=cosπ/6·cosα+sinπ/6·sinα;⇒cosπ/6=√3/2;sinπ/6=1/2;cos²α=1-sin²α;⇒cosα=√(1-sin²α);cosα=√(1-64/289)=√(289-64)/289=√225/289=⁺₋15/17;⇒cosα=-15/17;cos(π/6-α)=√3/2·-15/17)+1/2·8/17=-15√3/34+8/34=(8-15√3)/34;