–6 sin 58°
________
sin 29° sin 61
sin61 ° =sin(90 ° –29 ° )=cos29 °
Тогда
sin29 ° ·sin61 ° =sin29 ° ·cos29 ° =
по формуле sin2 α =2sin α ·cos α ⇒ sin α ·cos α =(1/2)sin2 α
=(1/2)sin(2·29 ° )=(1/2)sin58 °
[m]\frac{-6sin58^{o}}{sin29^{o}\cdot cos61^{o}}=\frac{-6sin58^{o}}{\frac{1}{2}sin58^{o}}=\frac{-6}{\frac{1}{2}}=-12[/m]
О т в е т. –12