sin\alpha cos\beta = \dfrac{1}{2}( + sin(\alpha-\beta)); cos\alpha cos\beta = \dfrac{1}{2}( + cos(\alpha-\beta)); sin\alpha sin\beta = \dfrac{1}{2}( - cos(\alpha+\beta)).

Задание

Заполни пропуски

\(sin\alpha cos\beta = \dfrac{1}{2}(\) [ ] \(+ sin(\alpha-\beta))\) ;

\(cos\alpha cos\beta = \dfrac{1}{2}(\) [ ] \(+ cos(\alpha-\beta))\) ;

\(sin\alpha sin\beta = \dfrac{1}{2}(\) [ ] \(- cos(\alpha+\beta))\) .