Страница 406 номер 1294, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(\sin\alpha + \sin\beta - \sin\gamma = 4\sin\frac{\alpha}{2}\sin\frac{\beta}{2}\cos\frac{\gamma}{2}\)
\(4\sin\dfrac{\alpha}{2}\sin\dfrac{\beta}{2}\cos\dfrac{\gamma}{2} = 4\sin\dfrac{\alpha}{2}\sin\dfrac{\beta}{2}\cos\left( \dfrac{\pi}{2} - \dfrac{\alpha + \beta}{2} \right) =\)
\(= 4\sin\dfrac{\alpha}{2}\sin\dfrac{\beta}{2}\sin\left( \dfrac{\alpha}{2} + \dfrac{\beta}{2} \right) =\)
\(= 4\sin\dfrac{\alpha}{2}\sin\dfrac{\beta}{2}\left( \sin\dfrac{\alpha}{2}\cos\dfrac{\beta}{2} + \sin\dfrac{\beta}{2}\cos\dfrac{\alpha}{2} \right) =\)
\(= 4\sin^{2}\dfrac{\alpha}{2}\sin\dfrac{\beta}{2}\cos\dfrac{\beta}{2} + 4\sin\dfrac{\alpha}{2}\sin^{2}\dfrac{\beta}{2}\cos\dfrac{\alpha}{2} =\)
\(= \sin\beta\left( 1 - \cos\alpha \right) + \sin\alpha\left( 1 - \cos\beta \right) =\)
\(= \sin\beta - \sin\beta\cos\alpha + \sin\alpha - \sin\alpha\cos\beta =\)
\(= \sin\beta + \sin\alpha - \sin(\alpha + \beta) =\)
\(= \sin\beta + \sin\alpha - \sin\left( \pi - (\alpha + \beta) \right) = \sin\alpha + \sin\beta - \sin\gamma;\)

2) \(\sin{2\alpha} + \sin{2\beta} + \sin{2\gamma} = 4\sin\alpha\sin\beta\sin\gamma\)
\(\sin{2\alpha} + \sin{2\beta} + \sin{2\gamma} = 2\sin(\alpha + \beta)\cos(\alpha - \beta) + \sin{2\gamma} =\)
\(= 2\sin(\pi - \gamma)\cos(\alpha - \beta) + 2\sin\gamma\cos\gamma =\)
\(= 2\sin\gamma\left( \cos(\alpha - \beta) + \cos\gamma \right) =\)
\(= 2\sin\gamma \cdot 2\cos\dfrac{\alpha - \beta + \gamma}{2}\cos\dfrac{\alpha - \beta - \gamma}{2} =\)
\(= 2\sin\gamma \cdot 2\cos\dfrac{\pi - 2\beta}{2}\cos\dfrac{\pi - 2\beta - 2\gamma}{2} =\)
\(= 4\sin\gamma\cos\left( \dfrac{\pi}{2} - \beta \right)\cos\left( \dfrac{\pi}{2} - (\beta + \gamma) \right) =\)
\(= 4\sin\gamma\sin\beta\sin(\pi - \alpha) = 4\sin\gamma\sin\beta\sin\alpha.\)