Страница 148 номер 492, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(\frac{\sin(\alpha + \beta)}{\sin(\alpha - \beta)} = \frac{\operatorname{tg} \alpha + \operatorname{tg} \beta}{\operatorname{tg} \alpha - \operatorname{tg} \beta}\)
\(\dfrac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\sin \alpha \cos \beta - \cos \alpha \sin \beta} = \dfrac{\operatorname{tg} \alpha + \operatorname{tg} \beta}{\operatorname{tg} \alpha - \operatorname{tg} \beta},\)
\(\dfrac{\frac{\sin \alpha \cos \beta}{\cos \alpha \cos \beta} + \frac{\cos \alpha \sin \beta}{\cos \alpha \cos \beta}}{\dfrac{\sin \alpha \cos \beta}{\cos \alpha \cos \beta} - \dfrac{\cos \alpha \sin \beta}{\cos \alpha \cos \beta}} = \dfrac{\operatorname{tg} \alpha + \operatorname{tg} \beta}{\operatorname{tg} \alpha - \operatorname{tg} \beta},\)
\(\dfrac{\operatorname{tg} \alpha + \operatorname{tg} \beta}{\operatorname{tg} \alpha - \operatorname{tg} \beta} = \dfrac{\operatorname{tg} \alpha + \operatorname{tg} \beta}{\operatorname{tg} \alpha - \operatorname{tg} \beta};\)

2) \(\frac{\cos(\alpha - \beta)}{\cos(\alpha + \beta)} = \frac{\operatorname{ctg} \alpha \cdot \operatorname{ctg} \beta + 1}{\operatorname{ctg} \alpha \cdot \operatorname{ctg} \beta - 1}\)
\(\dfrac{\cos \alpha \cos \beta + \sin \alpha \sin \beta}{\cos \alpha \cos \beta - \sin \alpha \sin \beta} = \dfrac{\operatorname{ctg} \alpha \cdot \operatorname{ctg} \beta + 1}{\operatorname{ctg} \alpha \cdot \operatorname{ctg} \beta - 1},\)
\(\dfrac{\frac{\cos \alpha \cos \beta}{\sin \alpha \sin \beta} + \frac{\sin \alpha \sin \beta}{\sin \alpha \sin \beta}}{\dfrac{\cos \alpha \cos \beta}{\sin \alpha \sin \beta} - \dfrac{\sin \alpha \sin \beta}{\sin \alpha \sin \beta}} = \dfrac{\operatorname{ctg} \alpha \cdot \operatorname{ctg} \beta + 1}{\operatorname{ctg} \alpha \cdot \operatorname{ctg} \beta - 1},\)
\(\dfrac{\operatorname{ctg} \alpha \cdot \operatorname{ctg} \beta + 1}{\operatorname{ctg} \alpha \cdot \operatorname{ctg} \beta - 1} = \dfrac{\operatorname{ctg} \alpha \cdot \operatorname{ctg} \beta + 1}{\operatorname{ctg} \alpha \cdot \operatorname{ctg} \beta - 1};\)

3) \(\cos\left(\frac{\pi}{4} + \alpha\right) = \frac{\sqrt{2}}{2} (\cos \alpha - \sin \alpha)\)
\(\cos \dfrac{\pi}{4} \cos \alpha - \sin \dfrac{\pi}{4} \sin \alpha = \dfrac{\sqrt{2}}{2} (\cos \alpha - \sin \alpha),\)
\(\dfrac{\sqrt{2}}{2} \cos \alpha - \dfrac{\sqrt{2}}{2} \sin \alpha = \dfrac{\sqrt{2}}{2} (\cos \alpha - \sin \alpha);\)

4) \(\frac{\cos(\alpha + \beta)}{\cos \alpha \sin \beta} = \operatorname{ctg} \beta - \operatorname{tg} \alpha\)
\(\dfrac{\cos \alpha \cos \beta - \sin \alpha \sin \beta}{\cos \alpha \sin \beta} = \operatorname{ctg} \beta - \operatorname{tg} \alpha,\)
\(\dfrac{\cos \beta}{\sin \beta} - \dfrac{\sin \alpha}{\cos \alpha} = \operatorname{ctg} \beta - \operatorname{tg} \alpha;\)

5) \(\cos \alpha \cos \beta = \frac{1}{2} (\cos(\alpha + \beta) + \cos(\alpha - \beta))\)
\(\cos \alpha \cos \beta = \dfrac{1}{2} (\cos \alpha \cos \beta - \sin \alpha \sin \beta + \cos \alpha \cos \beta + \sin \alpha \sin \beta),\)
\(\cos \alpha \cos \beta = \dfrac{1}{2} \cdot 2 \cos \alpha \cos \beta;\)

6) \(\sin \alpha \sin \beta = \frac{1}{2} (\cos(\alpha - \beta) - \cos(\alpha + \beta))\)
\(\sin \alpha \sin \beta = \dfrac{1}{2} (\cos \alpha \cos \beta + \sin \alpha \sin \beta - \cos \alpha \cos \beta + \sin \alpha \sin \beta),\)
\(\sin \alpha \sin \beta = \dfrac{1}{2} \cdot 2 \sin \alpha \sin \beta.\)