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МатематикаГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 492
Задание 492
\[\boxed{\mathbf{492.}}\]
\[1)\ \frac{\sin(a + \beta)}{\sin(a - \beta)} = \frac{tg\ a + tg\ \beta}{tg\ a - tg\ \beta}\]
\[Преобразуем\ левую\ часть\ \]
\[равенства:\]
\[\frac{\sin(a + \beta)}{\sin(a - \beta)} =\]
\[= \frac{\sin a \bullet \cos\beta + \cos a \bullet \sin\beta}{\sin a \bullet \cos\beta - \cos a \bullet \sin\beta} =\]
\[= \frac{\frac{\sin a \bullet \cos\beta}{\cos a \bullet \cos\beta} + \frac{\cos a \bullet \sin\beta}{\cos a \bullet \cos\beta}}{\frac{\sin a \bullet \cos\beta}{\cos a \bullet \cos\beta} - \frac{\cos a \bullet \sin\beta}{\cos a \bullet \cos\beta}} =\]
\[= \frac{\frac{\sin a}{\cos a} + \frac{\sin\beta}{\cos\beta}}{\frac{\sin a}{\cos a} - \frac{\sin\beta}{\cos\beta}} = \frac{tg\ a + tg\ \beta}{tg\ a - tg\ \beta}\]
\[\frac{tg\ a + tg\ \beta}{tg\ a - tg\ \beta} = \frac{tg\ a + tg\ \beta}{tg\ a - tg\ \beta}\]
\[Что\ и\ требовалось\ доказать.\]
\[2)\ \frac{\cos(a - \beta)}{\cos(a + \beta)} =\]
\[= \frac{ctg\ a \bullet ctg\ \beta + 1}{ctg\ a \bullet ctg\ \beta - 1}\]
\[Преобразуем\ левую\ часть\ \]
\[равенства:\]
\[\frac{\cos(a - \beta)}{\cos(a + \beta)} =\]
\[= \frac{\cos a \bullet \cos\beta + \sin a \bullet \sin\beta}{\cos a \bullet \cos\beta - \sin a \bullet \sin\beta} =\]
\[= \frac{\frac{\cos a \bullet \cos\beta}{\sin a \bullet \sin\beta} + \frac{\sin a \bullet \sin\beta}{\sin a \bullet \sin\beta}}{\frac{\cos a \bullet \cos\beta}{\sin a \bullet \sin\beta} - \frac{\sin a \bullet \sin\beta}{\sin a \bullet \sin\beta}} =\]
\[= \frac{ctg\ a \bullet tg\ \beta + 1}{ctg\ a \bullet tg\ \beta - 1}\]
\[\frac{ctg\ a \bullet tg\ \beta + 1}{ctg\ a \bullet tg\ \beta - 1} =\]
\[= \frac{ctg\ a \bullet tg\ \beta + 1}{ctg\ a \bullet tg\ \beta - 1}\]
\[Что\ и\ требовалось\ доказать.\]
\[3)\cos\left( \frac{\pi}{4} + a \right) =\]
\[= \frac{\sqrt{2}}{2}\left( \cos a - \sin a \right)\]
\[Преобразуем\ левую\ часть\ \]
\[равенства:\]
\[\cos\left( \frac{\pi}{4} + a \right) =\]
\[= \cos\frac{\pi}{4} \bullet \cos a - \sin\frac{\pi}{4} \bullet \sin a =\]
\[= \frac{\sqrt{2}}{2}\cos a - \frac{\sqrt{2}}{2}\sin a =\]
\[= \frac{\sqrt{2}}{2}\left( \cos a - \sin a \right)\]
\[\frac{\sqrt{2}}{2}\left( \cos a - \sin a \right) =\]
\[= \frac{\sqrt{2}}{2}\left( \cos a - \sin a \right)\]
\[Что\ и\ требовалось\ доказать.\]
\[4)\ \frac{\cos(a + \beta)}{\cos a \bullet \sin\beta} = ctg\ \beta - tg\ a\]
\[Преобразуем\ левую\ часть\ \]
\[равенства:\]
\[\frac{\cos(a + \beta)}{\cos a \bullet \sin\beta} =\]
\[= \frac{\cos a \bullet \cos\beta - \sin a \bullet \sin\beta}{\cos a \bullet \sin\beta} =\]
\[= \frac{\cos\beta}{\sin\beta} - \frac{\sin a}{\cos a} = ctg\ \beta - tg\ a\]
\[ctg\ \beta - tg\ a = ctg\ \beta - tg\ a\]
\[Что\ и\ требовалось\ доказать.\]
\[5)\cos a \bullet \cos\beta =\]
\[= \frac{1}{2}\left( \cos(a + \beta) + \cos(a - \beta) \right)\]
\[Преобразуем\ правую\ часть\ \]
\[тождества:\]
\[\frac{1}{2}\left( \cos(a + \beta) + \cos(a - \beta) \right) =\]
\[= \frac{1}{2} \bullet 2\cos a \bullet \cos\beta =\]
\[= \cos a \bullet \cos\beta\]
\[\cos a \bullet \cos\beta = \cos a \bullet \cos\beta\]
\[Что\ и\ требовалось\ доказать.\]
\[6)\sin a \bullet \sin\beta =\]
\[= \frac{1}{2}\left( \cos(a - \beta) - \cos(a + \beta) \right)\]
\[Преобразуем\ правую\ часть\ \]
\[тождества:\]
\[\frac{1}{2}\left( \cos(a - \beta) - \cos(a + \beta) \right) =\]
\[= \frac{1}{2} \bullet 2\sin a \bullet \sin\beta = \sin a \bullet \sin\beta\]
\[\sin a \bullet \sin\beta = \sin a \bullet \sin\beta\]
\[Что\ и\ требовалось\ доказать.\]
