Страница 160 номер 532, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Доказать тождество (532—533). 1) \(\sin\left(\frac{\pi}{4} + \alpha\right) - \cos\left(\frac{\pi}{4} - \alpha\right) = 0\); 2) \(\cos\left(\frac{\pi}{6} - \alpha\right) - \sin\left(\frac{\pi}{3} + \alpha\right) = 0\); 3) \(\frac{\sin\left(\frac{3\pi}{2} - \alpha\right)}{\operatorname{tg} (\pi + \alpha)} \cdot \frac{\operatorname{ctg}\left(\frac{\pi}{2} + \alpha\right)}{\operatorname{tg}\left(\alpha - \frac{3\pi}{2}\right)} = -\sin \alpha\).
1) \(\sin\left(\frac{\pi}{4} + \alpha\right) - \cos\left(\frac{\pi}{4} - \alpha\right) = 0\)
\(\sin\left(\dfrac{\pi}{4} + \alpha\right) - \cos\left(\dfrac{\pi}{4} - \alpha\right) =\)
\(= \sin\left(\dfrac{\pi}{2} - \left(\dfrac{\pi}{4} - \alpha\right)\right) - \cos\left(\dfrac{\pi}{4} - \alpha\right) =\)
\(= \cos\left(\dfrac{\pi}{4} - \alpha\right) - \cos\left(\dfrac{\pi}{4} - \alpha\right) = 0;\)
2) \(\cos\left(\frac{\pi}{6} - \alpha\right) - \sin\left(\frac{\pi}{3} + \alpha\right) = 0\)
\(\cos\left(\dfrac{\pi}{6} - \alpha\right) - \sin\left(\dfrac{\pi}{3} + \alpha\right) =\)
\(= \cos\left(\dfrac{\pi}{6} - \alpha\right) - \sin\left(\dfrac{\pi}{2} - \left(\dfrac{\pi}{6} - \alpha\right)\right) =\)
\(= \cos\left(\dfrac{\pi}{6} - \alpha\right) - \cos\left(\dfrac{\pi}{6} - \alpha\right) = 0;\)
3) \(\frac{\sin\left(\frac{3\pi}{2} - \alpha\right)}{\operatorname{tg}(\pi + \alpha)} \cdot \frac{\operatorname{ctg}\left(\frac{\pi}{2} + \alpha\right)}{\operatorname{tg}\left(\alpha - \frac{3\pi}{2}\right)} = -\sin \alpha\)
\(\dfrac{\sin\left(\frac{3\pi}{2} - \alpha\right)}{\operatorname{tg}(\pi + \alpha)} \cdot \dfrac{\operatorname{ctg}\left(\frac{\pi}{2} + \alpha\right)}{\operatorname{tg}\left(\alpha - \dfrac{3\pi}{2}\right)} = \dfrac{-\cos \alpha}{\operatorname{tg} \alpha} \cdot \dfrac{-\operatorname{tg} \alpha}{-\operatorname{ctg} \alpha} = -\sin \alpha.\)
