Страница 165 номер 551, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Упростить выражение (550—551). 1) \(\frac{\sin\left(\frac{\pi}{4} + \alpha\right) - \cos\left(\frac{\pi}{4} + \alpha\right)}{\sin\left(\frac{\pi}{4} + \alpha\right) + \cos\left(\frac{\pi}{4} + \alpha\right)}\); 2) \(\frac{\sin\left(\frac{\pi}{4} - \alpha\right) + \cos\left(\frac{\pi}{4} - \alpha\right)}{\sin\left(\frac{\pi}{4} - \alpha\right) - \cos\left(\frac{\pi}{4} - \alpha\right)}\).
1) \(\frac{\sin\left(\frac{\pi}{4} + \alpha\right) - \cos\left(\frac{\pi}{4} + \alpha\right)}{\sin\left(\frac{\pi}{4} + \alpha\right) + \cos\left(\frac{\pi}{4} + \alpha\right)} = \frac{\sin\frac{\pi}{4}\cos\alpha + \sin\alpha\cos\frac{\pi}{4} - \cos\frac{\pi}{4}\cos\alpha + \sin\frac{\pi}{4}\sin\alpha}{\sin\frac{\pi}{4}\cos\alpha + \sin\alpha\cos\frac{\pi}{4} + \cos\frac{\pi}{4}\cos\alpha - \sin\frac{\pi}{4}\sin\alpha} =\)
\(= \dfrac{2 \cdot \frac{\sqrt{2}}{2}\sin\alpha}{2 \cdot \dfrac{\sqrt{2}}{2}\cos\alpha} = \operatorname{tg}\alpha;\)
2) \(\frac{\sin\left(\frac{\pi}{4} - \alpha\right) + \cos\left(\frac{\pi}{4} - \alpha\right)}{\sin\left(\frac{\pi}{4} - \alpha\right) - \cos\left(\frac{\pi}{4} - \alpha\right)} = \frac{\sin\frac{\pi}{4}\cos\alpha - \sin\alpha\cos\frac{\pi}{4} + \cos\frac{\pi}{4}\cos\alpha + \sin\frac{\pi}{4}\sin\alpha}{\sin\frac{\pi}{4}\cos\alpha - \sin\alpha\cos\frac{\pi}{4} - \cos\frac{\pi}{4}\cos\alpha - \sin\frac{\pi}{4}\sin\alpha} =\)
\(= \dfrac{\frac{2\sqrt{2}}{2}\cos\alpha}{-\dfrac{2\sqrt{2}}{2}\sin\alpha} = -\operatorname{ctg}\alpha.\)
