Страница 143 номер 475, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(\cos \left(-\frac{\pi}{6}\right)\sin \left(-\frac{\pi}{3}\right) + \operatorname{tg} \left(-\frac{\pi}{4}\right) = -\cos \frac{\pi}{6} \sin \frac{\pi}{3} - \operatorname{tg} \frac{\pi}{4} =\)
\(= -\dfrac{\sqrt{3}}{2} \cdot \dfrac{\sqrt{3}}{2} - 1 = -\dfrac{7}{4};\)

2) \(\frac{1 + \operatorname{tg}^{2} \left(-\frac{\pi}{6}\right)}{1 + \operatorname{ctg}^{2} \left(-\frac{\pi}{6}\right)} = \frac{1 + \operatorname{tg}^{2} \frac{\pi}{6}}{1 + \operatorname{ctg}^{2} \frac{\pi}{6}} = \frac{1 + \frac{1}{3}}{1 + 3} = \frac{4}{3} : 4 = \frac{1}{3};\)

3) \(2\sin \left(-\frac{\pi}{6}\right)\cos \left(-\frac{\pi}{6}\right) + \operatorname{tg} \left(-\frac{\pi}{3}\right) + \sin^{2} \left(-\frac{\pi}{4}\right) =\)
\(= -2\sin \dfrac{\pi}{6} \cos \dfrac{\pi}{6} - \operatorname{tg} \dfrac{\pi}{3} + \sin^{2} \dfrac{\pi}{4} =\)
\(= -2 \cdot \dfrac{1}{2} \cdot \dfrac{\sqrt{3}}{2} - \sqrt{3} + \left(\dfrac{\sqrt{2}}{2}\right)^{2} = -\dfrac{3\sqrt{3}}{2} + \dfrac{1}{2};\)

4) \(\cos(-\pi) + \operatorname{ctg} \left(-\frac{\pi}{2}\right) - \sin \left(-\frac{3\pi}{2}\right) + \operatorname{ctg} \left(-\frac{\pi}{4}\right) =\)
\(= \cos \pi - \operatorname{ctg} \dfrac{\pi}{2} + \sin \dfrac{3\pi}{2} - \operatorname{ctg} \dfrac{\pi}{4} = -1 - 0 - 1 - 1 = -3;\)

5) \(\frac{3 - \sin^{2} \left(-\frac{\pi}{3}\right) - \cos^{2} \left(-\frac{\pi}{3}\right)}{2\cos \left(-\frac{\pi}{4}\right)} = \frac{3 - \sin^{2} \frac{\pi}{3} - \cos^{2} \frac{\pi}{3}}{2\cos \frac{\pi}{4}} =\)
\(= \frac{3 - \left(\sin^{2} \frac{\pi}{3} + \cos^{2} \frac{\pi}{3}\right)}{2\cos \frac{\pi}{4}} = \frac{3 - 1}{2 \cdot \frac{\sqrt{2}}{2}} = \frac{2}{\sqrt{2}} = \sqrt{2};\)

6) \(2\sin \left(-\frac{\pi}{6}\right) + 3 + 7{,}5 \operatorname{tg} (-\pi) + \frac{1}{8}\cos \frac{3\pi}{2} =\)
\(= -2\sin \dfrac{\pi}{6} + 3 - 7{,}5 \operatorname{tg} \pi + \dfrac{1}{8}\cos \dfrac{3\pi}{2} = -1 + 3 - 0 + 0 = 2.\)