Упр.13.20 ГДЗ Мордкович Семенов 11 класс (Алгебра)
- Найдите площадь треугольника, образованного касательной к графику функции $$y=\sqrt{2x^2+7}$$ в точке $$x=1$$ и биссектрисами координатных углов.
- Вычислите:
а) $$2+\log_{\sqrt{2}}\left(\sin\frac{\pi}{8}\right)+\log_{\sqrt{2}}\left(\cos\frac{\pi}{8}\right)$$;
б) $$\log_{\frac{1}{2}}\left(\cos\frac{\pi}{6}-\sin\frac{\pi}{6}\right)+\log_{\frac{1}{2}}\left(\cos\frac{\pi}{6}+\sin\frac{\pi}{6}\right)$$;
в) $$\log_{3}\left(2\tg\frac{\pi}{6}\right)-\log_{3}\left(1-\tg^2\frac{\pi}{6}\right)$$;
г) $$\log_{\frac{\sqrt{3}}{2}}\left(\cos\frac{\pi}{12}-\sin\frac{\pi}{12}\right)+\log_{\frac{\sqrt{3}}{2}}\left(\cos\frac{\pi}{12}+\sin\frac{\pi}{12}\right)$$;
д) $$\log_{\frac{1}{2}}\left(\cos\frac{\pi}{12}\right)+\log_{\frac{1}{2}}\left(\sin\frac{\pi}{12}\right)-1$$;
е) $$1+\log_{2}\left(\tg\frac{\pi}{8}\right)-\log_{2}\left(1-\tg^2\frac{\pi}{8}\right)$$.
Используем формулы приведения и тождества:
$$\sin 2\alpha = 2\sin \alpha \cos \alpha,$$
$$\cos 2\alpha = \cos^2 \alpha — \sin^2 \alpha,$$
$$1-\tg^2 \alpha = \frac{\cos 2\alpha}{\cos^2 \alpha}.$$
$$2+\log_{\sqrt{2}}\left(\sin \frac{\pi}{8}\right)+\log_{\sqrt{2}}\left(\cos \frac{\pi}{8}\right)$$
$$=\log_{\sqrt{2}}2+\log_{\sqrt{2}}\left(\sin \frac{\pi}{8}\cos \frac{\pi}{8}\right)$$
$$=\log_{\sqrt{2}}\left(2\sin \frac{\pi}{8}\cos \frac{\pi}{8}\right)=\log_{\sqrt{2}}\left(\sin \frac{\pi}{4}\right)$$
$$=\log_{\sqrt{2}}\frac{\sqrt{2}}{2}=\log_{\sqrt{2}}(\sqrt{2})^{-1}=-1.$$
$$\log_{\frac12}\left(\cos \frac{\pi}{6}-\sin \frac{\pi}{6}\right)+\log_{\frac12}\left(\cos \frac{\pi}{6}+\sin \frac{\pi}{6}\right)$$
$$=\log_{\frac12}\left(\cos^2 \frac{\pi}{6}-\sin^2 \frac{\pi}{6}\right)=\log_{\frac12}\left(\cos \frac{\pi}{3}\right)$$
$$=\log_{\frac12}\frac12=1.$$
$$\log_3\left(2\tg \frac{\pi}{6}\right)-\log_3\left(1-\tg^2 \frac{\pi}{6}\right)$$
$$=\log_3\frac{2\tg \frac{\pi}{6}}{1-\tg^2 \frac{\pi}{6}}=\log_3\left(\tg \frac{\pi}{3}\right)$$
$$=\log_3\sqrt{3}=\log_3 3^{1/2}=\frac12.$$
$$\log_{\frac{\sqrt{3}}{2}}\left(\cos \frac{\pi}{12}-\sin \frac{\pi}{12}\right)+\log_{\frac{\sqrt{3}}{2}}\left(\cos \frac{\pi}{12}+\sin \frac{\pi}{12}\right)$$
$$=\log_{\frac{\sqrt{3}}{2}}\left(\cos^2 \frac{\pi}{12}-\sin^2 \frac{\pi}{12}\right)=\log_{\frac{\sqrt{3}}{2}}\left(\cos \frac{\pi}{6}\right)$$
$$=\log_{\frac{\sqrt{3}}{2}}\frac{\sqrt{3}}{2}=1.$$
$$\log_{\frac12}\left(\cos \frac{\pi}{12}\right)+\log_{\frac12}\left(\sin \frac{\pi}{12}\right)-1$$
$$=\log_{\frac12}\left(\cos \frac{\pi}{12}\right)+\log_{\frac12}\left(\sin \frac{\pi}{12}\right)-\log_{\frac12}\frac12$$
$$=\log_{\frac12}\left(2\cos \frac{\pi}{12}\sin \frac{\pi}{12}\right)=\log_{\frac12}\left(\sin \frac{\pi}{6}\right)$$
$$=\log_{\frac12}\frac12=1.$$
$$1+\log_2\left(\tg \frac{\pi}{8}\right)-\log_2\left(1-\tg^2 \frac{\pi}{8}\right)$$
$$=\log_2 2+\log_2\frac{\tg \frac{\pi}{8}}{1-\tg^2 \frac{\pi}{8}}=\log_2\frac{2\tg \frac{\pi}{8}}{1-\tg^2 \frac{\pi}{8}}$$
$$=\log_2\left(\tg \frac{\pi}{4}\right)=\log_2 1=0.$$
Ответ
а) $$-1$$; б) $$1$$; в) $$\frac12$$; г) $$1$$; д) $$1$$; е) $$0$$.
