Упр.43.19 ГДЗ Мордкович 10-11 класс (Алгебра)
Вычислите:
- а) $$\log_{\sqrt{2}}\left(\sin\frac{\pi}{8}\right)+\log_{\sqrt{2}}\left(2\cos\frac{\pi}{8}\right)$$;
- 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) $$\log_{\frac{1}{2}}\left(2\sin\frac{\pi}{12}\right)+\log_{\frac{1}{2}}\left(\cos\frac{\pi}{12}\right)$$;
- г) $$\log_{\frac{\sqrt{3}}{2}}\left(\cos\frac{\pi}{12}-\sin\frac{\pi}{12}\right)+\log_{\frac{\sqrt{3}}{2}}\left(\cos\frac{\pi}{12}+\sin\frac{\pi}{12}\right)$$.
а) $$\log_{\sqrt{2}}\left(\sin\frac{\pi}{8}\right)+\log_{\sqrt{2}}\left(2\cos\frac{\pi}{8}\right)=\log_{\sqrt{2}}\left(2\sin\frac{\pi}{8}\cos\frac{\pi}{8}\right)$$
$$=\log_{\sqrt{2}}\left(\sin\frac{\pi}{4}\right)=\log_{\sqrt{2}}\frac{1}{\sqrt{2}}=\log_{\sqrt{2}}(\sqrt{2})^{-1}=-1.$$
б) $$\log_{\frac12}\left(\cos\frac{\pi}{6}+\sin\frac{\pi}{6}\right)+\log_{\frac12}\left(\cos\frac{\pi}{6}-\sin\frac{\pi}{6}\right)$$
$$=\log_{\frac12}\left(\left(\cos\frac{\pi}{6}+\sin\frac{\pi}{6}\right)\left(\cos\frac{\pi}{6}-\sin\frac{\pi}{6}\right)\right)$$
$$=\log_{\frac12}\left(\cos^2\frac{\pi}{6}-\sin^2\frac{\pi}{6}\right)=\log_{\frac12}\left(\cos\frac{\pi}{3}\right)=\log_{\frac12}\frac12=1.$$
в) $$\log_{\frac12}\left(2\sin\frac{\pi}{12}\right)+\log_{\frac12}\left(\cos\frac{\pi}{12}\right)$$
$$=\log_{\frac12}\left(2\sin\frac{\pi}{12}\cos\frac{\pi}{12}\right)=\log_{\frac12}\left(\sin\frac{\pi}{6}\right)=\log_{\frac12}\frac12=1.$$
г) $$\log_{\frac{\sqrt{3}}{2}}\left(\cos\frac{\pi}{12}-\sin\frac{\pi}{12}\right)+\log_{\frac{\sqrt{3}}{2}}\left(\cos\frac{\pi}{12}+\sin\frac{\pi}{12}\right)$$
$$=\log_{\frac{\sqrt{3}}{2}}\left(\left(\cos\frac{\pi}{12}-\sin\frac{\pi}{12}\right)\left(\cos\frac{\pi}{12}+\sin\frac{\pi}{12}\right)\right)$$
$$=\log_{\frac{\sqrt{3}}{2}}\left(\cos^2\frac{\pi}{12}-\sin^2\frac{\pi}{12}\right)=\log_{\frac{\sqrt{3}}{2}}\left(\cos\frac{\pi}{6}\right)=\log_{\frac{\sqrt{3}}{2}}\frac{\sqrt{3}}{2}=1.$$
Ответ: а) $$-1$$; б) $$1$$; в) $$1$$; г) $$1$$.
