Упр.16.19 ГДЗ Мордковича 11 класс профильный уровень (Алгебра)
- 16.19. 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) $$\log_{\sqrt2}\!\left(\sin\frac{\pi}{8}\right)+\log_{\sqrt2}\!\left(2\cos\frac{\pi}{8}\right)=\log_{\sqrt2}\!\left(2\sin\frac{\pi}{8}\cos\frac{\pi}{8}\right).$$
Так как $$2\sin\frac{\pi}{8}\cos\frac{\pi}{8}=\sin\frac{\pi}{4}=\frac{1}{\sqrt2},$$ то
$$\log_{\sqrt2}\!\left(\frac{1}{\sqrt2}\right)=\log_{\sqrt2}\!\left((\sqrt2)^{-1}\right)=-1.$$
б) $$\log_{1/2}\!\left(\cos\frac{\pi}{6}+\sin\frac{\pi}{6}\right)+\log_{1/2}\!\left(\cos\frac{\pi}{6}-\sin\frac{\pi}{6}\right)$$
$$=\log_{1/2}\!\left(\left(\cos\frac{\pi}{6}+\sin\frac{\pi}{6}\right)\left(\cos\frac{\pi}{6}-\sin\frac{\pi}{6}\right)\right)$$
$$=\log_{1/2}\!\left(\cos^2\frac{\pi}{6}-\sin^2\frac{\pi}{6}\right)=\log_{1/2}\!\left(\cos\frac{\pi}{3}\right)=\log_{1/2}\!\left(\frac12\right)=1.$$
в) $$\log_{1/2}\!\left(2\sin\frac{\pi}{12}\right)+\log_{1/2}\!\left(\cos\frac{\pi}{12}\right)$$
$$=\log_{1/2}\!\left(2\sin\frac{\pi}{12}\cos\frac{\pi}{12}\right)=\log_{1/2}\!\left(\sin\frac{\pi}{6}\right)=\log_{1/2}\!\left(\frac12\right)=1.$$
г) $$\log_{\sqrt3/2}\!\left(\cos\frac{\pi}{12}-\sin\frac{\pi}{12}\right)+\log_{\sqrt3/2}\!\left(\cos\frac{\pi}{12}+\sin\frac{\pi}{12}\right)$$
$$=\log_{\sqrt3/2}\!\left(\left(\cos\frac{\pi}{12}-\sin\frac{\pi}{12}\right)\left(\cos\frac{\pi}{12}+\sin\frac{\pi}{12}\right)\right)$$
$$=\log_{\sqrt3/2}\!\left(\cos^2\frac{\pi}{12}-\sin^2\frac{\pi}{12}\right)=\log_{\sqrt3/2}\!\left(\cos\frac{\pi}{6}\right)=\log_{\sqrt3/2}\!\left(\frac{\sqrt3}{2}\right)=1.$$
Ответ: а) $$-1$$; б) $$1$$; в) $$1$$; г) $$1$$.
