Упр.43.19 ГДЗ Мордкович 10-11 класс (Алгебра)

а) logкорень(2) (sin пи/8) + logкорень(2) (2cos пи/8);
б) log1/2 (cos пи/6 + sin пи/6) + log1/2 (cos пи/6 — sin пи/6);
в) log1/2 (2sin пи/12) + log1/2 (cos пи/12);
г) logкорень(3)/2 (cos пи/12 — sin пи/12) + logкорень(3)/2 (cos пи/12 + sin пи/12).

1. а)

   $$\log_{\sqrt{2}}\left(\sin\frac{\pi}{8}\right)+\log_{\sqrt{2}}\left(2\cos\frac{\pi}{8}\right) =\log_{\sqrt{2}}\left(2\sin\frac{\pi}{8}\cos\frac{\pi}{8}\right)$$
   $$=\log_{\sqrt{2}}\left(\sin\frac{\pi}{4}\right) =\log_{\sqrt{2}}\left(\frac{1}{\sqrt{2}}\right) =\log_{\sqrt{2}}\left((\sqrt{2})^{-1}\right)=-1.$$

2. б)

   $$\log_{\frac12}\left(\cos\frac{\pi}{6}+\sin\frac{\pi}{6}\right)+\log_{\frac12}\left(\cos\frac{\pi}{6}-\sin\frac{\pi}{6}\right)$$
   $$=\log_{\frac12}\left(\left(\cos\frac{\pi}{6}+\sin\frac{\pi}{6}\right)\left(\cos\frac{\pi}{6}-\sin\frac{\pi}{6}\right)\right)$$
   $$=\log_{\frac12}\left(\cos^2\frac{\pi}{6}-\sin^2\frac{\pi}{6}\right) =\log_{\frac12}\left(\cos\frac{\pi}{3}\right) =\log_{\frac12}\left(\frac12\right)=1.$$

3. в)

   $$\log_{\frac12}\left(2\sin\frac{\pi}{12}\right)+\log_{\frac12}\left(\cos\frac{\pi}{12}\right)$$
   $$=\log_{\frac12}\left(2\sin\frac{\pi}{12}\cos\frac{\pi}{12}\right) =\log_{\frac12}\left(\sin\frac{\pi}{6}\right)$$
   $$=\log_{\frac12}\left(\frac12\right)=1.$$

4. г)

   $$\log_{\frac{\sqrt{3}}{2}}\left(\cos\frac{\pi}{12}-\sin\frac{\pi}{12}\right)+\log_{\frac{\sqrt{3}}{2}}\left(\cos\frac{\pi}{12}+\sin\frac{\pi}{12}\right)$$
   $$=\log_{\frac{\sqrt{3}}{2}}\left(\left(\cos\frac{\pi}{12}-\sin\frac{\pi}{12}\right)\left(\cos\frac{\pi}{12}+\sin\frac{\pi}{12}\right)\right)$$
   $$=\log_{\frac{\sqrt{3}}{2}}\left(\cos^2\frac{\pi}{12}-\sin^2\frac{\pi}{12}\right) =\log_{\frac{\sqrt{3}}{2}}\left(\cos\frac{\pi}{6}\right)$$
   $$=\log_{\frac{\sqrt{3}}{2}}\left(\frac{\sqrt{3}}{2}\right)=1.$$

Ответ

а) $$-1$$; б) $$1$$; в) $$1$$; г) $$1$$.