Упр.13.20 ГДЗ Мордкович Семенов 11 класс (Алгебра)

а) 2+log_v2(sin(?/8))+log_v2(cos(?/8));
б) log_(1/2)(cos(?/6)-sin(?/6))+log_(1/2)(cos(?/6)+sin(?/6));
в) log_3(2tg(?/6))-log_3(1-tg^2(?/6));
г) log_(v3/2)(cos(?/12)-sin(?/12))+log_(v3/2)(cos(?/12)+sin(?/12));
д) log_(1/2)(cos(?/12))+log_(1/2)(sin(?/12))-1;
е) 1+log_2(tg(?/8))-log_2(1-tg^2(?/8)).

Используем формулы приведения и тождества:

$$\sin 2\alpha = 2\sin \alpha \cos \alpha,$$

$$\cos 2\alpha = \cos^2 \alpha — \sin^2 \alpha,$$

$$1-\tg^2 \alpha = \frac{\cos 2\alpha}{\cos^2 \alpha}.$$

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

   $$=\log_{\sqrt{2}}2+\log_{\sqrt{2}}\left(\sin \frac{\pi}{8}\cos \frac{\pi}{8}\right)$$

   $$=\log_{\sqrt{2}}\left(2\sin \frac{\pi}{8}\cos \frac{\pi}{8}\right)=\log_{\sqrt{2}}\left(\sin \frac{\pi}{4}\right)$$

   $$=\log_{\sqrt{2}}\frac{\sqrt{2}}{2}=\log_{\sqrt{2}}(\sqrt{2})^{-1}=-1.$$

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

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

   $$=\log_{\frac12}\frac12=1.$$

3. $$\log_3\left(2\tg \frac{\pi}{6}\right)-\log_3\left(1-\tg^2 \frac{\pi}{6}\right)$$

   $$=\log_3\frac{2\tg \frac{\pi}{6}}{1-\tg^2 \frac{\pi}{6}}=\log_3\left(\tg \frac{\pi}{3}\right)$$

   $$=\log_3\sqrt{3}=\log_3 3^{1/2}=\frac12.$$

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

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

   $$=\log_{\frac{\sqrt{3}}{2}}\frac{\sqrt{3}}{2}=1.$$

5. $$\log_{\frac12}\left(\cos \frac{\pi}{12}\right)+\log_{\frac12}\left(\sin \frac{\pi}{12}\right)-1$$

   $$=\log_{\frac12}\left(\cos \frac{\pi}{12}\right)+\log_{\frac12}\left(\sin \frac{\pi}{12}\right)-\log_{\frac12}\frac12$$

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

   $$=\log_{\frac12}\frac12=1.$$

6. $$1+\log_2\left(\tg \frac{\pi}{8}\right)-\log_2\left(1-\tg^2 \frac{\pi}{8}\right)$$

   $$=\log_2 2+\log_2\frac{\tg \frac{\pi}{8}}{1-\tg^2 \frac{\pi}{8}}=\log_2\frac{2\tg \frac{\pi}{8}}{1-\tg^2 \frac{\pi}{8}}$$

   $$=\log_2\left(\tg \frac{\pi}{4}\right)=\log_2 1=0.$$

Ответ

а) $$-1$$; б) $$1$$; в) $$\frac12$$; г) $$1$$; д) $$1$$; е) $$0$$.