Упр.438 ГДЗ Алимов 10-11 класс (Алгебра)

Найти значение выражения:

1. а) $$\sin\frac{\pi}{4}\cdot\cos\frac{\pi}{4}-\sin\frac{\pi}{3}\cdot\cos\frac{\pi}{6}$$;
   б) $$2\tan^2\frac{\pi}{3}-\cot^2\frac{\pi}{6}-\sin\frac{\pi}{6}\cdot\cos\frac{\pi}{3}$$;
   в) $$\left(\tan\frac{\pi}{4}-\cot\frac{\pi}{3}\right)\left(\cot\frac{\pi}{4}+\tan\frac{\pi}{6}\right)$$;
   г) $$2\cos^2\frac{\pi}{6}-\sin^2\frac{\pi}{3}+\tan\frac{\pi}{6}\cdot\cot\frac{\pi}{3}$$.

1. $$\sin \frac{\pi}{4}\cos \frac{\pi}{4}-\sin \frac{\pi}{3}\cos \frac{\pi}{6}$$

   $$\sin \frac{\pi}{4}=\frac{\sqrt{2}}{2},\quad \cos \frac{\pi}{4}=\frac{\sqrt{2}}{2},\quad \sin \frac{\pi}{3}=\frac{\sqrt{3}}{2},\quad \cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}$$

   $$\sin \frac{\pi}{4}\cos \frac{\pi}{4}-\sin \frac{\pi}{3}\cos \frac{\pi}{6}=\frac{\sqrt{2}}{2}\cdot \frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{3}}{2}=\frac{1}{2}-\frac{3}{4}=-\frac{1}{4}$$

2. $$2\tg^2 \frac{\pi}{3}-\ctg^2 \frac{\pi}{6}-\sin \frac{\pi}{6}\cos \frac{\pi}{3}$$

   $$\tg \frac{\pi}{3}=\sqrt{3},\quad \ctg \frac{\pi}{6}=\sqrt{3},\quad \sin \frac{\pi}{6}=\frac{1}{2},\quad \cos \frac{\pi}{3}=\frac{1}{2}$$

   $$2\tg^2 \frac{\pi}{3}-\ctg^2 \frac{\pi}{6}-\sin \frac{\pi}{6}\cos \frac{\pi}{3}=2\cdot (\sqrt{3})^2-(\sqrt{3})^2-\frac{1}{2}\cdot \frac{1}{2}=6-3-\frac{1}{4}=\frac{11}{4}$$

3. $$\left(\tg \frac{\pi}{4}-\ctg \frac{\pi}{3}\right)\left(\ctg \frac{\pi}{4}+\tg \frac{\pi}{6}\right)$$

   $$\tg \frac{\pi}{4}=1,\quad \ctg \frac{\pi}{3}=\frac{1}{\sqrt{3}},\quad \ctg \frac{\pi}{4}=1,\quad \tg \frac{\pi}{6}=\frac{1}{\sqrt{3}}$$

   $$\left(\tg \frac{\pi}{4}-\ctg \frac{\pi}{3}\right)\left(\ctg \frac{\pi}{4}+\tg \frac{\pi}{6}\right)=\left(1-\frac{1}{\sqrt{3}}\right)\left(1+\frac{1}{\sqrt{3}}\right)=1-\frac{1}{3}=\frac{2}{3}$$

4. $$2\cos^2 \frac{\pi}{6}-\sin^2 \frac{\pi}{3}+\tg \frac{\pi}{6}\cdot \ctg \frac{\pi}{3}$$

   $$\cos \frac{\pi}{6}=\frac{\sqrt{3}}{2},\quad \sin \frac{\pi}{3}=\frac{\sqrt{3}}{2},\quad \tg \frac{\pi}{6}=\frac{1}{\sqrt{3}},\quad \ctg \frac{\pi}{3}=\frac{1}{\sqrt{3}}$$

   $$2\cos^2 \frac{\pi}{6}-\sin^2 \frac{\pi}{3}+\tg \frac{\pi}{6}\cdot \ctg \frac{\pi}{3}=2\cdot \left(\frac{\sqrt{3}}{2}\right)^2-\left(\frac{\sqrt{3}}{2}\right)^2+\frac{1}{\sqrt{3}}\cdot \frac{1}{\sqrt{3}}$$

   $$=2\cdot \frac{3}{4}-\frac{3}{4}+\frac{1}{3}=\frac{3}{4}+\frac{1}{3}=\frac{9}{12}+\frac{4}{12}=\frac{13}{12}$$

Ответ: 1) $$-\frac{1}{4}$$; 2) $$\frac{11}{4}$$; 3) $$\frac{2}{3}$$; 4) $$\frac{13}{12}$$.