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

1) sin пи/4*cos пи/4 — sin пи/3*cos пи/6;
2) 2tg^2 пи/3 — ctg^2 пи/6 — sin пи/6* cos пи/3;
3) (tg пи/4 — ctg пи/3) ( ctg пи/4 + tg пи/6);
4) 2cos^2 пи/6 — sin^2 пи/3 + tg пи/6* ctg пи/3.

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

   $$\sin \frac{\pi}{4}\cos \frac{\pi}{4}=\frac{\sqrt2}{2}\cdot \frac{\sqrt2}{2}=\frac12,$$
   $$\sin \frac{\pi}{3}\cos \frac{\pi}{6}=\frac{\sqrt3}{2}\cdot \frac{\sqrt3}{2}=\frac34.$$

   $$\frac12-\frac34=-\frac14.$$

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

   $$\tg \frac{\pi}{3}=\sqrt3,\quad \ctg \frac{\pi}{6}=\sqrt3,\quad \sin \frac{\pi}{6}=\frac12,\quad \cos \frac{\pi}{3}=\frac12.$$

   $$2\cdot (\sqrt3)^2-(\sqrt3)^2-\frac12\cdot \frac12=2\cdot 3-3-\frac14=\frac{11}{4}.$$

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

   $$\tg \frac{\pi}{4}=1,\quad \ctg \frac{\pi}{3}=\frac{1}{\sqrt3},\quad \ctg \frac{\pi}{4}=1,\quad \tg \frac{\pi}{6}=\frac{1}{\sqrt3}.$$

   $$(1-\frac{1}{\sqrt3})(1+\frac{1}{\sqrt3})=1-\frac13=\frac23.$$

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

   $$\cos \frac{\pi}{6}=\frac{\sqrt3}{2},\quad \sin \frac{\pi}{3}=\frac{\sqrt3}{2},\quad \tg \frac{\pi}{6}=\frac{1}{\sqrt3},\quad \ctg \frac{\pi}{3}=\frac{1}{\sqrt3}.$$

   $$2\left(\frac{\sqrt3}{2}\right)^2-\left(\frac{\sqrt3}{2}\right)^2+\frac{1}{\sqrt3}\cdot \frac{1}{\sqrt3} =2\cdot \frac34-\frac34+\frac13 =\frac34+\frac13 =\frac{13}{12}.$$

Ответ

1) $$-\frac14$$; 2) $$\frac{11}{4}$$; 3) $$\frac23$$; 4) $$\frac{13}{12}$$.