Упр.1045 ГДЗ Макарычев Миндюк Феоктистов 9 класс (Алгебра)

Используем значения тригонометрических функций для основных углов.

1. $$\frac{1}{2}\sin\frac{\pi}{4}\cos\frac{\pi}{4} =\frac{1}{2}\cdot\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{2}}{2} =\frac{1}{4}$$
2. $$\frac{1}{3}\tg\frac{\pi}{3}\ctg\frac{\pi}{6} =\frac{1}{3}\cdot\sqrt{3}\cdot\sqrt{3} =1$$
3. $$4\tg\frac{\pi}{6}\sin\frac{\pi}{3} =4\cdot\frac{\sqrt{3}}{3}\cdot\frac{\sqrt{3}}{2} =2$$
4. $$8\cos\frac{\pi}{2}\ctg\frac{\pi}{3} =8\cdot 0\cdot\frac{1}{\sqrt{3}} =0$$
5. $$\tg\frac{\pi}{6}\left(\cos\frac{\pi}{4}+\sin\frac{\pi}{4}\right) =\frac{\sqrt{3}}{3}\left(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\right) =\frac{\sqrt{6}}{3}$$
6. $$\sin\frac{\pi}{3}\cos\frac{\pi}{3}-\tg\frac{\pi}{6} =\frac{\sqrt{3}}{2}\cdot\frac{1}{2}-\frac{\sqrt{3}}{3} =\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{3} =-\frac{\sqrt{3}}{12}$$
7. $$\cos\frac{\pi}{6}-\tg\frac{\pi}{6}+\ctg\frac{\pi}{6} =\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{3}+\sqrt{3} =\frac{7\sqrt{3}}{6}$$
8. $$\sin\frac{\pi}{4}\left(\tg\frac{\pi}{6}-\cos\frac{\pi}{3}\right) =\frac{\sqrt{2}}{2}\left(\frac{\sqrt{3}}{3}-\frac{1}{2}\right) =\frac{\sqrt{2}(2\sqrt{3}-3)}{12} =\frac{2\sqrt{6}-3\sqrt{2}}{12}$$