Упр.438 ГДЗ Алимов 10-11 класс (Алгебра)
Найти значение выражения:
- $$\sin\frac{\pi}{4}\cdot\cos\frac{\pi}{4}-\sin\frac{\pi}{3}\cdot\cos\frac{\pi}{6}$$
- $$2\operatorname{tg}^2\frac{\pi}{3}-\operatorname{ctg}^2\frac{\pi}{6}-\sin\frac{\pi}{6}\cdot\cos\frac{\pi}{3}$$
- $$(\operatorname{tg}\frac{\pi}{4}-\operatorname{ctg}\frac{\pi}{3})(\operatorname{ctg}\frac{\pi}{4}+\operatorname{tg}\frac{\pi}{6})$$
- $$2\cos^2\frac{\pi}{6}-\sin^2\frac{\pi}{3}+\operatorname{tg}\frac{\pi}{6}\cdot\operatorname{ctg}\frac{\pi}{3}$$
1) $$\sin \frac{\pi}{4}\cos \frac{\pi}{4}-\sin \frac{\pi}{3}\cos \frac{\pi}{6}$$
$$\sin \frac{\pi}{4}\cos \frac{\pi}{4}-\sin \frac{\pi}{3}\cos \frac{\pi}{6}=\frac{\sqrt2}{2}\cdot\frac{\sqrt2}{2}-\frac{\sqrt3}{2}\cdot\frac{\sqrt3}{2}=\frac12-\frac34=-\frac14$$
Ответ: $$-\frac14$$
2) $$2\tg^2 \frac{\pi}{3}-\ctg^2 \frac{\pi}{6}-\sin \frac{\pi}{6}\cos \frac{\pi}{3}$$
$$2\tg^2 \frac{\pi}{3}-\ctg^2 \frac{\pi}{6}-\sin \frac{\pi}{6}\cos \frac{\pi}{3}=2\cdot (\sqrt3)^2-(\sqrt3)^2-\frac12\cdot\frac12=6-3-\frac14=\frac{11}{4}$$
Ответ: $$\frac{11}{4}$$
3) $$(\tg \frac{\pi}{4}-\ctg \frac{\pi}{3})(\ctg \frac{\pi}{4}+\tg \frac{\pi}{6})$$
$$\left(\tg \frac{\pi}{4}-\ctg \frac{\pi}{3}\right)\left(\ctg \frac{\pi}{4}+\tg \frac{\pi}{6}\right)=\left(1-\frac{1}{\sqrt3}\right)\left(1+\frac{1}{\sqrt3}\right)=1-\frac13=\frac23$$
Ответ: $$\frac23$$
4) $$2\cos^2 \frac{\pi}{6}-\sin^2 \frac{\pi}{3}+\tg \frac{\pi}{6}\cdot \ctg \frac{\pi}{3}$$
$$2\cos^2 \frac{\pi}{6}-\sin^2 \frac{\pi}{3}+\tg \frac{\pi}{6}\cdot \ctg \frac{\pi}{3}=2\cdot \left(\frac{\sqrt3}{2}\right)^2-\left(\frac{\sqrt3}{2}\right)^2+\frac{1}{\sqrt3}\cdot\frac{1}{\sqrt3}$$
$$=\frac32-\frac34+\frac13=\frac{13}{12}$$
Ответ: $$\frac{13}{12}$$
