Упр.529 ГДЗ Алимов 10-11 класс (Алгебра)
Вычислить:
- $$\cos 750^\circ$$;
- $$\sin 1140^\circ$$;
- $$\tg 405^\circ$$;
- $$\cos 840^\circ$$;
- $$\sin \frac{47\pi}{6}$$;
- $$\tg \frac{25\pi}{4}$$;
- $$\ctg \frac{27\pi}{4}$$;
- $$\cos \frac{21\pi}{4}$$.
Используем формулы приведения и периодичность тригонометрических функций.
$$\cos 750^\circ=\cos(720^\circ+30^\circ)=\cos 30^\circ=\frac{\sqrt{3}}{2}$$
$$\sin 1140^\circ=\sin(1080^\circ+60^\circ)=\sin 60^\circ=\frac{\sqrt{3}}{2}$$
$$\tg 405^\circ=\tg(360^\circ+45^\circ)=\tg 45^\circ=1$$
$$\cos 840^\circ=\cos(720^\circ+120^\circ)=\cos 120^\circ=\cos(180^\circ-60^\circ)=-\cos 60^\circ=-\frac{1}{2}$$
$$\sin \frac{47\pi}{6}=\sin\left(8\pi-\frac{\pi}{6}\right)=\sin\left(-\frac{\pi}{6}\right)=-\sin \frac{\pi}{6}=-\frac{1}{2}$$
$$\tg \frac{25\pi}{4}=\tg\left(6\pi+\frac{\pi}{4}\right)=\tg \frac{\pi}{4}=1$$
$$\ctg \frac{27\pi}{4}=\ctg\left(7\pi-\frac{\pi}{4}\right)=\ctg\left(-\frac{\pi}{4}\right)=-\ctg \frac{\pi}{4}=-1$$
$$\cos \frac{21\pi}{4}=\cos\left(4\pi+\frac{5\pi}{4}\right)=\cos \frac{5\pi}{4}=\cos\left(\pi+\frac{\pi}{4}\right)=-\cos \frac{\pi}{4}=-\frac{\sqrt{2}}{2}$$
Ответ: $$\frac{\sqrt{3}}{2},\ \frac{\sqrt{3}}{2},\ 1,\ -\frac{1}{2},\ -\frac{1}{2},\ 1,\ -1,\ -\frac{\sqrt{2}}{2}$$









