Упр.676 ГДЗ Никольский Потапов 9 класс (Алгебра)

Задача

Вычислите:
а) $$\operatorname{tg}135^\circ$$;
б) $$\operatorname{ctg}120^\circ$$;
в) $$\operatorname{tg}210^\circ$$;
г) $$\operatorname{ctg}330^\circ$$;
д) $$12\sin120^\circ\operatorname{tg}300^\circ\operatorname{ctg}225^\circ$$;
е) $$20\sin330^\circ\cos(-240^\circ)\operatorname{tg}120^\circ-2\cos150^\circ\operatorname{tg}(-135^\circ)$$;
ж) $$\cos20^\circ+\cos40^\circ+\cos60^\circ+\ldots+\cos160^\circ+\cos180^\circ$$;
з) $$\operatorname{tg}20^\circ+\operatorname{tg}40^\circ+\operatorname{tg}60^\circ+\ldots+\operatorname{tg}160^\circ+\operatorname{tg}180^\circ$$.

Подробный ответ

Воспользуемся формулами приведения и значениями тригонометрических функций для углов $$30^\circ$$, $$45^\circ$$, $$60^\circ$$.

$$\tg 135^\circ=\tg(180^\circ-45^\circ)=-\tg 45^\circ=-1.$$
$$\ctg 120^\circ=\ctg(180^\circ-60^\circ)=-\ctg 60^\circ=-\frac{1}{\sqrt{3}}=-\frac{\sqrt{3}}{3}.$$
$$\tg 210^\circ=\tg(180^\circ+30^\circ)=\tg 30^\circ=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}.$$
$$\ctg 330^\circ=\ctg(360^\circ-30^\circ)=-\ctg 30^\circ=-\sqrt{3}.$$
$$12\sin 120^\circ\cdot \tg 300^\circ\cdot \ctg 225^\circ$$
$$=12\sin(90^\circ+30^\circ)\cdot \tg(360^\circ-60^\circ)\cdot \ctg(180^\circ+45^\circ)$$
$$=12\cos 30^\circ\cdot(-\tg 60^\circ)\cdot \ctg 45^\circ$$
$$=12\cdot \frac{\sqrt{3}}{2}\cdot(-\sqrt{3})\cdot 1=-18.$$
$$20\sin 330^\circ\cdot \cos(-240^\circ)\cdot \tg 120^\circ-2\cos 150^\circ\cdot \tg(-135^\circ)$$
$$=20\sin(360^\circ-30^\circ)\cdot \cos(180^\circ+60^\circ)\cdot \tg(180^\circ-60^\circ)-2\cos(180^\circ-30^\circ)\cdot \tg(180^\circ-45^\circ)$$
$$=20\cdot(-\sin 30^\circ)\cdot(-\cos 60^\circ)\cdot(-\tg 60^\circ)-2\cdot(-\cos 30^\circ)\cdot(-\tg 45^\circ)$$
$$=20\cdot\left(-\frac12\right)\cdot\left(-\frac12\right)\cdot(-\sqrt{3})-2\cdot\left(-\frac{\sqrt{3}}{2}\right)\cdot(-1)$$
$$=-5\sqrt{3}+\sqrt{3}=-4\sqrt{3}.$$
$$\cos 20^\circ+\cos 40^\circ+\cos 60^\circ+\dots+\cos 160^\circ+\cos 180^\circ$$
$$=(\cos 20^\circ+\cos 160^\circ)+(\cos 40^\circ+\cos 140^\circ)+(\cos 60^\circ+\cos 120^\circ)+(\cos 80^\circ+\cos 100^\circ)+\cos 180^\circ$$
$$=(\cos 20^\circ-\cos 20^\circ)+(\cos 40^\circ-\cos 40^\circ)+(\cos 60^\circ-\cos 60^\circ)+(\cos 80^\circ-\cos 80^\circ)-1=-1.$$
$$\tg 20^\circ+\tg 40^\circ+\tg 60^\circ+\dots+\tg 160^\circ+\tg 180^\circ$$
$$=(\tg 20^\circ+\tg 160^\circ)+(\tg 40^\circ+\tg 140^\circ)+(\tg 60^\circ+\tg 120^\circ)+(\tg 80^\circ+\tg 100^\circ)+\tg 180^\circ$$
$$=(\tg 20^\circ-\tg 20^\circ)+(\tg 40^\circ-\tg 40^\circ)+(\tg 60^\circ-\tg 60^\circ)+(\tg 80^\circ-\tg 80^\circ)+0=0.$$