Упр.676 ГДЗ Никольский Потапов 9 класс (Алгебра)

a) tg 135°; 6) ctg 120°; в) tg210°; г) ctg 330°;
д) 12 sin 120° tg 300° ctg 225°;
е) 20 sin 330° cos (-240°) tg 120° — 2 cos 150° tg (-135°);
ж) cos 20° + cos40° + cos60° + … + cos 160° + cos 180°;
з) tg20° + tg40° + tg60° + … + tgl60° + tg 180°.

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

1. $$\tg 135^\circ=\tg(180^\circ-45^\circ)=-\tg 45^\circ=-1.$$
2. $$\ctg 120^\circ=\ctg(180^\circ-60^\circ)=-\ctg 60^\circ=-\frac{1}{\sqrt{3}}=-\frac{\sqrt{3}}{3}.$$
3. $$\tg 210^\circ=\tg(180^\circ+30^\circ)=\tg 30^\circ=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}.$$
4. $$\ctg 330^\circ=\ctg(360^\circ-30^\circ)=-\ctg 30^\circ=-\sqrt{3}.$$
5. $$12\sin 120^\circ\cdot \tg 300^\circ\cdot \ctg 225^\circ$$
   $$=12\sin(90^\circ+30^\circ)\cdot \tg(360^\circ-60^\circ)\cdot \ctg(180^\circ+45^\circ)$$
   $$=12\cos 30^\circ\cdot(-\tg 60^\circ)\cdot \ctg 45^\circ$$
   $$=12\cdot \frac{\sqrt{3}}{2}\cdot(-\sqrt{3})\cdot 1=-18.$$
6. $$20\sin 330^\circ\cdot \cos(-240^\circ)\cdot \tg 120^\circ-2\cos 150^\circ\cdot \tg(-135^\circ)$$
   $$=20\sin(360^\circ-30^\circ)\cdot \cos(180^\circ+60^\circ)\cdot \tg(180^\circ-60^\circ)-2\cos(180^\circ-30^\circ)\cdot \tg(180^\circ-45^\circ)$$
   $$=20\cdot(-\sin 30^\circ)\cdot(-\cos 60^\circ)\cdot(-\tg 60^\circ)-2\cdot(-\cos 30^\circ)\cdot(-\tg 45^\circ)$$
   $$=20\cdot\left(-\frac12\right)\cdot\left(-\frac12\right)\cdot(-\sqrt{3})-2\cdot\left(-\frac{\sqrt{3}}{2}\right)\cdot(-1)$$
   $$=-5\sqrt{3}+\sqrt{3}=-4\sqrt{3}.$$
7. $$\cos 20^\circ+\cos 40^\circ+\cos 60^\circ+\dots+\cos 160^\circ+\cos 180^\circ$$
   $$=(\cos 20^\circ+\cos 160^\circ)+(\cos 40^\circ+\cos 140^\circ)+(\cos 60^\circ+\cos 120^\circ)+(\cos 80^\circ+\cos 100^\circ)+\cos 180^\circ$$
   $$=(\cos 20^\circ-\cos 20^\circ)+(\cos 40^\circ-\cos 40^\circ)+(\cos 60^\circ-\cos 60^\circ)+(\cos 80^\circ-\cos 80^\circ)-1=-1.$$
8. $$\tg 20^\circ+\tg 40^\circ+\tg 60^\circ+\dots+\tg 160^\circ+\tg 180^\circ$$
   $$=(\tg 20^\circ+\tg 160^\circ)+(\tg 40^\circ+\tg 140^\circ)+(\tg 60^\circ+\tg 120^\circ)+(\tg 80^\circ+\tg 100^\circ)+\tg 180^\circ$$
   $$=(\tg 20^\circ-\tg 20^\circ)+(\tg 40^\circ-\tg 40^\circ)+(\tg 60^\circ-\tg 60^\circ)+(\tg 80^\circ-\tg 80^\circ)+0=0.$$

Ответ

а) $-1$; б) $-\frac{\sqrt{3}}{3}$; в) $\frac{\sqrt{3}}{3}$; г) $-\sqrt{3}$; д) $-18$; е) $-4\sqrt{3}$; ж) $-1$; з) $0$.