Упр.1307 ГДЗ Макарычев Миндюк 9 класс (Углубленный) (Алгебра)

а) sin(-405°)+cos(750°); в) sin(2,5?)+ctg(-3?/4);
б) cos(-780°)-tg(-225°); г) tg(-7?/3)-ctg(10?/3).

1. $$\sin(-405^\circ)+\cos(750^\circ)=\sin(-45^\circ)+\cos 30^\circ$$

   $$\sin(-45^\circ)=-\frac{\sqrt2}{2}, \qquad \cos 30^\circ=\frac{\sqrt3}{2}$$

   $$\sin(-405^\circ)+\cos(750^\circ)=\frac{\sqrt3-\sqrt2}{2}$$

2. $$\cos(-780^\circ)-\tg(-225^\circ)=\cos(-60^\circ)-\tg(-45^\circ)$$

   $$\cos(-60^\circ)=\frac12, \qquad \tg(-45^\circ)=-1$$

   $$\cos(-780^\circ)-\tg(-225^\circ)=\frac12-(-1)=\frac32$$

3. $$\sin\left(2{,}5\pi\right)+\ctg\left(-\frac{3\pi}{4}\right)=\sin\left(\frac{5\pi}{2}\right)+\ctg\left(-\frac{3\pi}{4}\right)$$

   $$\sin\left(\frac{5\pi}{2}\right)=1, \qquad \ctg\left(-\frac{3\pi}{4}\right)=1$$

   $$\sin\left(2{,}5\pi\right)+\ctg\left(-\frac{3\pi}{4}\right)=1+1=2$$

4. $$\tg\left(-\frac{7\pi}{3}\right)-\ctg\left(\frac{10\pi}{3}\right)=\tg\left(-\frac{\pi}{3}\right)-\ctg\left(\frac{\pi}{3}\right)$$

   $$\tg\left(-\frac{\pi}{3}\right)=-\sqrt3, \qquad \ctg\left(\frac{\pi}{3}\right)=\frac{\sqrt3}{3}$$

   $$\tg\left(-\frac{7\pi}{3}\right)-\ctg\left(\frac{10\pi}{3}\right)=-\sqrt3-\frac{\sqrt3}{3}=-\frac{4\sqrt3}{3}$$

Ответ

а) $$\frac{\sqrt3-\sqrt2}{2}$$; б) $$\frac32$$; в) $$2$$; г) $$-\frac{4\sqrt3}{3}$$.