378. Calculate: a) $\sin \left(-60^{\circ}\right)$; b) $\cos \left(-45^{\circ}\right)$; c) $\operatorname{tg}\left(-945^{\circ}\right)$.
Solution. a) Since $\sin x$ is an odd function, then $\sin \left(-60^{\circ}\right)=$ $=-\sin 60^{\circ}$. Therefore, $\sin \left(-60^{\circ}\right)=-\sin 60^{\circ}=-\sqrt{3} / 2$.

6) The function $\cos x$ is even; therefore, the minus sign can be omitted, i.e., $\cos \left(-45^{\circ}\right)=\cos 45^{\circ}$. Therefore, $\cos \left(-45^{\circ}\right)=\cos 45^{\circ}=\sqrt{2} / 2$.

c) We will use the properties of periodicity and oddness. Since the function $\operatorname{tg} x$ is odd, then $\operatorname{tg}\left(-945^{\circ}\right)=-\operatorname{tg} 945^{\circ}$. Further, the period of the function $\operatorname{tg} x$ is $180^{\circ}$, and therefore, $\operatorname{tg}\left(-945^{\circ}\right)=-\operatorname{tg} 945^{\circ}=$ $=-\operatorname{tg}\left(180^{\circ} \cdot 5+45^{\circ}\right)=-\operatorname{tg} 45^{\circ}=-1$.

379-382. Simplify the expressions: