Упр.1180 ГДЗ Никольский Потапов 9 класс (Алгебра)

а) cos(37°30′)cos(7°30′); б) sin(75°)sin(15°);
в) cos(15°)cos(75°); г) sin(5?/24)sin(?/24);
д) sin(15°)+tg(30°)cos(15°); е) tg(11?/12)+tg(5?/12);
ж) (tg(13°)+tg(17°))/(1-tg(13°)tg(17°)); з) (tg(1°)-tg(46°))/(1+tg(1°)tg(46°));
и) (tg(113°)+tg(7°))/(1-tg(113°)tg(7°)); к) (tg(150°)-tg(15°))/(1+tg(150°)tg(15°)).

1. $$\cos 37^\circ 30′ \cdot \cos 7^\circ 30’=\frac12\bigl(\cos(37^\circ 30′-7^\circ 30′)+\cos(37^\circ 30’+7^\circ 30′)\bigr)$$
   $$=\frac12(\cos 30^\circ+\cos 45^\circ)=\frac12\left(\frac{\sqrt3}{2}+\frac{\sqrt2}{2}\right)=\frac{\sqrt3+\sqrt2}{4}.$$
2. $$\sin 75^\circ \cdot \sin 15^\circ=\frac12\bigl(\cos(75^\circ-15^\circ)-\cos(75^\circ+15^\circ)\bigr)$$
   $$=\frac12(\cos 60^\circ-\cos 90^\circ)=\frac12\left(\frac12-0\right)=\frac14.$$
3. $$\cos 15^\circ \cdot \cos 75^\circ=\frac12\bigl(\cos(75^\circ-15^\circ)+\cos(75^\circ+15^\circ)\bigr)$$
   $$=\frac12(\cos 60^\circ+\cos 90^\circ)=\frac12\left(\frac12+0\right)=\frac14.$$
4. $$\sin \frac{5\pi}{24}\cdot \sin \frac{\pi}{24}=\frac12\left(\cos\left(\frac{5\pi}{24}-\frac{\pi}{24}\right)-\cos\left(\frac{5\pi}{24}+\frac{\pi}{24}\right)\right)$$
   $$=\frac12\left(\cos \frac{\pi}{6}-\cos \frac{\pi}{4}\right)=\frac12\left(\frac{\sqrt3}{2}-\frac{\sqrt2}{2}\right)=\frac{\sqrt3-\sqrt2}{4}.$$
5. $$\sin 15^\circ+\tg 30^\circ \cos 15^\circ=\sin 15^\circ+\frac{\sin 30^\circ}{\cos 30^\circ}\cos 15^\circ$$
   $$=\frac{\sin 15^\circ\cos 30^\circ+\cos 15^\circ\sin 30^\circ}{\cos 30^\circ}=\frac{\sin(15^\circ+30^\circ)}{\cos 30^\circ}$$
   $$=\frac{\sin 45^\circ}{\cos 30^\circ}=\frac{\frac{\sqrt2}{2}}{\frac{\sqrt3}{2}}=\frac{\sqrt6}{3}.$$
6. $$\tg \frac{11\pi}{12}+\tg \frac{5\pi}{12}=\frac{\sin \frac{11\pi}{12}}{\cos \frac{11\pi}{12}}+\frac{\sin \frac{5\pi}{12}}{\cos \frac{5\pi}{12}}$$
   $$=\frac{\sin \frac{11\pi}{12}\cos \frac{5\pi}{12}+\sin \frac{5\pi}{12}\cos \frac{11\pi}{12}}{\cos \frac{11\pi}{12}\cos \frac{5\pi}{12}}$$
   $$=\frac{\sin\left(\frac{11\pi}{12}+\frac{5\pi}{12}\right)}{\frac12\left(\cos\left(\frac{11\pi}{12}-\frac{5\pi}{12}\right)+\cos\left(\frac{11\pi}{12}+\frac{5\pi}{12}\right)\right)}$$
   $$=\frac{\sin \frac{4\pi}{3}}{\frac12(\cos \frac{\pi}{2}+\cos \frac{4\pi}{3})} =\frac{2\sin \frac{4\pi}{3}}{0+\cos \frac{4\pi}{3}} =2\tg \frac{4\pi}{3}=2\tg\left(\pi+\frac{\pi}{3}\right)=2\tg \frac{\pi}{3}=2\sqrt3.$$
7. $$\frac{\tg 13^\circ+\tg 17^\circ}{1-\tg 13^\circ \tg 17^\circ}=\tg(13^\circ+17^\circ)=\tg 30^\circ=\frac{\sqrt3}{3}.$$
8. $$\frac{\tg 1^\circ-\tg 46^\circ}{1+\tg 1^\circ \tg 46^\circ}=\tg(1^\circ-46^\circ)=\tg(-45^\circ)=-1.$$
9. $$\frac{\tg 113^\circ+\tg 7^\circ}{1-\tg 113^\circ \tg 7^\circ}=\tg(113^\circ+7^\circ)=\tg 120^\circ$$
   $$=\tg(180^\circ-60^\circ)=-\tg 60^\circ=-\sqrt3.$$
10. $$\frac{\tg 150^\circ-\tg 15^\circ}{1+\tg 150^\circ \tg 15^\circ}=\tg(150^\circ-15^\circ)=\tg 135^\circ$$
    $$=\tg(180^\circ-45^\circ)=-\tg 45^\circ=-1.$$

Ответ

а) $$\frac{\sqrt3+\sqrt2}{4}$$; б) $$\frac14$$; в) $$\frac14$$; г) $$\frac{\sqrt3-\sqrt2}{4}$$; д) $$\frac{\sqrt6}{3}$$; е) $$2\sqrt3$$; ж) $$\frac{\sqrt3}{3}$$; з) $$-1$$; и) $$-\sqrt3$$; к) $$-1$$.