Упр.538 ГДЗ Алимов 10-11 класс (Алгебра)

1) cos 105° + cos 75°;
2) sin 105 — sin 75;
3) cos 11пи/12 + cos 5пи/12;
4) cos 11пи/12- cos 5пи.12;
5) sin 7пи/12 — sin пи/12;
6) sin 105 + sin 165.

1. $$\cos 105^\circ+\cos 75^\circ=2\cos\frac{105^\circ+75^\circ}{2}\cos\frac{105^\circ-75^\circ}{2}$$
   $$=2\cos 90^\circ\cos 15^\circ=0.$$

2. $$\sin 105^\circ-\sin 75^\circ=2\cos\frac{105^\circ+75^\circ}{2}\sin\frac{105^\circ-75^\circ}{2}$$
   $$=2\cos 90^\circ\sin 15^\circ=0.$$

3. $$\cos\frac{11\pi}{12}+\cos\frac{5\pi}{12}=2\cos\frac{\frac{11\pi}{12}+\frac{5\pi}{12}}{2}\cos\frac{\frac{11\pi}{12}-\frac{5\pi}{12}}{2}$$
   $$=2\cos\frac{2\pi}{3}\cos\frac{\pi}{4}=2\cdot\left(-\frac12\right)\cdot\frac{\sqrt2}{2}=-\frac{\sqrt2}{2}.$$

4. $$\cos\frac{11\pi}{12}-\cos\frac{5\pi}{12}=-2\sin\frac{\frac{11\pi}{12}+\frac{5\pi}{12}}{2}\sin\frac{\frac{11\pi}{12}-\frac{5\pi}{12}}{2}$$
   $$=-2\sin\frac{2\pi}{3}\sin\frac{\pi}{4}=-2\cdot\frac{\sqrt3}{2}\cdot\frac{\sqrt2}{2}=-\frac{\sqrt6}{2}.$$

5. $$\sin\frac{7\pi}{12}-\sin\frac{\pi}{12}=2\cos\frac{\frac{7\pi}{12}+\frac{\pi}{12}}{2}\sin\frac{\frac{7\pi}{12}-\frac{\pi}{12}}{2}$$
   $$=2\cos\frac{\pi}{3}\sin\frac{\pi}{4}=2\cdot\frac12\cdot\frac{\sqrt2}{2}=\frac{\sqrt2}{2}.$$

6. $$\sin 105^\circ+\sin 165^\circ=2\sin\frac{105^\circ+165^\circ}{2}\cos\frac{105^\circ-165^\circ}{2}$$
   $$=2\sin 135^\circ\cos(-30^\circ)=2\cdot\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2}=\frac{\sqrt6}{2}.$$

Ответ

1) $$0$$; 2) $$0$$; 3) $$-\frac{\sqrt2}{2}$$; 4) $$-\frac{\sqrt6}{2}$$; 5) $$\frac{\sqrt2}{2}$$; 6) $$\frac{\sqrt6}{2}$$.