Упр.9.67 ГДЗ Никольский 10 класс (Алгебра)

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1. $$\sin \frac{11\pi}{24}\sin \frac{5\pi}{24}=\frac12\left(\cos\frac{11\pi-5\pi}{24}-\cos\frac{11\pi+5\pi}{24}\right)$$
   $$=\frac12\left(\cos\frac{\pi}{4}-\cos\frac{2\pi}{3}\right)=\frac12\left(\frac{\sqrt2}{2}+\frac12\right)=\frac{\sqrt2+1}{4}.$$

2. $$\cos \frac{13\pi}{24}\cos \frac{3\pi}{24}=\frac12\left(\cos\frac{13\pi-3\pi}{24}+\cos\frac{13\pi+3\pi}{24}\right)$$
   $$=\frac12\left(\cos\frac{5\pi}{12}+\cos\frac{\pi}{4}\right)=\frac12\left(\frac{\sqrt3-1}{2\sqrt2}+\frac{\sqrt2}{2}\right)=\frac{\sqrt2-\sqrt3}{4}.$$

3. $$\sin \frac{2\pi}{24}\cos \frac{\pi}{24}=\frac12\left(\sin\frac{2\pi+\pi}{24}+\sin\frac{2\pi-\pi}{24}\right)$$
   $$=\frac12\left(\sin\frac{\pi}{8}+\sin\frac{\pi}{24}\right)=\frac{\sqrt3+\sqrt2}{4}.$$

4. $$\cos 63^\circ\cos 27^\circ-\sin 12^\circ\sin 48^\circ$$
   $$=\frac12\left(\cos(63^\circ-27^\circ)+\cos(63^\circ+27^\circ)\right)-\frac12\left(\cos(12^\circ-48^\circ)-\cos(12^\circ+48^\circ)\right)$$
   $$=\frac12(\cos36^\circ+\cos90^\circ)-\frac12(\cos(-36^\circ)-\cos60^\circ)$$
   $$=\frac12\cos36^\circ-\frac12\cos36^\circ+\frac14=\frac14.$$

5. $$\cos \frac{11\pi}{56}\cos \frac{3\pi}{56}-\sin \frac{11\pi}{42}\sin \frac{17\pi}{42}$$
   $$=\frac12\left(\cos\frac{11\pi-3\pi}{56}+\cos\frac{11\pi+3\pi}{56}\right)-\frac12\left(\cos\frac{11\pi-17\pi}{42}-\cos\frac{11\pi+17\pi}{42}\right)$$
   $$=\frac12\left(\cos\frac{\pi}{7}+\cos\frac{\pi}{4}\right)-\frac12\left(\cos\frac{2\pi}{3}-\cos\frac{2\pi}{3}\right)$$
   $$=\frac12\cdot\frac{\sqrt2}{2}-\frac12\cdot\left(-\frac12\right)=\frac{\sqrt2-1}{4}.$$

6. $$\sin \frac{47\pi}{24}\cos \frac{\pi}{24}=\frac12\left(\sin\frac{47\pi+\pi}{24}+\sin\frac{47\pi-\pi}{24}\right)$$
   $$=\frac12\left(\sin 2\pi+\sin\frac{23\pi}{12}\right)=\frac12\left(0-\sin\frac{\pi}{12}\right)=\frac{\sqrt2-\sqrt6}{8}.$$

7. $$\cos \frac{23\pi}{24}\cos \frac{4\pi}{24}+\cos \frac{20\pi}{24}\cos \frac{6\pi}{24}$$
   $$=\frac12\left(\cos\frac{23\pi-4\pi}{24}+\cos\frac{23\pi+4\pi}{24}\right)+\frac12\left(\cos\frac{20\pi-6\pi}{24}+\cos\frac{20\pi+6\pi}{24}\right)$$
   $$=\frac12\left(\cos\frac{19\pi}{24}+\cos\frac{9\pi}{8}\right)+\frac12\left(\cos\frac{7\pi}{12}+\cos\frac{13\pi}{12}\right)$$
   $$=\frac{\sqrt2-\sqrt3}{4}.$$

8. $$\sin \frac{4\pi}{24}\cos \frac{\pi}{24}=\frac12\left(\sin\frac{5\pi}{24}+\sin\frac{3\pi}{24}\right)$$
   $$=\frac12\sin\frac{\pi}{3}+\frac12\sin\frac{\pi}{8}=\frac{\sqrt3+\sqrt2}{4}.$$

9. $$\cos 63^\circ\cos 27^\circ-\sin 12^\circ\sin 48^\circ=\frac14.$$

10. $$\cos \frac{11\pi}{56}\cos \frac{3\pi}{56}-\sin \frac{11\pi}{42}\sin \frac{17\pi}{42}=\frac{\sqrt2-1}{4}.$$

Ответ

а) $$\frac{\sqrt2+1}{4}$$; б) $$\frac{\sqrt2-\sqrt3}{4}$$; в) $$\frac{\sqrt3+\sqrt2}{4}$$; г) $$\frac14$$; д) $$\frac{\sqrt2-1}{4}$$; е) $$\frac{\sqrt2-\sqrt6}{8}$$; ж) $$\frac{\sqrt2-\sqrt3}{4}$$; з) $$\frac{\sqrt3+\sqrt2}{4}$$; и) $$\frac14$$; к) $$\frac{\sqrt2-1}{4}$$.