Упр.23.1 ГДЗ Мордкович 10-11 класс (Алгебра)

a) sin 23 * sin 32;
б) cos пи/12 * cos пи/8;
в) sin 14 * cos 16;
г) 2sin пи/8 * cos пи/5.

1. $$\sin 23^\circ \cdot \sin 32^\circ=\frac{1}{2}\bigl(\cos(23^\circ-32^\circ)-\cos(23^\circ+32^\circ)\bigr)$$
   $$=\frac{1}{2}\bigl(\cos 9^\circ-\cos 55^\circ\bigr).$$
2. $$\cos \frac{\pi}{12}\cdot \cos \frac{\pi}{8}=\frac{1}{2}\left(\cos\left(\frac{\pi}{12}-\frac{\pi}{8}\right)+\cos\left(\frac{\pi}{12}+\frac{\pi}{8}\right)\right)$$
   $$=\frac{1}{2}\left(\cos\frac{\pi}{24}+\cos\frac{5\pi}{24}\right).$$
3. $$\sin 14^\circ \cdot \cos 16^\circ=\frac{1}{2}\bigl(\sin(14^\circ+16^\circ)+\sin(14^\circ-16^\circ)\bigr)$$
   $$=\frac{1}{2}\bigl(\sin 30^\circ+\sin(-2^\circ)\bigr)$$
   $$=\frac{1}{2}\left(\frac{1}{2}-\sin 2^\circ\right)=\frac{1}{4}-\frac{1}{2}\sin 2^\circ.$$
4. $$2\sin \frac{\pi}{8}\cdot \cos \frac{\pi}{5}=2\cdot \frac{1}{2}\left(\sin\left(\frac{\pi}{8}+\frac{\pi}{5}\right)+\sin\left(\frac{\pi}{8}-\frac{\pi}{5}\right)\right)$$
   $$=\sin\frac{13\pi}{40}+\sin\left(-\frac{3\pi}{40}\right)=\sin\frac{13\pi}{40}-\sin\frac{3\pi}{40}.$$

Ответ

а) $$\frac{1}{2}\bigl(\cos 9^\circ-\cos 55^\circ\bigr)$$;
б) $$\frac{1}{2}\left(\cos\frac{\pi}{24}+\cos\frac{5\pi}{24}\right)$$;
в) $$\frac{1}{4}-\frac{1}{2}\sin 2^\circ$$;
г) $$\sin\frac{13\pi}{40}-\sin\frac{3\pi}{40}$$.