Упр.25.1 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

1) cos(15°)cos(5°); 3) 2sin(?/10)cos(?/40); 5) 2sin(a)sin(2a);
2) 2cos(3a)cos(2a); 4) sin(48°)sin(74°); 6) sin(60°+a)sin(60°-a).

1. $$\cos 15^\circ \cdot \cos 5^\circ=\frac12\bigl(\cos(15^\circ-5^\circ)+\cos(15^\circ+5^\circ)\bigr)$$
   $$=\frac12(\cos 10^\circ+\cos 20^\circ).$$
2. $$2\cos 3a \cdot \cos 2a=2\cdot \frac12\bigl(\cos(3a-2a)+\cos(3a+2a)\bigr)$$
   $$=\cos a+\cos 5a.$$
3. $$2\sin \frac{\pi}{10}\cdot \cos \frac{\pi}{40}=2\cdot \frac12\bigl(\sin\bigl(\frac{\pi}{10}-\frac{\pi}{40}\bigr)+\sin\bigl(\frac{\pi}{10}+\frac{\pi}{40}\bigr)\bigr)$$
   $$=\sin \frac{3\pi}{40}+\sin \frac{5\pi}{40}=\sin \frac{3\pi}{40}+\sin \frac{\pi}{8}.$$
4. $$\sin 48^\circ \cdot \sin 74^\circ=\frac12\bigl(\cos(74^\circ-48^\circ)-\cos(74^\circ+48^\circ)\bigr)$$
   $$=\frac12(\cos 26^\circ-\cos 122^\circ).$$
5. $$2\sin a \cdot \sin 2a=2\cdot \frac12\bigl(\cos(2a-a)-\cos(2a+a)\bigr)$$
   $$=\cos a-\cos 3a.$$
6. $$\sin(60^\circ+a)\cdot \sin(60^\circ-a)=\frac12\bigl(\cos((60^\circ+a)-(60^\circ-a))-\cos((60^\circ+a)+(60^\circ-a))\bigr)$$
   $$=\frac12(\cos 2a-\cos 120^\circ).$$
   Так как $$\cos 120^\circ=-\frac12,$$ то
   $$\sin(60^\circ+a)\cdot \sin(60^\circ-a)=\frac12\left(\cos 2a+\frac12\right)=\frac{2\cos 2a+1}{4}.$$

Ответ

1) $$\frac12(\cos 10^\circ+\cos 20^\circ)$$;
2) $$\cos a+\cos 5a$$;
3) $$\sin \frac{3\pi}{40}+\sin \frac{\pi}{8}$$;
4) $$\frac12(\cos 26^\circ-\cos 122^\circ)$$;
5) $$\cos a-\cos 3a$$;
6) $$\frac{2\cos 2a+1}{4}$$.