Упр.1478 ГДЗ Макарычев Миндюк 9 класс (Углубленный) (Алгебра)

а) cos(x)+sin(y); д) cos(3?)-sin(?);
б) sin(x)-cos(y); е) sin(?)+cos(3?);
в) sin(?)+cos(?); ж) sin(2?)-cos(?);
г) cos(?)-sin(?); з) cos(?)+sin(2?).

Используем формулы преобразования суммы в произведение:

$$\cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$$
$$\sin A-\sin B=2\cos\frac{A+B}{2}\sin\frac{A-B}{2}$$
$$\sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$$
$$\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$$

1. $$\cos x+\sin y=\cos x+\cos\left(\frac{\pi}{2}-y\right)$$

   $$=2\cos\left(\frac{x+\frac{\pi}{2}-y}{2}\right)\cos\left(\frac{x-\left(\frac{\pi}{2}-y\right)}{2}\right)$$

   $$=2\cos\left(\frac{x-y}{2}+\frac{\pi}{4}\right)\cos\left(\frac{x+y}{2}-\frac{\pi}{4}\right)$$

2. $$\sin x-\cos y=\sin x-\sin\left(\frac{\pi}{2}-y\right)$$

   $$=2\cos\left(\frac{x+\frac{\pi}{2}-y}{2}\right)\sin\left(\frac{x-\left(\frac{\pi}{2}-y\right)}{2}\right)$$

   $$=2\sin\left(\frac{x+y}{2}-\frac{\pi}{4}\right)\cos\left(\frac{x-y}{2}+\frac{\pi}{4}\right)$$

3. $$\sin a+\cos a=\sin a+\sin\left(\frac{\pi}{2}-a\right)$$

   $$=2\sin\left(\frac{a+\frac{\pi}{2}-a}{2}\right)\cos\left(\frac{a-\left(\frac{\pi}{2}-a\right)}{2}\right)$$

   $$=2\sin\frac{\pi}{4}\cos\left(a-\frac{\pi}{4}\right)=\sqrt{2}\cos\left(a-\frac{\pi}{4}\right)$$

4. $$\cos a-\sin a=\cos a-\cos\left(\frac{\pi}{2}-a\right)$$

   $$=-2\sin\left(\frac{a+\frac{\pi}{2}-a}{2}\right)\sin\left(\frac{a-\left(\frac{\pi}{2}-a\right)}{2}\right)$$

   $$=-2\sin\frac{\pi}{4}\sin\left(a-\frac{\pi}{4}\right)=\sqrt{2}\sin\left(\frac{\pi}{4}-a\right)$$

5. $$\cos 3a-\sin a=\cos 3a-\cos\left(\frac{\pi}{2}-a\right)$$

   $$=-2\sin\left(\frac{3a+\frac{\pi}{2}-a}{2}\right)\sin\left(\frac{3a-\left(\frac{\pi}{2}-a\right)}{2}\right)$$

   $$=2\sin\left(\frac{\pi}{4}+a\right)\sin\left(\frac{\pi}{4}-2a\right)$$

6. $$\sin a+\cos 3a=\sin a+\sin\left(\frac{\pi}{2}-3a\right)$$

   $$=2\sin\left(\frac{a+\frac{\pi}{2}-3a}{2}\right)\cos\left(\frac{a-\left(\frac{\pi}{2}-3a\right)}{2}\right)$$

   $$=2\sin\left(\frac{\pi}{4}-a\right)\cos\left(\frac{4a-\frac{\pi}{2}}{2}\right)=2\sin\left(\frac{\pi}{4}-a\right)\cos\left(2a-\frac{\pi}{4}\right)$$

7. $$\sin 2a-\cos a=\sin 2a-\sin\left(\frac{\pi}{2}-a\right)$$

   $$=2\cos\left(\frac{2a+\frac{\pi}{2}-a}{2}\right)\sin\left(\frac{2a-\left(\frac{\pi}{2}-a\right)}{2}\right)$$

   $$=2\cos\left(\frac{a}{2}+\frac{\pi}{4}\right)\sin\left(\frac{3a}{2}-\frac{\pi}{4}\right)$$

8. $$\cos a+\sin 2a=\sin\left(\frac{\pi}{2}-a\right)+\sin 2a$$

   $$=2\sin\left(\frac{\frac{\pi}{2}-a+2a}{2}\right)\cos\left(\frac{\frac{\pi}{2}-a-2a}{2}\right)$$

   $$=2\sin\left(\frac{\pi}{4}+\frac{a}{2}\right)\cos\left(\frac{\pi}{4}-\frac{3a}{2}\right)$$

Ответ

$$ \begin{aligned} &\text{а) } \cos x+\sin y=2\cos\left(\frac{x-y}{2}+\frac{\pi}{4}\right)\cos\left(\frac{x+y}{2}-\frac{\pi}{4}\right);\\ &\text{б) } \sin x-\cos y=2\sin\left(\frac{x+y}{2}-\frac{\pi}{4}\right)\cos\left(\frac{x-y}{2}+\frac{\pi}{4}\right);\\ &\text{в) } \sin a+\cos a=\sqrt{2}\cos\left(a-\frac{\pi}{4}\right);\\ &\text{г) } \cos a-\sin a=\sqrt{2}\sin\left(\frac{\pi}{4}-a\right);\\ &\text{д) } \cos 3a-\sin a=2\sin\left(\frac{\pi}{4}+a\right)\sin\left(\frac{\pi}{4}-2a\right);\\ &\text{е) } \sin a+\cos 3a=2\sin\left(\frac{\pi}{4}-a\right)\cos\left(2a-\frac{\pi}{4}\right);\\ &\text{ж) } \sin 2a-\cos a=2\cos\left(\frac{a}{2}+\frac{\pi}{4}\right)\sin\left(\frac{3a}{2}-\frac{\pi}{4}\right);\\ &\text{з) } \cos a+\sin 2a=2\sin\left(\frac{\pi}{4}+\frac{a}{2}\right)\cos\left(\frac{\pi}{4}-\frac{3a}{2}\right). \end{aligned} $$