Упр.1485 ГДЗ Макарычев Миндюк 9 класс (Углубленный) (Алгебра)

а) (sin(?)+sin(?))^2+(cos(?)+cos(?))^2=4cos^2((?-?)/2);
б) sin^2(x+y)-sin^2(x-y)=sin(2x)sin(2y);
в) cos^2(?+?)-cos^2(?-?)=-sin^2(?)sin(2?);
г) (sin(x)-sin(y))^2-(cos(x)-cos(y))^2=4sin^2((x-y)/2);
д) (sin(?)+sin(3?)+sin(5?))/(cos(?)+cos(3?)+cos(5?))=tg(3?);
е) sin(x)+sin(3x)+sin(5x)+sin(7x)=4cos(x)cos(2x)sin(4x).

1. а)

   $$\begin{aligned} (\sin a+\sin \beta)^2+(\cos a+\cos \beta)^2 &=\sin^2 a+\sin^2 \beta+2\sin a\sin \beta+\cos^2 a+\cos^2 \beta+2\cos a\cos \beta \\ &=2+2(\sin a\sin \beta+\cos a\cos \beta) \\ &=2+2\cos(a-\beta) \\ &=4\cdot \frac{1+\cos(a-\beta)}{2} \\ &=4\cos^2\frac{a-\beta}{2}. \end{aligned}$$

2. б)

   $$\begin{aligned} \sin^2(x+y)-\sin^2(x-y) &=\frac{1-\cos(2x+2y)}{2}-\frac{1-\cos(2x-2y)}{2} \\ &=\frac{\cos(2x-2y)-\cos(2x+2y)}{2} \\ &=\sin 2x\sin 2y. \end{aligned}$$

3. в)

   $$\begin{aligned} \cos^2(a+\beta)-\cos^2(a-\beta) &=\frac{1+\cos(2a+2\beta)}{2}-\frac{1+\cos(2a-2\beta)}{2} \\ &=\frac{\cos(2a+2\beta)-\cos(2a-2\beta)}{2} \\ &=-\sin 2a\sin 2\beta. \end{aligned}$$

4. г)

   $$\begin{aligned} (\sin x-\sin y)^2+(\cos x-\cos y)^2 &=\sin^2 x+\sin^2 y-2\sin x\sin y+\cos^2 x+\cos^2 y-2\cos x\cos y \\ &=2-2(\cos x\cos y+\sin x\sin y) \\ &=2-2\cos(x-y) \\ &=4\sin^2\frac{x-y}{2}. \end{aligned}$$

5. д)

   $$\begin{aligned} \frac{\sin a+\sin 3a+\sin 5a}{\cos a+\cos 3a+\cos 5a} &=\frac{(\sin a+\sin 5a)+\sin 3a}{(\cos a+\cos 5a)+\cos 3a} \\ &=\frac{2\sin 3a\cos 2a+\sin 3a}{2\cos 3a\cos 2a+\cos 3a} \\ &=\frac{\sin 3a(2\cos 2a+1)}{\cos 3a(2\cos 2a+1)} \\ &=\tg 3a. \end{aligned}$$

6. е)

   $$\begin{aligned} \sin x+\sin 3x+\sin 5x+\sin 7x &=(\sin x+\sin 7x)+(\sin 3x+\sin 5x) \\ &=2\sin 4x\cos 3x+2\sin 4x\cos x \\ &=2\sin 4x(\cos 3x+\cos x) \\ &=2\sin 4x\cdot 2\cos 2x\cos x \\ &=4\cos x\cos 2x\sin 4x. \end{aligned}$$

Ответ

Все тождества доказаны.