Упр.24.9 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

1)

$$\cos 3a-\cos 4a-\cos 5a+\cos 6a$$
$$=(\cos 6a+\cos 3a)-(\cos 5a+\cos 4a)$$
$$=2\cos \frac{9a}{2}\cos \frac{3a}{2}-2\cos \frac{9a}{2}\cos \frac{a}{2}$$
$$=2\cos \frac{9a}{2}\left(\cos \frac{3a}{2}-\cos \frac{a}{2}\right)$$
$$=2\cos \frac{9a}{2}\cdot\left(-2\sin \frac{a}{2}\sin a\right) =-4\sin \frac{a}{2}\sin a\cos \frac{9a}{2}.$$

2)

$$\cos\left(\frac{3\pi}{2}+4a\right)+\sin(3\pi-8a)-\sin(4\pi-12a)$$
$$=\sin 4a+\sin 8a+\sin 12a$$
$$=2\sin 6a\cos 2a+\sin 12a$$
$$=2\sin 6a\cos 2a+2\sin 6a\cos 6a$$
$$=2\sin 6a(\cos 2a+\cos 6a)$$
$$=2\sin 6a\cdot 2\cos 4a\cos 2a =4\cos 2a\cos 4a\sin 6a.$$

3)

$$\cos^2 a-\cos^2 \beta=(\cos a-\cos \beta)(\cos a+\cos \beta)$$
$$=-2\sin \frac{a+\beta}{2}\sin \frac{a-\beta}{2}\cdot 2\cos \frac{a+\beta}{2}\cos \frac{a-\beta}{2}$$
$$=-\sin(a+\beta)\sin(a-\beta) =\sin(a+\beta)\sin(\beta-a).$$

4)

$$\frac{\sin a+\sin 3a+\sin 5a+\sin 7a}{\cos a+\cos 3a+\cos 5a+\cos 7a}$$
$$=\frac{2\sin 2a\cos a+2\sin 6a\cos a}{2\cos 2a\cos a+2\cos 6a\cos a}$$
$$=\frac{\sin 2a+\sin 6a}{\cos 2a+\cos 6a}$$
$$=\frac{2\sin 4a\cos 2a}{2\cos 4a\cos 2a} =\tg 4a.$$

5)

$$\frac{2(\sin 2a+2\cos^2 a-1)}{\cos a-\sin a-\cos 3a+\sin 3a} = \frac{2(\sin 2a+\cos 2a)}{(\sin 3a-\sin a)-(\cos 3a-\cos a)}$$
$$=\frac{2(\sin 2a+\cos 2a)}{2\cos 2a\sin a+2\sin 2a\sin a}$$
$$=\frac{2(\sin 2a+\cos 2a)}{2\sin a(\sin 2a+\cos 2a)} =\frac{1}{\sin a}.$$

6)

$$\frac{1+\cos a+\cos 2a+\cos 3a}{\cos a+2\cos^2 a-1} = \frac{1+\cos a+\cos 2a+\cos 3a}{\cos a+\cos 2a}$$
$$=\frac{1+(2\cos^2 a-1)+2\cos 2a\cos a}{\cos a+\cos 2a}$$
$$=\frac{2\cos^2 a+2\cos 2a\cos a}{\cos a+\cos 2a} =\frac{2\cos a(\cos a+\cos 2a)}{\cos a+\cos 2a} =2\cos a.$$

7)

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

8)

$$\frac{(\sin a-\cos a)^2-1+\sin 4a}{\cos 2a+\cos 4a}$$
$$=\frac{\sin^2 a+\cos^2 a-2\sin a\cos a-1+\sin 4a}{\cos 2a+\cos 4a}$$
$$=\frac{-2\sin a\cos a+\sin 4a}{\cos 2a+\cos 4a} =\frac{\sin 4a-\sin 2a}{\cos 2a+\cos 4a}$$
$$=\frac{2\cos 3a\sin a}{2\cos 3a\cos a} =\tg a.$$

9)

$$\left(\frac{\sin a}{\sin 2a}-\frac{\cos a}{\cos 2a}\right)\cdot \frac{\cos a-\cos 7a}{\sin a}$$
$$=\frac{\sin a\cos 2a-\sin 2a\cos a}{\sin 2a\cos 2a}\cdot \frac{-2\sin 4a\sin(-3a)}{\sin a}$$
$$=\frac{\sin(a-2a)}{\frac12\sin 4a}\cdot \frac{2\sin 4a\sin 3a}{\sin a}$$
$$=\frac{-\sin a}{\frac12\sin 4a}\cdot \frac{2\sin 4a\sin 3a}{\sin a} =-4\sin 3a.$$

10)

$$\frac{(\cos a-\cos 3a)(\sin a+\sin 3a)}{1-\cos 4a}$$
$$=\frac{\left(-2\sin 2a\sin(-a)\right)\left(2\sin 2a\cos a\right)}{1-\cos 4a}$$
$$=\frac{4\sin^2 2a\sin a\cos a}{2\sin^2 2a} =2\sin a\cos a =\sin 2a.$$