Помогите пожалуйста решить 1,3,6 задания

Question 1

Question 2

$1)sin(\alpha +\beta)-2sin\beta cos\alpha=sin\alpha cos\beta +cos\alpha sin\beta -2sin\beta cos\alpha=$
$sin\alpha cos\beta -cos\alpha sin\beta =sin(\alpha -\beta )=sin(\frac{\pi}{2})=1$

$sin^2\alpha -\frac{cos(\frac{\pi}{2}+\alpha)sin(\pi-\alpha)}{tg(\pi-\alpha)ctg(\frac{3\pi}{2}+\alpha)}=sin^2\alpha-\frac{-sin\alpha*sin\alpha}{-tg\alpha*tg\alpha}=$
$=sin^2\alpha-\frac{-sin^2\alpha}{tg^2\alpha}=sin^2\alpha+\frac{sin^2\alpha}{tg^2\alpha}=sin^2\alpha+cos^2\alpha=1$

$3. 1)cos\alpha=\frac{sin\alpha}{tg\alpha}$
$\alpha \in (\frac{3\pi}{2};2\pi)$ , следоват. $\alpha \in$ 4 четверти, tg a<0<br> $tg\alpha=\frac{sin\alpha}{\sqrt{1-sin^2\alpha}}=\frac{\frac{-12}{13}}{\sqrt{1-(\frac{-12}{13})^2}}=\frac{-12}{13\sqrt{1-\frac{144}{169}}}=\frac{-12}{13\sqrt{\frac{25}{169}}}=$
$=\frac{-12\sqrt{169}}{13\sqrt{25}}=\frac{-12*13}{13*5}=-\frac{12}{5}=-2,4$
$cos\alpha=\frac{-12}{13}:\frac{-12}{5}=\frac{-12}{13}*\frac{5}{-12}=\frac{5}{13}$

$2)sin(2\alpha)=2cos\alpha sin\alpha=2*\frac{5}{13}*\frac{-12}{13}=$
$=-\frac{120}{169}$

$3)cos(2\alpha)=1-2sin^2\alpha=1-2*\frac{144}{169}=1-\frac{288}{169}=-\frac{119}{169}$

$6. cos\frac{\pi}{9}cos\frac{2\pi}{9}cos\frac{4\pi}{9}=\frac{1}{8}$
$\frac{2sin\frac{\pi}{9}*cos\frac{\pi}{9}*cos\frac{2\pi}{9}*cos\frac{4pi}{9}}{2sin\frac{\pi}{9}}=\frac{1}{8}$

$\frac{2sin\frac{2\pi}{9}*cos\frac{2\pi}{9}*cos\frac{4\pi}{9}}{4sin\frac{\pi}{9}}=\frac{1}{8}$

$\frac{2sin\frac{4\pi}{9}*cos\frac{4\pi}{9}}{8sin\frac{\pi}{9}}=\frac{1}{8}$

$\frac{sin\frac{8\pi}{9}}{8sin\frac{\pi}{9}}=\frac{1}{8}$

$\frac{sin\frac{\pi}{9}}{8sin\frac{\pi}{9}}=\frac{1}{8}$

$\frac{1}{8}=\frac{1}{8}$