Question

помогите решить эти два задания

Answer 1

$Cos\alpha =-\sqrt{1-Sin^{2} \alpha}=- \sqrt{1-(-\frac{15}{17}) ^{2}}= -\sqrt{1-\frac{225}{289} }=- \sqrt{\frac{64}{289}}=- \frac{8}{17}\\\\Sin\beta=- \sqrt{1-Cos^{2} \beta} =- \sqrt{1-(\frac{8}{17})^{2}}=- \sqrt{1-\frac{64}{289}}=- \sqrt{\frac{225}{289}}=- \frac{15}{17} \\\\Cos(\alpha+ \beta)=Cos \alpha Cos \beta-Sin \alpha Sin \beta=- \frac{8}{17}* \frac{8}{17}-(- \frac{15}{17})*(- \frac{15}{17})=- \frac{64}{289}- \frac{225}{289}=- \frac{289}{289}=-1$

$2)Sin\alpha= \sqrt{1-Cos^{2} \alpha} = \sqrt{1-(\frac{3}{5})^{2}}= \sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}= \frac{4}{5}\\\\Sin2\alpha =2Sin\alpha Cos \alpha = 2*\frac{4}{5} *\frac{3}{5}= \frac{24}{25} =0,96\\Cos2\alpha=2Cos ^{2} \alpha -1=2*(\frac{3}{5}) ^{2}-1=2* \frac{9}{25}-1= \frac{18}{25}-1=- \frac{7}{25}$

$tg2\alpha =\frac{Sin2\alpha} {Cos2\alpha}=  \frac{24}{25}:(- \frac{7}{25})=- \frac{24*25}{25*7}=- \frac{24}{7}=-3 \frac{3}{7}$