Question

ОЧЕНЬ СРОЧНО!!! ДАЮ 100 БАЛЛОВ!

Answer 1

Объяснение:

$tg(\alpha +\beta )=\frac{tg\alpha +tg\beta }{1-tg\alpha *tg\beta }.\\a)\ sin\alpha =\frac{1}{\sqrt{5} } \ \ \ \ \frac{\pi }{2} \leq \alpha \leq \pi \\ sin^2\alpha +cos^2\alpha =1\ \ \ \ \Rightarrow\\cos^2\alpha =1-sin^2\alpha=1-(\frac{1}{\sqrt{5} } )^2=1-\frac{1}{5} =\frac{4}{5}.\\ cos\alpha =б\sqrt{\frac{4}{5} } =б\frac{2}{\sqrt{5} }\ \ \ \ \ \frac{\pi }{2} \leq \alpha \leq \pi\ \ \ \ \ \Rightarrow\\ cos\alpha =-\frac{2}{\sqrt{5} } .$

$tg\alpha =\frac{sin\alpha }{cos\alpha } =\frac{\frac{1}{\sqrt{5} } }{-\frac{2}{\sqrt{5} } } =-\frac{1}{2}.$

$b)\ cos\beta =-\frac{1}{3}\ \ \ \ \ \pi \leq \beta \leq \frac{3}{2} \pi \ \ \ \ \Rightarrow\\ sin^2\beta +cos^2\beta =1\\sin^2\beta =1-cos^2\beta =1-(-\frac{1}{3} )^2=1-\frac{1}{9}=\frac{8}{9}.\\ sin\beta =б\sqrt{\frac{8}{9} }= б\frac{2\sqrt{2} }{3} .\ \ \ \ \pi \leq \beta \leq \frac{3}{2} \pi \ \ \ \ \Rightarrow\\ sin\beta =-\frac{2\sqrt{2} }{3} .\\tg\beta =\frac{sin\beta }{cos\beta }=\frac{-\frac{2\sqrt{2} }{3} }{-\frac{1}{3} }=2\sqrt{2}.$

$c)\ tg(\alpha +\beta )=\frac{-\frac{1}{2}+2\sqrt{2} }{1-(-\frac{1}{2}*2\sqrt{2}) } =\frac{\frac{-1+4\sqrt2 }{2} }{1+\sqrt{2} } =\frac{4\sqrt{2}-1 }{2*(\sqrt{2}+1 )} =\frac{(4\sqrt{2}-1)*(\sqrt{2}-1) }{2*(\sqrt{2}+1)*(\sqrt{2}-1) } =\\=\frac{4*2-4\sqrt{2}-\sqrt{2} +1 }{2*1} =\frac{9-5\sqrt{2} }{2} =4,5-2,5\sqrt{2} .$