Question

sin$alpha$=$frac{4}{5}$; $frac{pi }{2}$<$alpha$<$pi$ ; Знайти cos2$alpha$, sin2$alpha$,ctg2$alpha$,tg2$alpha$

Answer 1

$sina=frac{4}{5}; ,\\frac{pi}{2}<a<pi ; ; Rightarrow ; ; underline {cosa<0; !!!}; ,; tga<0; ,; ctga<0\\1); ; cos2a=1-2sin^2a=1-2cdot frac{16}{25}=frac{25-32}{25}=-frac{7}{25}\\2); ; sin2a=2, sinacdot underbrace {cosa}_{<0}=2cdot sinacdot (-sqrt{1-sin^2a})=\\=-2cdot frac{4}{5}cdot sqrt{1-frac{16}{25}}=-2cdot frac{4}{5}cdot sqrt{frac{9}{25}}=-2cdot frac{4}{5}cdot frac{3}{5}=-frac{24}{25}\\3); ; tg2a=frac{2tga}{1+tg^2a}\\tga<0; ,; ; tga=frac{sina}{cosa}=frac{frac{4}{5}}{-frac{3}{5}}=-frac{4}{3}$

$tg2a=frac{2cdot (-frac{4}{3})}{1+frac{16}{9}}=-frac{8cdot 9}{3cdot (9+16)}=-frac{8}{3cdot 25}=-frac{8}{75}\\4); ; ctg2a=frac{1}{tg2a}=-frac{75}{8}$