Question

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

Answer 1

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

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