Question

cho \(tan2a=-\frac{4}{3}\) với \(\frac{\pi}{2}< \alpha< \pi\). Tính \(cos2a\) và \(tan\left(2a+\frac{\pi}{4}\right)\)

Answer 1

\(\frac{\pi}{2}< a< \pi\Rightarrow\pi< 2a< 2\pi\)

Mà \(tan2a< 0\) \(\Rightarrow\frac{3\pi}{2}< 2a< 2\pi\Rightarrow cos2a>0\)

\(\Rightarrow cos2a=\frac{1}{\sqrt{1+tan^22a}}=\frac{3}{5}\)

\(tan\left(2a+\frac{\pi}{4}\right)=\frac{tan2a+tan\frac{\pi}{4}}{1-tan2a.tan\frac{\pi}{4}}=\frac{-\frac{4}{3}+1}{1+\frac{4}{3}}=...\)