Question

cho \(sin\alpha+cos\alpha=\frac{1}{2}\) với \(\frac{3\pi}{4}< \alpha< \pi\). tính \(tan2\alpha\)

Answer 1

\(\frac{3\pi}{4}< a< \pi\Rightarrow\frac{3\pi}{2}< 2a< 2\pi\Rightarrow cos2a>0\)

\(\left(sina+cosa\right)^2=\frac{1}{4}\Leftrightarrow sin^2a+cos^2a+2sina.cosa=\frac{1}{4}\)

\(\Leftrightarrow1+sin2a=\frac{1}{4}\Rightarrow sin2a=-\frac{3}{4}\)

\(\Rightarrow cos2a=\sqrt{1-sin^22a}=\frac{\sqrt{7}}{4}\)

\(\Rightarrow tan2a=\frac{sin2a}{cos2a}=-\frac{3\sqrt{7}}{7}\)