PART 5：倍角公式 例題2

已知 \(\sin \theta  = \frac{3}{5}\) ， \(\theta \) 為第2象限角，試求 \(\tan 2\theta  = ?\)

SOL:  

\(\sin \theta  = \frac{3}{5}  \Rightarrow \cos \theta  =  - \frac{4}{5}\) ，

\(\tan 2\theta  = \frac{{\sin 2\theta }}{{\cos 2\theta }}\)  \( = \frac{{2\sin \theta \cos \theta }}{{{{\cos }^2}\theta  - {{\sin }^2}\theta }}\)  \( = \frac{{2\left( {\frac{3}{5}} \right)\left( { - \frac{4}{5}} \right)}}{{\frac{{16 - 9}}{{25}}}}\)  \( = \frac{{ - 24}}{7}\)