Nếu \(5\sin \alpha  = 3\sin \left( {\alpha  + 2\beta } \right)\) thì :

* Hướng dẫn giải

Ta có 

\(\begin{array}{l}
5\sin \alpha  = 3\sin \left( {\alpha  + 2\beta } \right) \Leftrightarrow 5\sin \left[ {\left( {\alpha  + \beta } \right) - \beta } \right] = 3\sin \left[ {\left( {\alpha  + \beta } \right) + \beta } \right]\\
 \Leftrightarrow 5\sin \left( {\alpha  + \beta } \right)\cos \beta  - 5\cos \left( {\alpha  + \beta } \right)sin\beta  = 3\sin \left( {\alpha  + \beta } \right)\cos \beta  + 3\cos \left( {\alpha  + \beta } \right)sin\beta \\
 \Leftrightarrow 2\sin \left( {\alpha  + \beta } \right)\cos \beta  = 8\cos \left( {\alpha  + \beta } \right)sin\beta  \Leftrightarrow \frac{{\sin \left( {\alpha  + \beta } \right)}}{{\cos \left( {\alpha  + \beta } \right)}} = 4\frac{{\sin \beta }}{{\cos \beta }} \Leftrightarrow \tan \left( {\alpha  + \beta } \right) = 4\tan \beta 
\end{array}\)