Chứng minh rằng nếu \(\sin \left( {\alpha - \beta } \right) = \dfrac{1}{3}\sin \beta ,\) thì \(\tan \left( {\alpha - \beta } \right) = \dfrac{{\sin \alpha }}{{3 + \cos \alpha }}.\)
\(\begin{array}{l}3\sin \left( {\alpha - \beta } \right) = \sin \left( {\beta - \alpha + \alpha } \right)\\ = \sin \alpha \cos \left( {\alpha - \beta } \right) - \sin \left( {\alpha - \beta } \right)\cos \alpha \end{array}\)
từ đó ta có
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\(\left( {3 + \cos \alpha } \right)\sin \left( {\alpha - \beta } \right) = \sin \alpha \cos \left( {\alpha - \beta } \right)\,\,\,\left( * \right)\) vậy \(\tan \left( {\alpha - \beta } \right) = \dfrac{{\sin \alpha }}{{3 + \cos \alpha }}.\)
(Chú ý. \(\cos \left( {\alpha - \beta } \right) \ne 0\) vì nếu \(\cos \left( {\alpha - \beta } \right) = 0\) thì từ (*) ta suy ra \(\sin \left( {\alpha - \beta } \right) = 0\), vô lí).