Cho cot α = 2 . Tính B = sin α + 2 cos α sin 3 α − cos 3 α .

Vì \(\cot \alpha  = 2 \Rightarrow \sin \alpha  \ne 0\). Chia cả tử và mẫu cho \({\sin ^3}\alpha \) ta được:

\(B = \frac{{(\sin \alpha  + 2\cos \alpha )\frac{1}{{{{\sin }^3}\alpha }}}}{{\left( {{{\sin }^3}\alpha  - {{\cos }^3}\alpha } \right)\frac{1}{{{{\sin }^3}\alpha }}}} = \frac{{\frac{1}{{{{\sin }^2}\alpha }} + 2\cot \alpha  \cdot \frac{1}{{{{\sin }^2}\alpha }}}}{{1 - {{\cot }^3}\alpha }}\)\(\)

\( = \frac{{1 + {{\cot }^2}\alpha  + 2\cot \alpha \left( {1 + {{\cot }^2}\alpha } \right)}}{{1 - {{\cot }^3}\alpha }} = \frac{{2{{\cot }^3}\alpha  + {{\cot }^2}\alpha  + 2\cot \alpha  + 1}}{{1 - {{\cot }^3}\alpha }} =  - \frac{{25}}{7}{\rm{. }}\)