Nếu \(\tan \left( {\frac{x}{2}} \right) = \frac{a}{b}\) thì \({\rm{a}}\sin \left( x \right) + b\cos \left( x \right)\) bằng:

Ta có

\(\begin{array}{l}\tan \left( {\frac{x}{2}} \right) = \frac{a}{b} \Rightarrow \left\{ {_{\cos \left( x \right) = \frac{{1 - {{\left( {\frac{a}{b}} \right)}^2}}}{{1 + {{\left( {\frac{a}{b}} \right)}^2}}} = \frac{{{b^2} - {a^2}}}{{{a^2} + {b^2}}}}^{\sin \left( x \right) = \frac{{2\frac{a}{b}}}{{1 + {{\left( {\frac{a}{b}} \right)}^2}}} = \frac{{2\frac{a}{b}}}{{\frac{{{a^2} + {b^2}}}{{{b^2}}}}} = \frac{{2ab}}{{{a^2} + {b^2}}}}} \right.\\ \Rightarrow {\rm{a}}\sin \left( x \right) + b\cos \left( x \right) = a.\frac{{2ab}}{{{a^2} + {b^2}}} + b.\frac{{{b^2} - {a^2}}}{{{a^2} + {b^2}}}\\ = \frac{{2{a^2}b + {b^3} - {a^2}b}}{{{a^2} + {b^2}}} = \frac{{{a^2}b + {b^3}}}{{{a^2} + {b^2}}} = \frac{{b\left( {{a^2} + {b^2}} \right)}}{{{a^2} + {b^2}}} = b\end{array}\)