Nếu \(\tan \dfrac{x}{2} = \dfrac{a}{b}\) thì \(a\s...

B

- Hướng dẫn giải

Phương pháp giải:

\(\tan \dfrac{x}{2} = a \Rightarrow \left\{ \begin{array}{l}\sin x = \dfrac{{2a}}{{1 + {a^2}}}\\\cos x = \dfrac{{1 - {a^2}}}{{1 + {a^2}}}\end{array} \right.\)

Giải chi tiết:

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

Chọn B.