Sử dụng PP lượng giác hoá giải toán BĐT.

Đề bài: (Câu hỏi của bạn Nguyễn Trà My hỏi trên facebook Trợ Giúp Toán Học)\[Cho\,\left\{ \begin{array}{l}
x,y \in R\\
{x^2} + {y^2} = 1
\end{array} \right..Min,\,Max\,P = \frac{{2\left( {{x^2} + 6xy} \right)}}{{1 + 2xy + 2{y^2}}} = ?\]Giải:
\[\begin{array}{l}
Do\,\left\{ \begin{array}{l}
x,y \in R\\
{x^2} + {y^2} = 1
\end{array} \right. \Rightarrow Coi\,\left\{ \begin{array}{l}
x = sin\,\alpha \\
y = cos\,\alpha \\
\alpha  \in \left[ {0;2\pi } \right]
\end{array} \right.\\
 \Rightarrow P = \frac{{2\left( {si{n^2}\alpha  + 6sin\alpha cos\alpha } \right)}}{{1 + 2sin\alpha cos\alpha  + 2co{s^2}\alpha }} = \frac{{6\sin 2\alpha  - cos2\alpha  + 1}}{{\sin 2\alpha  + cos2\alpha  + 2}}\\
 \Leftrightarrow P\left( {\sin 2\alpha  + cos2\alpha  + 2} \right) = 6\sin 2\alpha  - cos2\alpha  + 1\\
 \Leftrightarrow \left( {P - 6} \right)\sin 2\alpha  + \left( {P + 1} \right)cos2\alpha  = 1 - 2P\\
 \Rightarrow \exists \alpha  \Leftrightarrow {\left( {P - 6} \right)^2} + {\left( {P + 1} \right)^2} \ge {\left( {1 - 2P} \right)^2} \Leftrightarrow {P^2} + 3P - 18 \le 0 \Leftrightarrow  - 6 \le P \le 3\\
 \bullet \,Min\,P =  - 6 \Leftrightarrow 12\sin 2\alpha  + 5cos2\alpha  = 13 \Leftrightarrow \left\{ \begin{array}{l}
t = \tan \alpha \\
\frac{{24t}}{{1 + {t^2}}} + \frac{{5\left( {1 - {t^2}} \right)}}{{1 + {t^2}}} = 13
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
t = \tan \alpha \\
{\left( {3t - 2} \right)^2} = 0
\end{array} \right.\\
 \Leftrightarrow t = \tan \alpha  = \frac{x}{y} = \frac{2}{3} \Leftrightarrow \frac{{{x^2}}}{4} = \frac{{{y^2}}}{9} = \frac{{{x^2} + {y^2}}}{{13}} = \frac{1}{{13}} \Rightarrow \left\{ \begin{array}{l}
x = \frac{{ \pm 2}}{{\sqrt {13} }}\\
y = \frac{{ \pm 3}}{{\sqrt {13} }}
\end{array} \right.\\
 \bullet MaxP = 3 \Leftrightarrow 3\sin 2\alpha  - 4cos2\alpha  = 5 \Leftrightarrow \left\{ \begin{array}{l}
t = \tan \alpha \\
\frac{{6t}}{{1 + {t^2}}} - \frac{{4\left( {1 - {t^2}} \right)}}{{1 + {t^2}}} = 5
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
t = \tan \alpha \\
{\left( {t - 3} \right)^2} = 0
\end{array} \right.\\
 \Leftrightarrow t = \tan \alpha  = \frac{x}{y} = 3 \Leftrightarrow \frac{{{x^2}}}{9} = \frac{{{y^2}}}{1} = \frac{{{x^2} + {y^2}}}{{10}} = \frac{1}{{10}} \Rightarrow \left\{ \begin{array}{l}
x = \frac{{ \pm 3}}{{\sqrt {10} }}\\
y = \frac{{ \pm 1}}{{\sqrt {13} }}
\end{array} \right.
\end{array}\]