Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\cos 2x = 1 - 2{\sin ^2}x \Rightarrow {\sin ^2}x = \frac{{1 - \cos 2x}}{2}\\
A = {\sin ^2}A + {\sin ^2}B + {\sin ^2}C\\
= \frac{{1 - \cos 2A}}{2} + \frac{{1 - \cos 2B}}{2} + {\sin ^2}C\\
= 1 - \frac{{\cos 2A + \cos 2B}}{2} + \left( {1 - {{\cos }^2}C} \right)\\
= 1 - \frac{{2.cos\frac{{2A + 2B}}{2}.\cos \frac{{2A - 2B}}{2}}}{2} + 1 - {\cos ^2}C\\
= 2 - {\cos ^2}C - \cos \left( {A + B} \right).\cos \left( {A - B} \right)\\
= 2 - {\cos ^2}C - \left( { - \cos \left( {180^\circ - \left( {A + B} \right)} \right)} \right).\cos \left( {A - B} \right)\\
= 2 - {\cos ^2}C + \cos C.\cos \left( {A - B} \right)\\
- 1 \le \cos \left( {A - B} \right) \le 1\\
TH1:\,\,\,\,\left\{ \begin{array}{l}
- 1 \le \cos C \le 0\\
- 1 \le \cos \left( {A - B} \right) \le 1
\end{array} \right. \Rightarrow \cos C.\cos \left( {A - B} \right) \le - \cos C\\
\Rightarrow A \le 2 - {\cos ^2}C - \cos C = \frac{9}{4} - \left( {{{\cos }^2}C + \cos C + \frac{1}{4}} \right) = \frac{9}{4} - {\left( {\cos C + \frac{1}{2}} \right)^2} \le \frac{9}{4}\\
TH2:\,\,\,\,\,\left\{ \begin{array}{l}
0 \le \cos C \le 1\\
- 1 \le \cos \left( {A - B} \right) \le 1
\end{array} \right. \Rightarrow \cos C.\cos \left( {A - B} \right) \le \cos C\\
\Rightarrow A \le 2 - {\cos ^2}C + \cos C = \frac{9}{4} - \left( {{{\cos }^2}C - \cos C + \frac{1}{4}} \right) = \frac{9}{4} - {\left( {\cos C - \frac{1}{2}} \right)^2} \le \frac{9}{4}
\end{array}\)
