Trong tam giác ABC chứng minh sin^2A+sin^2B+sin^2C $\geq$ $\frac{9}{4}$ 21/11/2021 Bởi Nevaeh Trong tam giác ABC chứng minh sin^2A+sin^2B+sin^2C $\geq$ $\frac{9}{4}$
Giải thích các bước giải: Ta có: \(\begin{array}{l}\cos 2x = 1 – 2{\sin ^2}x \Rightarrow {\sin ^2}x = \frac{{1 – \cos 2x}}{2}\\A = {\sin ^2}A + {\sin ^2}B + {\sin ^2}C\\ = \frac{{1 – \cos 2A}}{2} + \frac{{1 – \cos 2B}}{2} + {\sin ^2}C\\ = 1 – \frac{{\cos 2A + \cos 2B}}{2} + \left( {1 – {{\cos }^2}C} \right)\\ = 1 – \frac{{2.cos\frac{{2A + 2B}}{2}.\cos \frac{{2A – 2B}}{2}}}{2} + 1 – {\cos ^2}C\\ = 2 – {\cos ^2}C – \cos \left( {A + B} \right).\cos \left( {A – B} \right)\\ = 2 – {\cos ^2}C – \left( { – \cos \left( {180^\circ – \left( {A + B} \right)} \right)} \right).\cos \left( {A – B} \right)\\ = 2 – {\cos ^2}C + \cos C.\cos \left( {A – B} \right)\\ – 1 \le \cos \left( {A – B} \right) \le 1\\TH1:\,\,\,\,\left\{ \begin{array}{l} – 1 \le \cos C \le 0\\ – 1 \le \cos \left( {A – B} \right) \le 1\end{array} \right. \Rightarrow \cos C.\cos \left( {A – B} \right) \le – \cos C\\ \Rightarrow A \le 2 – {\cos ^2}C – \cos C = \frac{9}{4} – \left( {{{\cos }^2}C + \cos C + \frac{1}{4}} \right) = \frac{9}{4} – {\left( {\cos C + \frac{1}{2}} \right)^2} \le \frac{9}{4}\\TH2:\,\,\,\,\,\left\{ \begin{array}{l}0 \le \cos C \le 1\\ – 1 \le \cos \left( {A – B} \right) \le 1\end{array} \right. \Rightarrow \cos C.\cos \left( {A – B} \right) \le \cos C\\ \Rightarrow A \le 2 – {\cos ^2}C + \cos C = \frac{9}{4} – \left( {{{\cos }^2}C – \cos C + \frac{1}{4}} \right) = \frac{9}{4} – {\left( {\cos C – \frac{1}{2}} \right)^2} \le \frac{9}{4}\end{array}\) Bình luận
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\cos 2x = 1 – 2{\sin ^2}x \Rightarrow {\sin ^2}x = \frac{{1 – \cos 2x}}{2}\\
A = {\sin ^2}A + {\sin ^2}B + {\sin ^2}C\\
= \frac{{1 – \cos 2A}}{2} + \frac{{1 – \cos 2B}}{2} + {\sin ^2}C\\
= 1 – \frac{{\cos 2A + \cos 2B}}{2} + \left( {1 – {{\cos }^2}C} \right)\\
= 1 – \frac{{2.cos\frac{{2A + 2B}}{2}.\cos \frac{{2A – 2B}}{2}}}{2} + 1 – {\cos ^2}C\\
= 2 – {\cos ^2}C – \cos \left( {A + B} \right).\cos \left( {A – B} \right)\\
= 2 – {\cos ^2}C – \left( { – \cos \left( {180^\circ – \left( {A + B} \right)} \right)} \right).\cos \left( {A – B} \right)\\
= 2 – {\cos ^2}C + \cos C.\cos \left( {A – B} \right)\\
– 1 \le \cos \left( {A – B} \right) \le 1\\
TH1:\,\,\,\,\left\{ \begin{array}{l}
– 1 \le \cos C \le 0\\
– 1 \le \cos \left( {A – B} \right) \le 1
\end{array} \right. \Rightarrow \cos C.\cos \left( {A – B} \right) \le – \cos C\\
\Rightarrow A \le 2 – {\cos ^2}C – \cos C = \frac{9}{4} – \left( {{{\cos }^2}C + \cos C + \frac{1}{4}} \right) = \frac{9}{4} – {\left( {\cos C + \frac{1}{2}} \right)^2} \le \frac{9}{4}\\
TH2:\,\,\,\,\,\left\{ \begin{array}{l}
0 \le \cos C \le 1\\
– 1 \le \cos \left( {A – B} \right) \le 1
\end{array} \right. \Rightarrow \cos C.\cos \left( {A – B} \right) \le \cos C\\
\Rightarrow A \le 2 – {\cos ^2}C + \cos C = \frac{9}{4} – \left( {{{\cos }^2}C – \cos C + \frac{1}{4}} \right) = \frac{9}{4} – {\left( {\cos C – \frac{1}{2}} \right)^2} \le \frac{9}{4}
\end{array}\)