Toán (sin x) ^2018 < (sin x) ^2 (cos x) ^2018 < (cos x) ^2 Chứng minh 28/09/2021 By Hailey (sin x) ^2018 < (sin x) ^2 (cos x) ^2018 < (cos x) ^2 Chứng minh
\[\begin{array}{l} {\sin ^{2018}}x – {\sin ^2}x = {\sin ^2}x\left( {{{\sin }^{2016}}x – 1} \right)\\ 0 \le {\sin ^2}x \le 1 \Rightarrow 0 \le {\sin ^{2016}}x \le 1 \Rightarrow {\sin ^{2016}}x – 1 \le 0\\ \Rightarrow {\sin ^2}x\left( {{{\sin }^{2016}}x – 1} \right) \le 0 \Leftrightarrow {\sin ^{2018}}x – {\sin ^2}x \le 0\\ \Leftrightarrow {\sin ^{2018}}x \le {\sin ^2}x\\ Tuong\,tu:\\ {\cos ^{2018}}x – {\cos ^2}x = {\cos ^2}x\left( {{{\cos }^{2016}}x – 1} \right)\\ 0 \le {\cos ^2}x \le 1 \Rightarrow 0 \le {\cos ^{2016}}x \le 1 \Rightarrow {\cos ^{2016}}x – 1 \le 0\\ \Rightarrow {\cos ^2}x\left( {{{\cos }^{2016}}x – 1} \right) \le 0 \Leftrightarrow {\cos ^{2018}}x – {\cos ^2}x \le 0\\ \Leftrightarrow {\cos ^{2018}}x \le {\cos ^2}x \end{array}\] Trả lời
\[\begin{array}{l}
{\sin ^{2018}}x – {\sin ^2}x = {\sin ^2}x\left( {{{\sin }^{2016}}x – 1} \right)\\
0 \le {\sin ^2}x \le 1 \Rightarrow 0 \le {\sin ^{2016}}x \le 1 \Rightarrow {\sin ^{2016}}x – 1 \le 0\\
\Rightarrow {\sin ^2}x\left( {{{\sin }^{2016}}x – 1} \right) \le 0 \Leftrightarrow {\sin ^{2018}}x – {\sin ^2}x \le 0\\
\Leftrightarrow {\sin ^{2018}}x \le {\sin ^2}x\\
Tuong\,tu:\\
{\cos ^{2018}}x – {\cos ^2}x = {\cos ^2}x\left( {{{\cos }^{2016}}x – 1} \right)\\
0 \le {\cos ^2}x \le 1 \Rightarrow 0 \le {\cos ^{2016}}x \le 1 \Rightarrow {\cos ^{2016}}x – 1 \le 0\\
\Rightarrow {\cos ^2}x\left( {{{\cos }^{2016}}x – 1} \right) \le 0 \Leftrightarrow {\cos ^{2018}}x – {\cos ^2}x \le 0\\
\Leftrightarrow {\cos ^{2018}}x \le {\cos ^2}x
\end{array}\]