\[\begin{array}{l}
{\sin ^{2018}}x - {\sin ^2}x = {\sin ^2}x\left( {{{\sin }^{2016}}x - 1} \right)\\
0 \le {\sin ^2}x \le 1 \Rightarrow 0 \le {\sin ^{2016}}x \le 1 \Rightarrow {\sin ^{2016}}x - 1 \le 0\\
\Rightarrow {\sin ^2}x\left( {{{\sin }^{2016}}x - 1} \right) \le 0 \Leftrightarrow {\sin ^{2018}}x - {\sin ^2}x \le 0\\
\Leftrightarrow {\sin ^{2018}}x \le {\sin ^2}x\\
Tuong\,tu:\\
{\cos ^{2018}}x - {\cos ^2}x = {\cos ^2}x\left( {{{\cos }^{2016}}x - 1} \right)\\
0 \le {\cos ^2}x \le 1 \Rightarrow 0 \le {\cos ^{2016}}x \le 1 \Rightarrow {\cos ^{2016}}x - 1 \le 0\\
\Rightarrow {\cos ^2}x\left( {{{\cos }^{2016}}x - 1} \right) \le 0 \Leftrightarrow {\cos ^{2018}}x - {\cos ^2}x \le 0\\
\Leftrightarrow {\cos ^{2018}}x \le {\cos ^2}x
\end{array}\]
