(sinx)^2007 +(cosx)^2008 = 1
$\begin{align}
& {{(\operatorname{sinx})}^{2007}}+{{(cosx)}^{2008}}=1 \\
& {{(\operatorname{sinx})}^{2007}}\le {{\sin }^{2}}x \\
& {{(cosx)}^{2008}}\le {{\cos }^{2}}x \\
& {{(\operatorname{sinx})}^{2007}}+{{(cosx)}^{2008}}\le {{\sin }^{2}}x+{{\cos }^{2}}x=1. \\
& \\
& \\
& \\
\end{align}$
Dấu “=” xảy ra $\Leftrightarrow $ $\begin{align}
& \\
& {{(\operatorname{sinx})}^{2007}}={{\sin }^{2}}x,{{(cosx)}^{2008}}={{\cos }^{2}}x \\
& \\
& \\
\end{align}$
E giải tiếp nhé