Đáp án:
$\begin{array}{l}
2{\mathop{\rm cosx}\nolimits} + cos3x + cos5x\\
= 2cosx + 2.cos\dfrac{{3x + 5x}}{2}.\cos \dfrac{{3x - 5x}}{2}\\
= 2.\cos x + 2.\cos 4x.\cos \left( { - x} \right)\\
= 2.\cos x + 2.\cos x.\cos 4x\\
= 2.\cos x.\left( {\cos 4x + 1} \right)\\
= 2.\cos x.2.{\cos ^2}2x\\
= 4.\cos x.{\cos ^2}2x\\
\Rightarrow \dfrac{{{{\sin }^2}4x}}{{2{\mathop{\rm cosx}\nolimits} + cos3x + cos5x}}\\
= \dfrac{{{{\left( {2.\sin 2x.\cos 2x} \right)}^2}}}{{4.\cos x.{{\cos }^2}2x}}\\
= \dfrac{{4.{{\sin }^2}2x.{{\cos }^2}2x}}{{4.\cos x.{{\cos }^2}2x}}\\
= \dfrac{{{{\left( {2.\sin x.\cos x} \right)}^2}}}{{\cos x}}\\
= 4.{\sin ^2}x.\cos x\\
= 2.\sin x.2.\sin x.\cos x\\
= 2.\sin x.sin2x
\end{array}$
