Đáp án:
$\begin{array}{l}
1 - {\left( {{{\left( {{\mathop{\rm s}\nolimits} {\rm{inx}} + \cos x} \right)}^2} - 1} \right)^2}\\
= 1 - {\left( {{{\sin }^2}x + 2\sin x.\cos x + {{\cos }^2}x - 1} \right)^2}\\
= 1 - {\left( {2\sin x.\cos x + 1 - 1} \right)^2}\\
= 1 - {\left( {\sin 2x} \right)^2}\\
= {\cos ^2}2x\left( {do:{{\sin }^2}u + {{\cos }^2}u = 1} \right)
\end{array}$