Answer
See the explanation below.
Work Step by Step
Now, take the negative sign outside and use the identity, ${{\cos }^{2}}x-{{\sin }^{2}}x=\cos 2x$.
$\begin{align}
& {{\sin }^{2}}\frac{x}{2}-{{\cos }^{2}}\frac{x}{2}=-\left( -{{\sin }^{2}}\frac{x}{2}+{{\cos }^{2}}\frac{x}{2} \right) \\
& =-\left( {{\cos }^{2}}\frac{x}{2}-{{\sin }^{2}}\frac{x}{2} \right) \\
& =-\cos \left( 2\times \frac{x}{2} \right) \\
& =-\cos x
\end{align}$
Thus, the left side of the equation is equal to $-\cos x$.
Hence, the left side is equal to $-\cos x$.
