I think that we can establish the identity without too much work; it certainly isn't short though. Your statement is equivalent to

\(\displaystyle{{\cos}^{{2}}{4}}{x}−{{\cos}^{{2}}{2}}{x}=−{\sin{{6}}}{x}{\sin{{2}}}{x}\)

We'll start with left hand side; my strategy will be to make everything in terms of \(\displaystyle{\cos{{2}}}{x}\). I will make use of the following identities:

\(\displaystyle{\sin{{2}}}\theta\equiv{1}-{\cos{{2}}}\theta\)

\(\displaystyle{\cos{{2}}}\theta\equiv{2}{\cos{{2}}}\theta−{1}\)

\(\displaystyle{\sin{{3}}}\theta\equiv{3}{\sin{\theta}}−{4}{\sin{{3}}}\theta\)

Here we go:

\(\displaystyle{{\cos}^{{24}}{x}}−{{\cos}^{{22}}{x}}\equiv{\left({2}{{\cos}^{{22}}{x}}−{1}\right)}^{{2}}−{{\cos}^{{22}}{x}}\)

\(\displaystyle\equiv{4}{{\cos}^{{42}}{x}}−{5}{{\cos}^{{22}}{x}}+{1}\)

Now for the right hand side:

\(\displaystyle−{\sin{{6}}}{x}{\sin{{2}}}{x}\equiv−{\sin{{2}}}{x}{\left({3}{\sin{{2}}}{x}−{4}{{\sin}^{{3}}{2}}{x}\right)}\)

\(\displaystyle\equiv{4}{{\sin}^{{42}}{x}}−{3}{{\sin}^{{2}}{x}}\)

\(\displaystyle\equiv{4}{\left({1}−{{\cos}^{{22}}{x}}\right)}^{{2}}−{3}{\left({1}−{{\cos}^{{22}}{x}}\right)}\)

\(\displaystyle\equiv{4}{\left({{\cos}^{{42}}{x}}−{2}{{\cos}^{{2}}{x}}+{1}\right)}−{3}+{3}{{\cos}^{{22}}{x}}\)

\(\displaystyle\equiv{4}{{\cos}^{{42}}{x}}−{5}{{\cos}^{{22}}{x}}+{1}\)

Hence we have

\(\displaystyle{{\cos}^{{2}}{4}}{x}−{{\cos}^{{2}}{2}}{x}\equiv−{\sin{{6}}}{x}{\sin{{2}}}{x}\)

as required.

\(\displaystyle{{\cos}^{{2}}{4}}{x}−{{\cos}^{{2}}{2}}{x}=−{\sin{{6}}}{x}{\sin{{2}}}{x}\)

We'll start with left hand side; my strategy will be to make everything in terms of \(\displaystyle{\cos{{2}}}{x}\). I will make use of the following identities:

\(\displaystyle{\sin{{2}}}\theta\equiv{1}-{\cos{{2}}}\theta\)

\(\displaystyle{\cos{{2}}}\theta\equiv{2}{\cos{{2}}}\theta−{1}\)

\(\displaystyle{\sin{{3}}}\theta\equiv{3}{\sin{\theta}}−{4}{\sin{{3}}}\theta\)

Here we go:

\(\displaystyle{{\cos}^{{24}}{x}}−{{\cos}^{{22}}{x}}\equiv{\left({2}{{\cos}^{{22}}{x}}−{1}\right)}^{{2}}−{{\cos}^{{22}}{x}}\)

\(\displaystyle\equiv{4}{{\cos}^{{42}}{x}}−{5}{{\cos}^{{22}}{x}}+{1}\)

Now for the right hand side:

\(\displaystyle−{\sin{{6}}}{x}{\sin{{2}}}{x}\equiv−{\sin{{2}}}{x}{\left({3}{\sin{{2}}}{x}−{4}{{\sin}^{{3}}{2}}{x}\right)}\)

\(\displaystyle\equiv{4}{{\sin}^{{42}}{x}}−{3}{{\sin}^{{2}}{x}}\)

\(\displaystyle\equiv{4}{\left({1}−{{\cos}^{{22}}{x}}\right)}^{{2}}−{3}{\left({1}−{{\cos}^{{22}}{x}}\right)}\)

\(\displaystyle\equiv{4}{\left({{\cos}^{{42}}{x}}−{2}{{\cos}^{{2}}{x}}+{1}\right)}−{3}+{3}{{\cos}^{{22}}{x}}\)

\(\displaystyle\equiv{4}{{\cos}^{{42}}{x}}−{5}{{\cos}^{{22}}{x}}+{1}\)

Hence we have

\(\displaystyle{{\cos}^{{2}}{4}}{x}−{{\cos}^{{2}}{2}}{x}\equiv−{\sin{{6}}}{x}{\sin{{2}}}{x}\)

as required.