How do you verify ${{\sin }^{4}}x-{{\cos }^{4}}x=1-2{{\cos }^{2}}x$?

Hint: To verify the given trigonometric equation ${{\sin }^{4}}x-{{\cos }^{4}}x=1-2{{\cos }^{2}}x$, we need to prove the L.H.S is equal to R.H.S. To achieve it, we are going to use the algebraic property in the L.H.S of the given equation which is equal to: ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$. After applying this algebraic property, we are going to use the trigonometric identity i.e. ${{\sin }^{2}}x+{{\cos }^{2}}x=1$.

Complete step by step answer:
The trigonometric equation which we are asked to prove is as follows:
${{\sin }^{4}}x-{{\cos }^{4}}x=1-2{{\cos }^{2}}x$ ………. Eq. (1)
Now, we are trying to prove L.H.S = R.H.S for that, we are trying to change L.H.S in such a way so that it becomes equal to R.H.S. We are going to write ${{\sin }^{4}}x$ as ${{\left( {{\sin }^{2}}x \right)}^{2}}$ and ${{\cos }^{4}}x$ as ${{\left( {{\cos }^{2}}x \right)}^{2}}$ in the L.H.S of the above equation and we get,
$\Rightarrow {{\left( {{\sin }^{2}}x \right)}^{2}}-{{\left( {{\cos }^{2}}x \right)}^{2}}=1-2{{\cos }^{2}}x$…………. Eq. (2)
As you can see that L.H.S of the above equation is of the form ${{a}^{2}}-{{b}^{2}}$ so we can use the following identity:
${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$
In this identity substituting $a={{\sin }^{2}}x$ and $b={{\cos }^{2}}x$ we get,
$\Rightarrow {{\left( {{\sin }^{2}}x \right)}^{2}}-{{\left( {{\cos }^{2}}x \right)}^{2}}=\left( {{\sin }^{2}}x-{{\cos }^{2}}x \right)\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)$
The value of the expression $\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)$ written above is found by using the trigonometric identity which is equal to:
${{\sin }^{2}}x+{{\cos }^{2}}x=1$
$\begin{align}
  & \Rightarrow {{\left( {{\sin }^{2}}x \right)}^{2}}-{{\left( {{\cos }^{2}}x \right)}^{2}}=\left( {{\sin }^{2}}x-{{\cos }^{2}}x \right)\left( 1 \right) \\
 & \Rightarrow {{\left( {{\sin }^{2}}x \right)}^{2}}-{{\left( {{\cos }^{2}}x \right)}^{2}}=\left( {{\sin }^{2}}x-{{\cos }^{2}}x \right) \\
\end{align}$
Using above relation in eq. (2) we get,
$\begin{align}
  & \Rightarrow {{\left( {{\sin }^{2}}x \right)}^{2}}-{{\left( {{\cos }^{2}}x \right)}^{2}}=1-2{{\cos }^{2}}x \\
 & \Rightarrow \left( {{\sin }^{2}}x-{{\cos }^{2}}x \right)=1-2{{\cos }^{2}}x \\
\end{align}$
Now, rearranging the trigonometric identity which we have shown above is as follows:
$\begin{align}
  & {{\sin }^{2}}x+{{\cos }^{2}}x=1 \\
 & \Rightarrow {{\sin }^{2}}x=1-{{\cos }^{2}}x \\
\end{align}$
Substituting the above value of ${{\sin }^{2}}x$ in $\left( {{\sin }^{2}}x-{{\cos }^{2}}x \right)=1-2{{\cos }^{2}}x$ we get,
$\begin{align}
  & \Rightarrow \left( 1-{{\cos }^{2}}x-{{\cos }^{2}}x \right)=1-2{{\cos }^{2}}x \\
 & \Rightarrow \left( 1-2{{\cos }^{2}}x \right)=1-2{{\cos }^{2}}x \\
\end{align}$
As you can see that L.H.S = R.H.S so we have proved the given trigonometric equation.

Note:
A trick to prove the above trigonometric equation is that:
${{\sin }^{4}}x-{{\cos }^{4}}x=1-2{{\cos }^{2}}x$
In the L.H.S of the above equation, you can see that square of $\sin x\And \cos x$ is possible by applying the algebraic identity ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ and you might think why we reduce the above equation to in the square of $\sin x\And \cos x$ because we know the trigonometric identity in $\sin x\And \cos x$ which is equal to:
${{\sin }^{2}}x+{{\cos }^{2}}x=1$