How do you use the trig identity \[\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x\] to verify that \[\cos 2x=2{{\cos }^{2}}x-1\]?

Hint: We can solve this by simple trigonometric identities. First we have to substitute the given identity in the question in the given question. Then we have to apply suitable trigonometric identities to reach the result.

Complete step by step answer: First we have to know some basic trigonometric identities
\[{{\cos }^{2}}x+{{\sin }^{2}}x=1\]
 \[1+{{\tan }^{2}}x={{\sec }^{2}}x\]
\[1+{{\cot }^{2}}x={{\csc }^{2}}x\]

Now we will start proving our identity
Given question is ‘
\[\cos 2x=2{{\cos }^{2}}x-1\]
We have to prove this using the identity given is \[\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x\]
First we will start solving from our LHS side
The LHS side of our identity is \[\cos 2x\].
We have given an identity of \[\cos 2x\] in the question so we have to substitute that value in place of \[\cos 2x\].
Then we will get
\[\Rightarrow \cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x\]
Now we can see in the above trigonometric identities we have one identity having relation between \[\sin \] and \[\cos \].
I,e., \[{{\cos }^{2}}x+{{\sin }^{2}}x=1\]
We can rewrite it as
\[\Rightarrow {{\sin }^{2}}x=1-{{\cos }^{2}}x\]
From this we can substitute this value in place of \[{{\sin }^{2}}x\] in our equation.
After substituting \[{{\sin }^{2}}x\] we will get
 \[\Rightarrow \cos 2x={{\cos }^{2}}x-\left( 1-{{\cos }^{2}}x \right)\]
Now we have to remove the parentheses. We will get
\[\Rightarrow \cos 2x={{\cos }^{2}}x-1+{{\cos }^{2}}x\]
By simplifying it we will get
\[\Rightarrow \cos 2x=2{{\cos }^{2}}x-1\]
Which is our RHS
So starting with our LHS we derived RHS
\[\cos 2x=2{{\cos }^{2}}x-1\]
Hence we can say proved.

Note: We can also start solving from RHS and derive LHS using the identities we have. We can also prove this by starting with the identity given without the interference of our question. We can choose any one way to prove it but the main thing is to use appropriate identities.