How do you prove $2{\cos ^2}A - 1 = \cos \left( {2A} \right)$?

Hint: To verify the given equation, we will at first, take the complicated side of the question, simplify it using trigonometric identities until it gets transformed into the same expression as on the other side of the equation. Moreover, some of the algebraic identities can also be used to simplify expressions.

Complete step-by-step solution:
To prove: $2{\cos ^2}A - 1 = \cos \left( {2A} \right)$
Proof:
We consider the LHS first: $2{\cos ^2}A - 1$
Now, we simplify it and write as:
$ = {\cos ^2}A + {\cos ^2}A - 1$
$ = {\cos ^2}A + \left( {{{\cos }^2}A - 1} \right)$ ……………………..(1)
We know the trigonometric identity: \[{\sin ^2}A + {\cos ^2}A = 1\] ……………………..(2)
On rearranging equation (2), we get:
${\cos ^2}A - 1 = - {\sin ^2}A$ ……………………..(3)
Now, substituting the value of $\left( {{{\cos }^2}A - 1} \right)$ from equation (3) in the expression (1), we get:
$ = {\cos ^2}A + \left( { - {{\sin }^2}A} \right)$
Removing the brackets, we get:
$ = {\cos ^2}A - {\sin ^2}A$
Splitting the product terms in the above expression, we get:
$ = \left( {\cos A} \right)\left( {\cos A} \right) - \left( {\sin A} \right)\left( {\sin A} \right)$ …………………….(4)
Now, we have the trigonometric identity:
$\cos x\cos y - \sin x\sin y = \cos \left( {x + y} \right)$ ……………………….(5)
Replacing the ‘x’ and ‘y’ in equation (5) with ‘A’, we get:
$ \Rightarrow \left( {\cos A} \right)\left( {\cos A} \right) - \left( {\sin A} \right)\left( {\sin A} \right) = \cos \left( {A + A} \right)$ ……………………(6)
Thus, substituting the respective value from equation (6) in expression (5), we get:
$ = \cos \left( {A + A} \right)$
$ = \cos \left( {2A} \right)$, which is nothing but the expression on the other side of the expression, that is, on the right side of the expression.
Hence, proved.

Note: The alternative way of solving the same problem would be to consider the RHS first, and rewrite the angle as the sum of 2 equal angles. We then follow the same steps as described above, but now in the reverse order, as we will arrive at the LHS expression at the last step.

To prove a trigonometric equation, we should tend to start with the more complicated side of the equation, and keep simplifying it until it is transformed into the same expression as on the other side of the given equation. In some cases, we can also try to simplify both sides of the equation and arrive at a common expression to prove their equality. The various procedures of solving a trigonometric equation are: expanding the expressions, making use of the identities, factoring the expressions or simply using basic algebraic strategies to obtain the desired results.