Using the identity $\cos (A + B) = \cos A\cos B - \sin A\sin B$ , how do you prove that $\dfrac{1}{4}\cos (3A) = {\cos ^3}A - \dfrac{3}{4}\cos A$ .

Hint: To prove the relation $\dfrac{1}{4}\cos (3A) = {\cos ^3}A - \dfrac{3}{4}\cos A$ using the identity $\cos (A + B) = \cos A\cos B - \sin A\sin B$ .
To prove this, we will consider left- hand side i.e., $\cos (3A)$ and write $3A$ as $A + 2A$ , from where, we’ll get the values for $A$ and $B$ , then using the identity we’ll solve until it becomes equal to the right-hand side.

Formulas to be used:
Half angle formulae
$\sin 2A = 2\sin A\cos A$ ,
$\cos 2A = {\cos ^2}A - {\sin ^2}A$ ,
Also, ${\sin ^2}A + {\cos ^2}A = 1$ .

Complete step-by-step answer:
We are given with the identity $\cos (A + B) = \cos A\cos B - \sin A\sin B$ and using this identity we need to prove the relation $\dfrac{1}{4}\cos (3A) = {\cos ^3}A - \dfrac{3}{4}\cos A$ .
Let us consider the left- hand side i.e., $\cos 3A$ which can be written as $\cos (A + 2A)$ .
Now, we have $A = A$ and $B = 2A$ .
Now, using the given identity, we can write it as $\cos (A + 2A) = \cos A\cos 2A - \sin A\sin 2A$ .
Next, we know the identities $\sin 2A = 2\sin A\cos A$ and $\cos 2A = {\cos ^2}A - {\sin ^2}A$ .
So, we’ll replace $\sin 2A$ by $2\sin A\cos A$ and $\cos 2A$ by ${\cos ^2}A - {\sin ^2}A$ , then the equation becomes,
$\cos (A + 2A) = \cos A({\cos ^2}A - {\sin ^2}A) - \sin A(2\sin A\cos A)$ .
Now, on solving the brackets, we get,
$\cos (A + 2A) = {\cos ^3}A - \cos A{\sin ^2}A - 2{\sin ^2}A\cos A$ .
Solving further gives $\cos (A + 2A) = {\cos ^3}A - 3{\sin ^2}A\cos A$ .
Now, since all the functions on the right- hand side in the equation to be proved are ‘ $\cos ine$ ‘. Therefore, using the identity ${\sin ^2}A + {\cos ^2}A = 1$ we get, ${\sin ^2}A = 1 - {\cos ^2}A$ .
So, replacing ${\sin ^2}A$ by $1 - {\cos ^2}A$ , the above equation becomes, $\cos (A + 2A) = {\cos ^3}A - 2\cos A(1 - {\cos ^2}A)$ ,
i.e., on opening the brackets, we get, $\cos (A + 2A) = {\cos ^3}A - 2\cos A - 2{\cos ^3}A$ ,
which gives, $\cos (A + 2A) = 4{\cos ^3}A - 3\cos A$ .
Now, take $4$ common from right- hand side, we get,
$\cos (A + 2A) = 4\left( {{{\cos }^3}A - \dfrac{3}{4}\cos A} \right)$ .
Finally, dividing both sides by $4$ , we get,
$\dfrac{1}{4}\cos (A + 2A) = {\cos ^3}A - \dfrac{3}{4}\cos A$ ,
$\dfrac{1}{4}\cos (3A) = {\cos ^3}A - \dfrac{3}{4}\cos A$ .
Hence the given relation is proved using the given identity.

Note: The equation that we have proved is the same as $\cos (3A) = 4{\cos ^3}A - 3\cos A$ , which is a direct identity to remember.
$A + 2A$ is the same as $2A + A$ , so we can write it anyway.
You need to see whether the functions on the right- hand side are $\sin e$ or $\cos ine$ or any other function and according to that use the required identities.