Question

Prove that: $\cos A\cos \left( \dfrac{\pi }{3}-A \right)\cos \left( \dfrac{\pi }{3}+A \right)=\dfrac{1}{4}\cos 3A$

Answer 1

Now let us first look at the LHS side. We can with the help of the trigonometric identity simplify the $\cos \left( \dfrac{\pi }{3}-A \right)\cos \left( \dfrac{\pi }{3}+A \right)$ term. Now simplify further by substituting the trigonometric values and then again expand in such a way that we derive back to the formula of $\cos 3A$

Complete step by step solution:
The given trigonometric expression is, $\cos A\cos \left( \dfrac{\pi }{3}-A \right)\cos \left( \dfrac{\pi }{3}+A \right)=\dfrac{1}{4}\cos 3A$
Now consider the LHS side first.
Multiply and divide with 2 on the numerator and the denominator.
$\Rightarrow \dfrac{1}{2}\cos A\left[ 2\cos \left( \dfrac{\pi }{3}-A \right)\cos \left( \dfrac{\pi }{3}+A \right) \right]=\dfrac{1}{4}\cos 3A$
Now here we use this trigonometric identity.
$2\cos A\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)$
Upon substituting the values of A as $\dfrac{\pi }{3}-A$ and B as $\dfrac{\pi }{3}+A$ we get,
$\Rightarrow \dfrac{1}{2}\cos A\left[ \cos \left( \dfrac{\pi }{3}-A+\dfrac{\pi }{3}+A \right)+\cos \left( \dfrac{\pi }{3}-A-\dfrac{\pi }{3}-A \right) \right]=\dfrac{1}{4}\cos 3A$
Now simplify the expression further.
$\Rightarrow \dfrac{1}{2}\cos A\left[ \cos \left( \dfrac{2\pi }{3} \right)+\cos \left( -2A \right) \right]=\dfrac{1}{4}\cos 3A$
Now substitute the trigonometric value of $\cos \dfrac{2\pi }{3}$ as $-\dfrac{1}{2}$
$\Rightarrow \dfrac{1}{2}\cos A\left[ -\dfrac{1}{2}+\cos \left( -2A \right) \right]=\dfrac{1}{4}\cos 3A$
And also, $\cos \left( -2A \right)=\cos 2A$ because cosine is negative in the second and third quadrant and its periodicity is, $2\pi$
$\Rightarrow -\dfrac{1}{4}\cos A+\dfrac{1}{2}\cos A\cos 2A=\dfrac{1}{4}\cos 3A$
Now again consider the term, $\cos A\cos 2A$
By using the trigonometric identities, this can be written as,
The formula is,
$2\cos A\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)$
$\Rightarrow -\dfrac{1}{4}\cos A+\dfrac{1}{4}\left[ 2\cos A\cos 2A \right]=\dfrac{1}{4}\cos 3A$
$\Rightarrow -\dfrac{1}{4}\cos A+\dfrac{1}{4}\left[ \cos \left( 2A+A \right)+\cos \left( 2A-A \right) \right]=\dfrac{1}{4}\cos 3A$
Now upon simplification, we get,
$\Rightarrow -\dfrac{1}{4}\cos A+\dfrac{1}{4}\left[ \cos \left( 3A \right)+\cos \left( A \right) \right]=\dfrac{1}{4}\cos 3A$
Now open the brackets.
$\Rightarrow -\dfrac{1}{4}\cos A+\dfrac{1}{4}\cos \left( 3A \right)+\dfrac{1}{4}\cos A=\dfrac{1}{4}\cos 3A$
Upon evaluating and canceling the common terms we get,
$\Rightarrow \dfrac{1}{4}\cos 3A=\dfrac{1}{4}\cos 3A$
Hence proved.

Note: Whenever complex equations are given to solve one must always Firstly start from the complex side and then convert all the terms into $\sin \theta$ or $\cos \theta$ . Then combine them into single fractions. Now it is most likely to use Trigonometric identities for the transformations if there are any. Know when and where to apply the Subtraction-Addition formula.