Question

Prove the following equation:
${\cos ^2}2x - {\cos ^2}6x = \sin 4x\sin 8x$

Answer 1

Hint- We will be using the following trigonometric identities as L.H.S. has the term ${\cos ^2}2x$.
We know $\Rightarrow \cos 2x = 2{\cos ^2}x - 1$ …(1)
Replacing $x \to 2x$ in equation (1), then we get
$\Rightarrow \cos 4x = 2{\cos ^2}2x - 1$
$\Rightarrow {\cos ^2}2x = \dfrac{{\cos 4x + 1}}{2}$ …(2)
and replacing $x \to 6x$ in equation (1), then we get
$\Rightarrow \cos 12x = 2{\cos ^2}6x - 1$
$\Rightarrow {\cos ^2}6x = \dfrac{{\cos 12x + 1}}{2}$ ….(3)

Complete step-by-step answer:
According to question, we have
${\cos ^2}2x - {\cos ^2}6x = \sin 4x\sin 8x$ ….(4)
Considering LHS,
${\cos ^2}2x - {\cos ^2}6x$
Now replacing ${\cos ^2}2x$ and ${\cos ^2}6x$ from equation (2) and (3), we get
$\Rightarrow \dfrac{{\cos 4x + 1}}{2} - \dfrac{{\cos 12x + 1}}{2}$
$\Rightarrow \dfrac{1}{2}\cos 4x + \dfrac{1}{2} - \dfrac{1}{2}\cos 12x - \dfrac{1}{2}$
\Rightarrow \dfrac{1}{2}\left\\{ {\cos 4x - \cos 12x} \right\\}
Now using formula $\cos A - \cos B = 2\sin \dfrac{{A + B}}{2}\sin \dfrac{{B - A}}{2}$ we get
\dfrac{1}{2}\left\\{ {2\sin \dfrac{{4x + 12x}}{2}\sin \dfrac{{12x - 4x}}{2}} \right\\} \Rightarrow \sin \dfrac{{16x}}{2}\sin \dfrac{{8x}}{2}
$\Rightarrow \sin 4x\sin 8x$ = RHS
Hence, we proved that ${\cos ^2}2x - {\cos ^2}6x = \sin 4x\sin 8x$.

Note- We have to solve such questions based on proving by using trigonometric identities. Also, it can be proved from both sides, from LHS and from RHS both. Also, if it is not getting solved by simplifying one side, then in that case we will simplify both sides and try to prove the given expression.