Question

${ \sin }^{2} 6x - { \sin }^{2} = \sin 2x \: \sin10x$
Solve it:​

Answer 1

$\underline{\huge{Answer:-}}$

$\implies \: \large \mathcal{( \sin6x - \sin4x) \: ( \sin6x + \sin4x)}$

$\implies \: \large \mathcal{ \sin( x ) + \sin (y )= 2 \sin \: \frac{x + y}{2} \cos \: \frac{x - y}{2} }$

$\implies \: \large \mathcal{2 \sin( \frac{6x + 4x}{2}) \cos( \frac{6x - 4x}{2} ) }$

$\implies \: \large \mathcal{2 \sin(5x) \: \cos(x) }$

$\implies \: \large \mathcal{ \sin(x) - \sin(y) = 2 \cos( \frac{6x + 4x}{2} ) \sin( \frac{6x - 4x}{2} ) }$

$\implies \: \large \mathcal{2 \cos(5x) \: \: \sin(x) }$

$\implies \: \large \mathcal {(2 \cos(5x) \sin(x)) \: (2 \sin(5x). \cos(x)) }$

$\implies \: \large \mathcal{2 \sin(x) \: \cos(x) . \: 2 \sin(5x) . \cos(5x) }$

$\Large \fbox\mathcal{ \implies \: \sin(2x) . \sin(10x) }$

$\implies \large \mathcal{LHS = RHS}$

Hence, Proved...

Answer 2

Answer:

$\underline{\huge{Answer:-}}$

$\implies \: \large \mathcal{( \sin6x - \sin4x) \: ( \sin6x + \sin4x)}$

$\implies \: \large \mathcal{ \sin( x ) + \sin (y )= 2 \sin \: \frac{x + y}{2} \cos \: \frac{x - y}{2} }$

$\implies \: \large \mathcal{2 \sin( \frac{6x + 4x}{2}) \cos( \frac{6x - 4x}{2} ) }$

$\implies \: \large \mathcal{2 \sin(5x) \: \cos(x) }$

$\implies \: \large \mathcal{ \sin(x) - \sin(y) = 2 \cos( \frac{6x + 4x}{2} ) \sin( \frac{6x - 4x}{2} ) }$

$\implies \: \large \mathcal{2 \cos(5x) \: \: \sin(x) }$

$\implies \: \large \mathcal {(2 \cos(5x) \sin(x)) \: (2 \sin(5x). \cos(x)) }$

$\implies \: \large \mathcal{2 \sin(x) \: \cos(x) . \: 2 \sin(5x) . \cos(5x) }$

$\implies \large \mathcal{LHS = RHS}$

Hence,Proved...