Q (12) Prove the following

\small \sin^{2}6x - \sin^{2}4x = \sin2x\sin10x

Answers (1)

We know that 
a^{2} - b^{2} = (a+b)(a-b)

So, 
\sin^{2}6x - \sin^{2}4x =(\sin6x + \sin4x)(\sin6x - \sin4x)

Now,  we know that 


\sin A + \sin B = 2\sin \left ( \frac{A+B}{2} \right )\cos\left ( \frac{A-B}{2} \right )\\ \\ \sin A - \sin B = 2\cos \left ( \frac{A+B}{2} \right )\sin\left ( \frac{A-B}{2} \right )
By using these identities
sin6x + sin4x = 2sin5xcosx
sin6x - sin4x = 2cos5xsinx

\Rightarrow \sin^{2}6x - \sin^{2}4x = (2\cos5x\sin5x)(2\sin x\cos x)

Now, 

2sinAcosB = sin(A+B) + sin(A-B)
2cosAsinB = sin(A+B) - sin(A-B)

by using these identities

2cos5xsin5x = sin10x - 0
2sinxcosx = sin2x + 0

hence 
 \sin^{2}6x-\sin^{2}4x = \sin2x\sin10x
                                                                       

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