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Question: Prove the following \({{\sin }^{2}}6x-{{\sin }^{2}}4x=\sin 2x\sin 10x\) ....

Prove the following sin⁡26x−sin⁡24x=sin⁡2xsin⁡10x{{\sin }^{2}}6x-{{\sin }^{2}}4x=\sin 2x\sin 10x .

Explanation

Solution

We must use the identity for the expansion of difference of squares a2−b2=(a+b)(a−b){{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) to expand the LHS part of the problem. We can then use the formula sin⁡A+sin⁡B=2sin⁡A+B2cos⁡A−B2\sin A+\sin B=2\sin \dfrac{A+B}{2}\cos \dfrac{A-B}{2} and sin⁡A−sin⁡B=2cos⁡A+B2sin⁡A−B2\sin A-\sin B=2\cos \dfrac{A+B}{2}\sin \dfrac{A-B}{2} to expand sin⁡6x+sin⁡4x\sin 6x+\sin 4x and sin⁡6x−sin⁡4x\sin 6x-\sin 4x respectively. We must then substitute these values in the main LHS equation and use the identity sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta =2\sin \theta \cos \theta to get the required result.

Complete step by step answer:
We have to prove sin⁡26x−sin⁡24x=sin⁡2xsin⁡10x{{\sin }^{2}}6x-{{\sin }^{2}}4x=\sin 2x\sin 10x.
Here, we have
LHS = sin⁡26x−sin⁡24x{{\sin }^{2}}6x-{{\sin }^{2}}4x
We know the identity for the expansion of difference of squares, according to which
a2−b2=(a+b)(a−b){{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)
Using this identity in LHS part, we get
LHS = (sin⁡6x+sin⁡4x)(sin⁡6x−sin⁡4x)...(i)\left( \sin 6x+\sin 4x \right)\left( \sin 6x-\sin 4x \right)...\left( i \right)
We know the identity for the addition of sine, according to which
sin⁡A+sin⁡B=2sin⁡A+B2cos⁡A−B2\sin A+\sin B=2\sin \dfrac{A+B}{2}\cos \dfrac{A-B}{2}
Using this identity, we can write
sin⁡6x+sin⁡4x=2sin⁡6x+4x2cos⁡6x−4x2\sin 6x+\sin 4x=2\sin \dfrac{6x+4x}{2}\cos \dfrac{6x-4x}{2}
Simplifying the above, we get
sin⁡6x+sin⁡4x=2sin⁡5xcos⁡x...(ii)\sin 6x+\sin 4x=2\sin 5x\cos x...\left( ii \right)
We also know the identity for the subtraction of sine, according to which
sin⁡A−sin⁡B=2cos⁡A+B2sin⁡A−B2\sin A-\sin B=2\cos \dfrac{A+B}{2}\sin \dfrac{A-B}{2}
Using this identity, we can write
sin⁡6x−sin⁡4x=2cos⁡6x+4x2sin⁡6x−4x2\sin 6x-\sin 4x=2\cos \dfrac{6x+4x}{2}\sin \dfrac{6x-4x}{2}
Simplifying the above, we get
sin⁡6x−sin⁡4x=2cos⁡5xsin⁡x...(iii)\sin 6x-\sin 4x=2\cos 5x\sin x...\left( iii \right)
Using the values of equation (ii) and equation (iii) in equation (i), we get
LHS = (2sin⁡5xcos⁡x)(2cos⁡5xsin⁡x)\left( 2\sin 5x\cos x \right)\left( 2\cos 5x\sin x \right)
We can regroup the terms as,
LHS = (2sin⁡5xcos⁡5x)(2sin⁡xcos⁡x)\left( 2\sin 5x\cos 5x \right)\left( 2\sin x\cos x \right)
We know that the identity of sine of double angle is
sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta =2\sin \theta \cos \theta
We can now use this identity in the LHS part of our equation,
LHS = sin⁡10xsin⁡2x\sin 10x\sin 2x
Thus, LHS = RHS. Hence, proved sin⁡26x−sin⁡24x=sin⁡2xsin⁡10x{{\sin }^{2}}6x-{{\sin }^{2}}4x=\sin 2x\sin 10x .

Note: We can also use the identities for the expansion of sin⁡2x=2sin⁡xcos⁡x\sin 2x=2\sin x\cos x and for the expansion of sin⁡3x=3sin⁡x−4cos⁡3x\sin 3x=3\sin x-4{{\cos }^{3}}x repeatedly to simplify the equation and get the desired result. But that will be a very long process, and we must avoid using that method.