Question

Prove : cos²2x - cos²6x = sin4x sin8x​

Answer 1

Step-by-step explanation:

Proof :-

On taking LHS :-

Cos² 2X - Cos² 6X

=> (Cos 2X)² - (Cos 6X)²

We know that

(a+b)(a-b) = a²-b²

Where, a = Cos 2X and b = Cos 6X

=>(Cos 2X+Cos 6X)(Cos 2X - Cos 6X) --------- (1)

We know that

Cos A + Cos B

= 2 Cos (A+B)/2 Cos (A-B)/2

Then ,Cos 2X + Cos 6X becomes

=2 Cos (2X+6X)/2 Cos (2X-6X)/2

Where , A = 2X and B = 6X

= 2 Cos (8X/2) Cos (-4X)/2

= 2 Cos 4X Cos (-2X)

We know that

Cos (-A) = Cos A

Therefore,

Cos 2X+Cos 6X = 2 Cos 4X Cos 2X --------(2)

and

We know that

Cos A - Cos B

= - 2 Sin (A+B)/2 Sin (A-B)/2

Then , Cos 2X - Cos 6X becomes

= -2 Sin (2X+6X)/2 Sin (2X-6X)/2

= -2 Sin(8X/2) Sin (-4X)/2

= - 2 Sin 4X Sin (-2X)

We know that

Sin (-A) = - Sin A

Therefore,

Cos 2X - Cos 6X = -(-2 Sin 4X Sin 2X)

Cos 2X - Cos 6X = 2 Sin 4X Sin 2X ------(3)

On multiplying (2) & (3) then

(Cos 2X+Cos 6X) ( Cos 2X - Cos 6X)

= (2 Cos 4X Cos 2X)(2 Sin 4X Sin 2X)

On rearranging them then

= (2 Sin 4X Cos 4X ) (2 Sin 2X Cos 2X)

On applying the formula

Sin 2A = 2 Sin A Cos A

Where, A = 4X in the first expression and A = 2X in the second expression

= [ Sin 2(4X) Sin 2(2X)]

= Sin 8X Sin 4X

= RHS

=> LHS = RHS

Cos² 2X - Cos² 6X = Sin 8X Sin 4X

Hence, Proved.

Used formulae:-

→ Cos A + Cos B

= 2 Cos (A+B)/2 Cos (A-B)/2

→ Cos A - Cos B

= - 2 Sin (A+B)/2 Sin (A-B)/2

→ Sin (-A) = - Sin A

→ Cos (-A) = Cos A

→ Sin 2A = 2 Sin A Cos A

→ (a+b)(a-b) = a²-b²

Answer 2

$\large\underline{\sf{Solution-}}$

Consider, LHS

$\rm \: {cos}^{2}2x - {cos}^{2}6x$

can be rewritten as

$\rm \: = \: \dfrac{1}{2}\bigg(2{cos}^{2}2x - 2 {cos}^{2}6x\bigg)$

We know,

$\boxed{\sf{ \: {2cos}^{2}x = 1 + cos2x \: }}$

So, on using this result, we get

$\rm \: = \: \dfrac{1}{2}\bigg((1 + cos4x) - (1 + cos12x)\bigg)$

$\rm \: = \: \dfrac{1}{2}\bigg(1 + cos4x - 1 - cos12x\bigg)$

$\rm \: = \: \dfrac{1}{2}\bigg(cos4x - cos12x\bigg)$

Now, we know that,

$\boxed{\sf{ \:cosx - cosy \: = \: 2sin\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{y - x}{2} \bigg] \: }}$

So, using this result, we get

$\rm \: = \: \dfrac{1}{2}\bigg(2sin\bigg[\dfrac{4x + 12x}{2} \bigg]sin\bigg[\dfrac{12x - 4x}{2} \bigg]\bigg)$

$\rm \: = \:sin\bigg[\dfrac{16x}{2} \bigg]sin\bigg[\dfrac{8x}{2} \bigg]$

$\rm \: = \: sin8x \: sin4x$

Hence,

$\rm\implies \:\boxed{\sf{ \: {cos}^{2}2x - {cos}^{2}6x = \: sin4x \: sin8x \: \: }}$

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Short Cut Formula

$\boxed{\sf{ \: {cos}^{2}x - {cos}^{2} y = sin(x + y) \: sin(y - x) \: }}$

So, using this result,

$\rm \: {cos}^{2}2x - {cos}^{2}6x$

$\rm \: = \: sin(2x + 6x) \: sin(6x - 2x)$

$\rm \: = \: sin8x \: sin4x$

Hence,

$\rm\implies \:\boxed{\sf{ \: {cos}^{2}2x - {cos}^{2}6x = \: sin4x \: sin8x \: \: }}$

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Additional Information

$\boxed{\sf{ \: {sin}^{2}x - {sin}^{2}y = sin(x + y) \: sin(x - y) \: }}$

$\boxed{\sf{ \: {cos}^{2}x - {sin}^{2}y = cos(x + y) \: cos(x - y) \: }}$

$\boxed{\sf{ \:cosx + cosy \: = \: 2cos\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg] \: }}$

$\boxed{\sf{ \:sinx + siny \: = \: 2sin\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg] \: }}$

$\boxed{\sf{ \:sinx - siny \: = \: 2cos\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{x - y}{2} \bigg] \: }}$