Question

13. Cos ² 2x
Cos ²2x- Cos26x = Sin 4o. Sin 8 x
​

Answer 1

Appropriate Question :-

Prove that

$\rm :\longmapsto\: {cos}^{2}2x - {cos}^{2}6x = sin4x \: sin8x$

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

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

$\: \: \rm= \: \:(1 - {sin}^{2}2x) - (1 - {sin}^{2}6x)$

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

$\: \: \rm= \: \:1 - {sin}^{2}2x -1 + {sin}^{2}6x$

$\: \: \rm= \: \: {sin}^{2}6x - {sin}^{2}2x$

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

$\: \: \rm= \: \:sin(8x) \: sin(4x)$

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

Hence,

$\bf :\longmapsto\: {cos}^{2}2x - {cos}^{2}6x = sin4x \: sin8x$

Alter Method :-

Consider,

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

$\: \: \rm= \: \:\dfrac{1 + cos2(2x)}{2} - \dfrac{1 + cos2(6x)}{2}$

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

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

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

$\: \: \rm= \: \:\dfrac{cos4x - cos12x}{2}$

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

$\boxed{ \sf{ \because\:cosx - cosy = - \: 2 \: sin\bigg(\dfrac{x + y}{2} \bigg)sin\bigg(\dfrac{x - y}{2} \bigg)}}$

$\: \: \rm= \: \: - \: sin\bigg(\dfrac{ - 8x}{2} \bigg)sin\bigg(\dfrac{16x}{2} \bigg)$

$\: \: \rm= \: \: - \: sin( - 4x) \: sin(8x)$

$\: \: \rm= \: \: \: sin( 4x) \: sin(8x)$

Hence,

$\bf :\longmapsto\: {cos}^{2}2x - {cos}^{2}6x = sin4x \: sin8x$

Additional Information:-

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

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

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

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

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