Pls and answer both parts in detail

1.	Pls and answer both parts in detail
Answer» $\large\underline{\sf{Given \:Question - }}$ Prove that $\sf \: \: (1). \: \: sin2x + 2sin4x + sin6x = 4 {cos}^{2}xsin4x$ $\sf \: \: (2). \: \: cos\bigg(\dfrac{3\pi}{4} + x \bigg) - cos\bigg(\dfrac{3\pi}{4} - x\bigg) = - \: \sqrt{2} \: sinx$ $\large\underline{\sf{Solution-1}}$ Consider, $\rm :\longmapsto\:sin2x + 2sin4x + sin6x$ can be re-arranged as $\rm \: = \:(sin6x + sin2x) + 2sin4x$ We know, $\boxed{ \bf{ \: sinx + siny = 2sin\bigg(\dfrac{x + y}{2} \bigg)cos\bigg(\dfrac{x - y}{2} \bigg)}}$ Using this identity, we get $\rm \: = \:2sin\bigg(\dfrac{6x + 2x}{2} \bigg)cos\bigg(\dfrac{6x - 2x}{2} \bigg) + 2sin4x$ $\rm \: = \:2sin\bigg(\dfrac{8x}{2} \bigg)cos\bigg(\dfrac{4x}{2} \bigg) + 2sin4x$ $\rm \: = \:2sin4xcos2x + 2sin4x$ $\rm \: = \:2sin4x(1 + cos2x)$ We know, $\boxed{ \bf{ \: 1 + cos2x = {2cos}^{2}x}}$ $\rm \: = \:2sin4x(2 {cos}^{2}x)$ $\rm \: = \:4sin4x \: {cos}^{2}x$ HENCE, $\boxed{ \bf{ \: sin2x + 2sin4x + sin6x = 4 {cos}^{2}xsin4x}}$ $\red{\large\underline{\sf{Solution-2}}}$ Consider, $\rm :\longmapsto\: cos\bigg(\dfrac{3\pi}{4} + x \bigg) - cos\bigg(\dfrac{3\pi}{4} - x\bigg)$ We know, $\boxed{ \bf{ \: cosx - cosy = - \: 2sin\bigg(\dfrac{x + y}{2} \bigg)sin\bigg(\dfrac{x - y}{2} \bigg)}}$ Using this identity, we get $\rm \: = \: - 2sin\bigg(\dfrac{\dfrac{3\pi}{4} + x + \dfrac{3\pi}{4} - x}{2} \bigg)sin\bigg(\dfrac{\dfrac{3\pi}{4} + x - \dfrac{3\pi}{4} + x}{2} \bigg)$ $\rm \: = \: - 2sin\bigg(\dfrac{\dfrac{3\pi}{4} + \dfrac{3\pi}{4} }{2} \bigg)sin\bigg(\dfrac{x + x}{2} \bigg)$ $\rm \: = \: - 2sin\bigg(\dfrac{2 \times \dfrac{3\pi}{4} }{2} \bigg)sin\bigg(\dfrac{2x}{2} \bigg)$ $\rm \: = \: - 2sin\bigg(\dfrac{3\pi}{4} \bigg)sinx$ $\rm \: = \: - 2sin\bigg(\pi - \dfrac{\pi}{4} \bigg)sinx$ $\rm \: = \: - 2sin\bigg(\dfrac{\pi}{4} \bigg)sinx$ $\rm \: = \: - 2 \times \dfrac{1}{ \sqrt{2} } \times sinx$ $\rm \: = \: - \sqrt{2} \times \sqrt{2} \times \dfrac{1}{ \sqrt{2} } \times sinx$ $\rm \: = \: - \sqrt{2} \: sinx$ Hence, $\boxed{ \bf{ \: cos\bigg(\dfrac{3\pi}{4} + x \bigg) - cos\bigg(\dfrac{3\pi}{4} - x\bigg) = - \: \sqrt{2} \: sinx}}$ ADDITIONAL INFORMATION :- $\boxed{ \bf{ \: sinx - siny = 2cos\bigg(\dfrac{x + y}{2} \bigg)sin\bigg(\dfrac{x - y}{2} \bigg)}}$ $\boxed{ \bf{ \: cosx + cosy = 2cos\bigg(\dfrac{x + y}{2} \bigg)cos\bigg(\dfrac{x - y}{2} \bigg)}}$ $\boxed{ \bf{ \: 2cosxcosy = cos(x + y) + cos(x - y)}}$ $\boxed{ \bf{ \: 2sinxsiny = cos(x - y) - cos(x + y)}}$ $\boxed{ \bf{ \: 2sinxcosy = sin(x + y) + sin(x - y)}}$

Answer»

$\large\underline{\sf{Given \:Question - }}$

Prove that

$\sf \: \: (1). \: \: sin2x + 2sin4x + sin6x = 4 {cos}^{2}xsin4x$

$\sf \: \: (2). \: \: cos\bigg(\dfrac{3\pi}{4} + x \bigg) - cos\bigg(\dfrac{3\pi}{4} - x\bigg) = - \: \sqrt{2} \: sinx$

$\large\underline{\sf{Solution-1}}$

Consider,

$\rm :\longmapsto\:sin2x + 2sin4x + sin6x$

can be re-arranged as

$\rm \: = \:(sin6x + sin2x) + 2sin4x$

We know,

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

Using this identity, we get

$\rm \: = \:2sin\bigg(\dfrac{6x + 2x}{2} \bigg)cos\bigg(\dfrac{6x - 2x}{2} \bigg) + 2sin4x$

$\rm \: = \:2sin\bigg(\dfrac{8x}{2} \bigg)cos\bigg(\dfrac{4x}{2} \bigg) + 2sin4x$

$\rm \: = \:2sin4xcos2x + 2sin4x$

$\rm \: = \:2sin4x(1 + cos2x)$

We know,

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

$\rm \: = \:2sin4x(2 {cos}^{2}x)$

$\rm \: = \:4sin4x \: {cos}^{2}x$

HENCE,

$\boxed{ \bf{ \: sin2x + 2sin4x + sin6x = 4 {cos}^{2}xsin4x}}$

$\red{\large\underline{\sf{Solution-2}}}$

Consider,

$\rm :\longmapsto\: cos\bigg(\dfrac{3\pi}{4} + x \bigg) - cos\bigg(\dfrac{3\pi}{4} - x\bigg)$

We know,

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

Using this identity, we get

$\rm \: = \: - 2sin\bigg(\dfrac{\dfrac{3\pi}{4} + x + \dfrac{3\pi}{4} - x}{2} \bigg)sin\bigg(\dfrac{\dfrac{3\pi}{4} + x - \dfrac{3\pi}{4} + x}{2} \bigg)$