Question

11.
$prove \: cos ((3\pi \div 4) + x) - ((3\pi \div 4) - x = - \sqrt{2 \sin(x) }$
​

Answer 1

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

Consider, LHS

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

We know,

Using Transformation Formulas, we have

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

On using this identity, we get

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

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

$\rm \: = \: \: - \: 2 \: sin\dfrac{3\pi}{4} \: sinx$

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

We know,

$\blue{ \boxed{\bf \:sin(\pi - x) = sinx}}$

Using this identity, we get

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

$\rm \: = \: \: - 2 \times \dfrac{1}{ \sqrt{2} } \times sinx$

$\rm \: = \: \: - \sqrt{2} sinx$

Hence,

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

Additional Information :-

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

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

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

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

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

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