Question

cos40° -- sin40° = V2sin5°​

Answer 1

$\underline{\bold{Given \:Question - }}$

Prove that

$\red{\rm :\longmapsto\:cos40 \degree \: - \: sin40\degree = \sqrt{2}sin5\degree}$

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

Consider, LHS

$\rm :\longmapsto\:cos40\degree - sin40\degree$

We know,

$\boxed{ \rm{ cos(90\degree - x) = sinx}}$

So, using this identity, we get

$\rm \: = \: \: cos(90\degree - 50\degree) - sin40\degree$

$\rm \: = \: \: sin50\degree - sin40\degree$

Now, we know that,

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

So, using this identity, we get

$\rm \: = \: \: 2cos\bigg(\dfrac{50\degree + 40\degree}{2} \bigg)sin\bigg(\dfrac{50\degree - 40\degree}{2} \bigg)$

$\rm \: = \: \: 2cos\bigg(\dfrac{90\degree }{2} \bigg)sin\bigg(\dfrac{10\degree }{2} \bigg)$

$\rm \: = \: \: 2 \: cos45\degree \: sin5\degree$

$\rm \: = \: \: 2 \: \times \dfrac{1}{ \sqrt{2} } \times \: sin5\degree$

$\rm \: = \: \: \sqrt{2} \times \sqrt{2} \: \times \dfrac{1}{ \sqrt{2} } \times \: sin5\degree$

$\rm \: = \: \: \sqrt{2}sin5\degree$

Hence, Proved

Additional Information :-

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

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

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

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

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

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