Question

Solve the above with clear steps.

Answer 1

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

Convert the following in to product :

$\rm :\longmapsto\:sin50\degree - cos80\degree$

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

Given Trigonometric function is ,

$\rm :\longmapsto\:sin50\degree - cos80\degree$

can be rewritten as

$\rm \:  =  \:  \: \:sin50\degree - cos(90\degree - 10\degree )$

We know that,

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

So, using this result, we get

$\rm \:  =  \:  \: \:sin50\degree - sin10\degree$

Now, we know that

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

So, using this identity, we get

$\rm \:  =  \:  \: 2cos\bigg(\dfrac{50\degree + 10\degree }{2} \bigg)sin\bigg(\dfrac{50\degree - 10\degree }{2} \bigg)$

$\rm \:  =  \:  \: 2cos\bigg(\dfrac{60\degree }{2} \bigg)sin\bigg(\dfrac{40\degree }{2} \bigg)$

$\rm \:  =  \:  \: 2cos30\degree sin20\degree$

$\rm \:  =  \:  \: 2 \times \dfrac{ \sqrt{3} }{2} \times sin20\degree$

$\rm \:  =  \:  \: \sqrt{3} \: sin20\degree$

Hence,

$\bf :\longmapsto\:sin50\degree - cos80\degree = \sqrt{3} \: sin20\degree$

Additional Information :-

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

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

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

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

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

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