Question

sin70 +sin50 =√3 cos10​

Answer 1

$\large\underline{\sf{To\:prove - }}$

$\red{\rm :\longmapsto\:sin70\degree + sin50\degree = \sqrt{3} \: cos10\degree }$

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

Consider LHS

$\red{\rm :\longmapsto\:sin70\degree + sin50\degree }$

We know that

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

So, using this result, we get

$\rm \: = \:2sin\bigg[\dfrac{70\degree + 50\degree }{2} \bigg]cos\bigg[\dfrac{70\degree - 50\degree }{2} \bigg]$

$\rm \: = \:2sin\bigg[\dfrac{120\degree }{2} \bigg]cos\bigg[\dfrac{20\degree}{2} \bigg]$

$\rm \: = \:2sin60\degree cos10\degree$

$\rm \: = \:2 \times \dfrac{ \sqrt{3} }{2} \times cos10\degree$

$\rm \: = \: \sqrt{3} \: cos10\degree$

Hence,

$\red {\boxed{ \bf{ \: {\bf :\longmapsto\:sin70\degree + sin50\degree = \sqrt{3} \: cos10\degree }}}}$

Additional Information :-

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

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

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

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

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

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