Question

sin24° + cos6° =

[ Hint : write cos 6°= sin84° and apply cos 6°= sin84°​

Answer 1

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

Evaluate

$\rm :\longmapsto\:sin24\degree + cos6\degree$

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

Given Trigonometric function is

$\rm :\longmapsto\:sin24\degree + cos6\degree$

$\rm \: = \: sin24\degree + cos(90\degree - 84\degree )$

$\rm \: = \: sin24\degree + sin84\degree$

$\rm \: = \: sin84\degree + sin24\degree$

We know,

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

So, using this identity, we get

$\rm \: = \: 2sin\bigg[\dfrac{84\degree + 24\degree }{2} \bigg]cos\bigg[\dfrac{84\degree - 24\degree }{2} \bigg]$

$\rm \: = \: 2sin[54\degree ] \: cos[30\degree ]$

We know,

$\red{ \boxed{ \sf{ \:sin54\degree = \frac{ \sqrt{5} + 1 }{4}}}} \: \: and \: \: \: \red{ \boxed{ \sf{ \:cos30\degree = \frac{ \sqrt{3} }{2}}}}$

So, on substituting these values, we get

$\rm \: = \: 2 \times \dfrac{ \sqrt{5} + 1}{4} \times \dfrac{ \sqrt{3} }{2}$

$\rm \: = \: \dfrac{ \sqrt{3}( \sqrt{5} + 1) }{4}$

$\rm \: = \: \dfrac{ \sqrt{15} + \sqrt{3} }{4}$

So,

$\red{ \boxed{ \sf{ \:sin24\degree + cos6\degree = \: \dfrac{ \sqrt{15} + \sqrt{3} }{4}}}}$

Additional Information :-

$\red{ \boxed{ \sf{ \:sin18\degree = \frac{ \sqrt{5} - 1}{4}}}}$

$\red{ \boxed{ \sf{ \:sin36\degree = \frac{ \sqrt{10 - 2 \sqrt{5} } }{4}}}}$

$\red{ \boxed{ \sf{ \:sin72\degree = \frac{ \sqrt{10 + 2 \sqrt{5} } }{4}}}}$

$\red{ \boxed{ \sf{ \:cos18\degree = \frac{ \sqrt{10 + 2 \sqrt{5} } }{4}}}}$

$\red{ \boxed{ \sf{ \:cos72\degree = \frac{ \sqrt{5} - 1}{4}}}}$

$\red{ \boxed{ \sf{ \:cos36\degree = \frac{ \sqrt{5} + 1}{4}}}}$