Question

cos40 + sin 40=\sqrt{ 2 cos 5}

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

We are aware of the Trigonometric identity that

$\displaystyle \: cos \: C + cos \: D =2 \: cos \frac{C + D}{2} \: cos \frac{C - D}{2}$

TO PROVE

$cos {40}^{ \circ} + sin {40}^{ \circ} = \sqrt{2} \: cos {5}^{ \circ}$

PROOF

$cos {40}^{ \circ} + sin {40}^{ \circ}$

$= cos {40}^{ \circ} + sin({90}^{ \circ} - {50}^{ \circ} )$

$= cos {40}^{ \circ} + cos {50}^{ \circ}$

$\displaystyle \: = 2 \: cos \frac{ {40}^{ \circ} + {50}^{ \circ} }{2} \: \: cos \frac{ {40}^{ \circ} - {50}^{ \circ} }{2}$

$= 2 \: cos {45}^{ \circ} cos ( - {5}^{ \circ} \: )$

$=2 \: cos {45}^{ \circ} cos {5}^{ \circ} \:$

$\displaystyle \: =2 \: \times \frac{1}{ \sqrt{2} } \times cos {5}^{ \circ} \:$

$\displaystyle \: = \sqrt{2} cos {5}^{ \circ} \:$

Hence proved