Question

$\frac{cos5+sin25}{cos25-sin5}$

Answer 1

$\frac{cos5+sin25}{cos25-sin5} = \frac{sin85+sin25}{sin65-sin5}$
$= \frac{sin2 \frac{85+25}{2} *cos\frac{85-25}{2} }{sin2 \frac{65-5}{2} * cos \frac{65+5}{2} }$$\frac{sin55*cos30}{sin30*cos35} = \frac{sin55*cos30}{sin30*sin55}$
$\frac{ \sqrt{3} }{2} * \frac{2}{1} = \sqrt{3}$

Answer 2

Folosim relația :

cos a = sin (90° - a)

și, referitor la enunț,  vom avea:

cos5° = sin(90° - 5°) = sin85°

cos25°= sin(90°- 25°) =sin65°

Raportul din enunț se va scrie:

$\it \dfrac{sin85^o+sin25^o}{sin65^o - sin5^o}$

Acum, folosim relațiile :

sina + sinb = 2sin(a+b)/2 cos(a-b)/2

sina - sinb = 2sin(a-b)/2 cos(a+b)/2

și raportul devine:


$\it \dfrac{2sin\dfrac{85^o+25^o}{2}cos\dfrac{85^o-25^o}{2}}{2sin\dfrac{65^o-5^o}{2}cos\dfrac{65^o+5^o}{2}} = \dfrac{sin55^ocos30^o}{sin30^ocos35^o}$

Știm că avem  :

cos30° = sin(90°- 30°) = sin60°

cos35° = sin(90° - 35°) = sin55°

 Raportul devine :



$\it \dfrac{sin55^o sin60^o}{sin30^osin55^o} = \dfrac{sin60^o}{sin30^o} = \dfrac{\dfrac{\sqrt3}{2}}{\dfrac{1}{2}} = \dfrac{\sqrt3}{2} : \dfrac{1}{2} = \dfrac{\sqrt3}{2} \cdot \dfrac{2}{1} = \sqrt3$