Question

aratati ca sin de pi/12=radical din 6-radical din 2/4

Answer 1

$sin^2(x)+cos^2(x) = 1$
$sin^2( \frac{ \pi }{12}) + cos^2 ( \frac{ \pi }{12})=1$

sin(2x) = 2sin(x)cos(x)
$2*sin( \frac{ \pi }{12})*cos( \frac{\pi}{12} ) = sin(2* \frac{\pi}{12}) = sin( \frac{\pi}{6} ) = 1/2$

x = cos (pi/12) si y = sin(pi/12)
Atat x cat si y sunt pozitivi (pi/12 fiind in primul cadran)
x > y (cos e descrescator & sin e crescator)

$\left \{ {{x^2+y^2=1} \atop {2xy= \frac{1}{2} }} \right.$ | (+)

$x^2 + 2xy+y^2 = \frac{3}{2}$
$(x+y)^2 = \frac{3}{2}$ (Ambele sunt pozitive) 
=> $x+y = \frac{ \sqrt{6} }{2}$


Revenim la sistem
$\left \{ {{x^2+y^2=1} \atop {2xy= \frac{1}{2} }} \right.$ | (-)
$x^2 -2xy+y^2 = \frac{1}{2}$
=> x-y= $\frac{ \sqrt{2} }{2}$

$\left \{ {{x+y= \frac{ \sqrt{6} }{2} } \atop {x-y= \frac{ \sqrt{2} }{2} }} \right.$ - Daca scadem si adunam gasim solutiile

2x = $\frac{ \sqrt{6} + \sqrt{2} }{2}$
2y = $\frac{ \sqrt{6} - \sqrt{2} }{2}$
De aici rezulta

$sin ( \frac{ \pi }{12}) = y = \frac{ \sqrt{6} - \sqrt{2} }{4}$ - Exact ce trebuia demonstrat