[tex]sin^2(x)+cos^2(x) = 1[/tex]
[tex]sin^2( \frac{ \pi }{12}) + cos^2 ( \frac{ \pi }{12})=1 [/tex]
sin(2x) = 2sin(x)cos(x)
[tex]2*sin( \frac{ \pi }{12})*cos( \frac{\pi}{12} ) = sin(2* \frac{\pi}{12}) = sin( \frac{\pi}{6} ) = 1/2[/tex]
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)
[tex] \left \{ {{x^2+y^2=1} \atop {2xy= \frac{1}{2} }} \right. [/tex] | (+)
[tex]x^2 + 2xy+y^2 = \frac{3}{2} [/tex]
[tex](x+y)^2 = \frac{3}{2} [/tex] (Ambele sunt pozitive)
=> [tex]x+y = \frac{ \sqrt{6} }{2} [/tex]
Revenim la sistem
[tex] \left \{ {{x^2+y^2=1} \atop {2xy= \frac{1}{2} }} \right. [/tex] | (-)
[tex]x^2 -2xy+y^2 = \frac{1}{2} [/tex]
=> x-y= [tex] \frac{ \sqrt{2} }{2} [/tex]
[tex] \left \{ {{x+y= \frac{ \sqrt{6} }{2} } \atop {x-y= \frac{ \sqrt{2} }{2} }} \right.[/tex] - Daca scadem si adunam gasim solutiile
2x = [tex] \frac{ \sqrt{6} + \sqrt{2} }{2} [/tex]
2y = [tex] \frac{ \sqrt{6} - \sqrt{2} }{2} [/tex]
De aici rezulta
[tex]sin ( \frac{ \pi }{12}) = y = \frac{ \sqrt{6} - \sqrt{2} }{4} [/tex] - Exact ce trebuia demonstrat
