(Bélgica) Trigonometria
Enviado: 24 Jun 2016, 10:15
Considere [tex3]t=\frac{\sen x }{1+\cos x }[/tex3] , prove que [tex3]\cos x =\frac{1-t^2}{1+t^2}[/tex3]
[tex3]t^2 = \frac{\sin^2 x}{(1+\cos x)^2} \therefore t^2 (1+ 2\cos x + \cos^2 x) = 1 - \cos^2 x \\ (1+t^2) \cos^2 x +2t^2 \cos x+(t^2-1) =0 \\ \cos x = \frac{-2t^2 \pm \sqrt{4t^2-4(t^2+1)(t^2-1)}}{2(1+t^2)}= \frac{-2t^2 \pm \sqrt{4t^2-4(t^4-1)}}{2(1+t^2)} \\ \cos x= \frac{-2t^2 \pm 2}{2(1+t^2)}[/tex3]
Como [tex3]|\cos x| \leq 1[/tex3],
[tex3]\cos x = \frac{1-t^2}{1+t^2}[/tex3]