(ITAJUBÁ-1959) Trigonometria
Enviado: 11 Out 2016, 22:39
Provar a igualdade:
[tex3]arc \tan\frac{1}{2}+arc \tan\frac{1}{3}=\frac{\pi }{4}[/tex3]
[tex3]arc \tan\frac{1}{2}+arc \tan\frac{1}{3}=\frac{\pi }{4}[/tex3]
[tex3]\arctan\frac{1}{2}=\frac{\pi }{4}-\arctan\frac{1}{3}[/tex3]
[tex3]\tg \( \arctan\frac{1}{2}\)=\tg \( \frac{\pi }{4}-\arctan\frac{1}{3}\)[/tex3]
[tex3]\frac{1}{2}=\frac{\tg \(\frac{\pi}{4}\) -\tg \(\arctan\frac{1}{3}\) }{1 +\tg \(\frac{\pi}{4}\) \cdot \tg \(\arctan\frac{1}{3}\) }[/tex3]
[tex3]\frac{1}{2}=\frac{1 -\frac{1}{3}}{1 +\frac{1}{3} }[/tex3]
[tex3]\frac{1}{2}=\frac{1}{2}[/tex3]