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2014-05-11T23:17:40-03:00
Tem uma fórmula na trigonometria que diz o seguinte:

\boxed{\mathrm{tg}\hspace{0,2mm}(a+b)=\frac{\mathrm{tg}\hspace{0,2mm}a+\mathrm{tg}\hspace{0,2mm}b}{1-\mathrm{tg}\hspace{0,2mm}a.\mathrm{tg}\hspace{0,2mm}b}}

Fazendo a=\frac{\pi}{18} e b=\frac{\pi}{9} teremos:

\mathrm{tg}\hspace{0,2mm}\left(\frac{\pi}{18}+\frac{\pi}{9}\right)=\frac{\mathrm{tg}\hspace{0,2mm}\frac{\pi}{18}+\mathrm{tg}\hspace{0,2mm}\frac{\pi}{9}}{1-\mathrm{tg}\hspace{0,2mm}\frac{\pi}{18}.\mathrm{tg}\hspace{0,2mm}\frac{\pi}{9}}

\mathrm{tg}\hspace{0,2mm}\left(\frac{3\pi}{18}\right)=\frac{\mathrm{tg}\hspace{0,2mm}\frac{\pi}{18}+\mathrm{tg}\hspace{0,2mm}\frac{\pi}{9}}{1-\mathrm{tg}\hspace{0,2mm}\frac{\pi}{18}.\mathrm{tg}\hspace{0,2mm}\frac{\pi}{9}}

\mathrm{tg}\hspace{0,2mm}\left(\frac{\pi}{6}\right)=\frac{\mathrm{tg}\hspace{0,2mm}\frac{\pi}{18}+\mathrm{tg}\hspace{0,2mm}\frac{\pi}{9}}{1-\mathrm{tg}\hspace{0,2mm}\frac{\pi}{18}.\mathrm{tg}\hspace{0,2mm}\frac{\pi}{9}}

\boxed{\boxed{\frac{\mathrm{tg}\hspace{0,2mm}\frac{\pi}{18}+\mathrm{tg}\hspace{0,2mm}\frac{\pi}{9}}{1-\mathrm{tg}\hspace{0,2mm}\frac{\pi}{18}.\mathrm{tg}\hspace{0,2mm}\frac{\pi}{9}}=\frac{\sqrt3}{3}}}
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