Função circular inversa

por Paduan Seg 31 Jul 2017, 22:52

arctg 1/3 + arctg 1/5 + arctg 1/7 + arctg 1/8

Obrigado!!

por **Giovana Martins** Dom 17 Fev 2019, 04:50

\\y=\underset{a}{\underbrace{arctg \left ( \frac{1}{3} \right )}} + \underset{b}{\underbrace{arctg \left ( \frac{1}{5} \right )}} +\underset{c}{\underbrace{arctg \left ( \frac{1}{7} \right )}}+ \underset{d}{\underbrace{arctg \left ( \frac{1}{8} \right )}}=a+b+c+d\\\\tg(y)=tg[(a+b)+(c+d)]=\frac{tg(a+b)+tg(c+d)}{1-tg(a+b)tg(c+d)}\\\\tg(y)=\frac{\left [ \frac{tg(a)+tg(b)}{1-tg(a)tg(b)} \right ]+\left [ \frac{tg(c)+tg(d)}{1-tg(c)tg(d)} \right ] }{1-\left [ \frac{tg(a)+tg(b)}{1-tg(a)tg(b)} \right ]\left [ \frac{tg(c)+tg(d)}{1-tg(c)tg(d)} \right ]}\to tg(y)=\frac{ \frac{\frac{1}{3}+\frac{1}{5}}{1-\frac{1}{15}}+ \frac{\frac{1}{7}+\frac{1}{8}}{1-\frac{1}{56}}}{1-\left ( \frac{\frac{1}{3}+\frac{1}{5}}{1-\frac{1}{15}} \right )\left ( \frac{\frac{1}{7}+\frac{1}{8}}{1-\frac{1}{56}} \right )}=1\\\\ \therefore \ \boxed {arctg \left (\frac{1}{3} \right ) + arctg \left (\frac{1}{5} \right ) + arctg \left (\frac{1}{7} \right ) + arctg \left (\frac{1}{8} \right )=\frac{\pi }{4}}

Função circular inversa

Função circular inversa

Re: Função circular inversa

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