Provar igualdade trigonométrica inversa

por **Emanuel Dias** Sáb 09 Fev 2019, 17:37

Provar a igualdade:

arc sin(\frac{1}{\sqrt{5}})+arccos(\frac{3}{\sqrt{10}})=\frac{\pi }{4}

O exemplo que vi sobre questões assim não é nem um pouco claro.

por **Giovana Martins** Sáb 09 Fev 2019, 18:16

\\x=arcos(y)\to cos(x)=y\ \wedge\ x=arcsen(y)\to sen(x)=y\\\\\underset{a}{\underbrace{arcsen\left ( \frac{1}{\sqrt{5}} \right )}}+\underset{b}{\underbrace{arccos\left ( \frac{3}{\sqrt{10}} \right )}}=\frac{\pi }{4}\\\\\therefore \ sen(a)=\frac{1}{\sqrt{5}}\ \wedge\ cos(b)=\frac{3}{\sqrt{10}}\ \therefore \ cos(a)=\frac{2}{\sqrt{5}}\ \wedge\ sen(b)=\frac{1}{\sqrt{10}}\\\\a+b=\frac{\pi }{4}\to cos(a+b)=cos\left ( \frac{\pi }{4} \right )\to cos(a)cos(b)-sen(a)sen(b)=\frac{\sqrt{2}}{2}\\\\\left ( \frac{2}{\sqrt{5}} \right )\left ( \frac{3}{\sqrt{10}} \right )-\left ( \frac{1}{\sqrt{5}} \right )\left ( \frac{1}{\sqrt{10}} \right )=\frac{\sqrt{2}}{2}\to\frac{6-1}{5\sqrt{2}}=\frac{\sqrt{2}}{2}\to \frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{2}\ (Ok)\\\\\therefore \ \boxed {arcsen\left ( \frac{1}{\sqrt{5}} \right )+arccos\left ( \frac{3}{\sqrt{10}} \right )=\frac{\pi }{4}}

por **Emanuel Dias** Sáb 09 Fev 2019, 18:23

Giovana Martins escreveu:
\\x=arcos(y)\to cos(x)=y\ \wedge\ x=arcsen(y)\to sen(x)=y\\\\\underset{a}{\underbrace{arcsen\left ( \frac{1}{\sqrt{5}} \right )}}+\underset{b}{\underbrace{arccos\left ( \frac{3}{\sqrt{10}} \right )}}=\frac{\pi }{4}\\\\\therefore \ sen(a)=\frac{1}{\sqrt{5}}\ \wedge\ cos(b)=\frac{3}{\sqrt{10}}\ \therefore \ cos(a)=\frac{2}{\sqrt{5}}\ \wedge\ sen(b)=\frac{1}{\sqrt{10}}\\\\a+b=\frac{\pi }{4}\to cos(a+b)=cos\left ( \frac{\pi }{4} \right )\to cos(a)cos(b)-sen(a)sen(b)=\frac{\sqrt{2}}{2}\\\\\left ( \frac{2}{\sqrt{5}} \right )\left ( \frac{3}{\sqrt{10}} \right )-\left ( \frac{1}{\sqrt{5}} \right )\left ( \frac{1}{\sqrt{10}} \right )=\frac{\sqrt{2}}{2}\to\frac{6-1}{5\sqrt{2}}=\frac{\sqrt{2}}{2}\to \frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{2}\ (Ok)\\\\\therefore \ \boxed {arcsen\left ( \frac{1}{\sqrt{5}} \right )+arccos\left ( \frac{3}{\sqrt{10}} \right )=\frac{\pi }{4}}

Bem mais simples que a forma em que o Iezzi apresenta. Obrigado novamente Giovana.