Question

calcule:
cos[arctan 2 + arctan 3]

Answer 1

$\sf cos(arc~tg(2)+arc~tg(3))\\\sf arc~tg(2)=a\implies tg(a)=2\\\sf arc~tg(3)=b\implies tg(b)=3\\\sf cos(arc~tg(2)+arc~tg(3))=cos(a+b)=cos(a)\cdot cos(b)- sen(a)\cdot sen(b)$

$\underline{\rm c\acute alculos~auxiliares:}\\\sf 1+tg^2(a)=sec^2(a)\\\sf sec^2(a)=1+2^2\\\sf sec^2(a)=1+4=5\\\sf sec(a)=\sqrt{5}\implies \boxed{\sf cos(a)=\dfrac{1}{\sqrt{5}}=\dfrac{\sqrt{5}}{5}} \\\sf sec^2(b)=1+tg^2(b)\\\sf sec^2(b)=1+3^2\\\sf sec^2(b)=1+9=10\implies sec(b)=\sqrt{10}\\\boxed{\sf cos(b)=\dfrac{1}{\sqrt{10}}=\dfrac{\sqrt{10}}{10}}$

$\sf sen(a)=cos(a)\cdot tg(a)\\\boxed{\sf sen(a)=\dfrac{\sqrt{5}}{5}\cdot2=\dfrac{2\sqrt{5}}{5}}\\\sf sen(b)=cos(b)\cdot tg(b)\\\boxed{\sf sen(b)=\dfrac{\sqrt{10}}{10}\cdot 3=\dfrac{3\sqrt{10}}{10}}$

$\sf cos(a+b)=cos(a)\cdot cos(b)-sen(a)\cdot sen(b)\\\sf cos(a+b)=\dfrac{\sqrt{5}}{5}\cdot\dfrac{\sqrt{10}}{10}-\dfrac{2\sqrt{5}}{5}\cdot\dfrac{3\sqrt{10}}{10}\\\sf cos(a+b)=\dfrac{5\sqrt{2}}{50}-\dfrac{30\sqrt{2}}{50}\\\sf cos(a+b)=-\dfrac{\diagup\!\!\!\!\!\!25\sqrt{2}}{\diagup\!\!\!\!\!50}=-\dfrac{\sqrt{2}}{2}$

$\boxed{\boxed{\boxed{\boxed{\sf cos( arc~tg(2)+arc~tg(3))=-\dfrac{\sqrt{2}}{2}}}}}\blue{\checkmark}$