Question

Sendo cosx = 2/3, calcule cos(2x). sen(2x)

Answer 1

Supondo "x" no 1º quadrante, vamos começar utilizando a identidade trigonométrica (sen²x+cos²x=1) para determinar o valor de sen(x).

$sen^2x~+~cos^2x~=~1\\\\\\sen^2x~+~\left(\dfrac{2}{3}\right)^2~=~1\\\\\\sen^2x~+~\dfrac{4}{9}~=~1\\\\\\sen^2x~=~1~-~\dfrac{4}{9}\\\\\\sen^2x~=~\dfrac{5}{9}\\\\\\sen(x)~=~\pm\sqrt{\dfrac{5}{9}}\\\\\\sen(x)~=~\pm\dfrac{\sqrt{5}}{3}\\\\\\Como~estamos~considerando~''x''~no~1^oquadrante,~seno~ser\acute{a}~positivo\\\\\\\boxed{sen(x)~=~\dfrac{\sqrt{5}}{3}}$

Com o valor de sen(x), basta utilizarmos as relações para o seno e cosseno da soma de dois arcos para calcular os valores de cos(2x) e sen(2x).

$\boxed{cos(a+b)~=~cos(a)\cdot cos(b)~-~sen(a)\cdot sen(b)}\\\\\\cos(2x)~=~cos(x+x)\\\\\\cos(2x)~=~cos(x)\cdot cos(x)~-~sen(x)\cdot sen(x)\\\\\\cos(2x)~=~\dfrac{2}{3}\cdot \dfrac{2}{3}~-~\dfrac{\sqrt{5}}{3}\cdot \dfrac{\sqrt{5}}{3}\\\\\\cos(2x)~=~\dfrac{4}{9}~-~\dfrac{5}{9}\\\\\\\boxed{cos(2x)~=\,-\dfrac{1}{9}}$

$\boxed{sen(a+b)~=~sen(a)\cdot cos(b)~+~sen(b)\cdot cos(a)}\\\\\\sen(2x)~=~sen(x+x)\\\\\\sen(2x)~=~sen(x)\cdot cos(x)~+~sen(x)\cdot cos(x)\\\\\\sen(2x)~=~\dfrac{2}{3}\cdot\dfrac{\sqrt{5}}{3}~+~\dfrac{\sqrt{5}}{3}\cdot\dfrac{2}{3}\\\\\\sen(2x)~=~\dfrac{2\sqrt{5}}{9}~+~\dfrac{2\sqrt{5}}{9}\\\\\\\boxed{sen(2x)~=~\dfrac{4\sqrt{5}}{9}}$

Por fim, podemos calcular a expressão solicitada:

$cos(2x)\cdot sen(2x)~=\,-\dfrac{1}{9}\cdot\dfrac{4\sqrt{5}}{9}\\\\\\\boxed{cos(2x)\cdot sen(2x)~=\,-\dfrac{4\sqrt{5}}{81}}$