Question

Seja f: R-R uma função definida por
$f(x) = \sin(2x) \times \cos(2x)$
a) determine o domínio, o período e o conjunto imagem de f.
b) obtenha o valor de
$f( \frac{\pi}{48} )$
​

Answer 1

$\large\boxed{\begin{array}{l}\underline{\sf identidade~de~Prostaferese}\\\sf sen(a)\cdot cos(b)=\dfrac{1}{2}\cdot[ sen(a+b)+sen(a-b)]\end{array}}$

$\large\boxed{\begin{array}{l}\sf f(x)=sen(2x)\cdot cos(2x)\\\sf f(x)=\dfrac{1}{2}\cdot[sen(2x+2x)+sen(2x-2x)]\\\sf f(x)=\dfrac{1}{2}\cdot[ sen(4x)+sen(0)]\\\\\sf f(x)=\dfrac{1}{2}\cdot sen(4x)\\\tt a)~\sf Dom~f(x)=x\in\mathbb{R}\\\sf Im=\bigg[-\dfrac{1}{2},\dfrac{1}{2}\bigg]\\\sf P=\dfrac{2\pi\div2}{4\div2}\\\sf P=\dfrac{\pi}{2}\end{array}}$

$\large\boxed{\begin{array}{l}\tt b)~\sf f\bigg(\dfrac{\pi}{48}\bigg)=\dfrac{1}{2}\cdot\bigg(\diagdown\!\!\!\!4\cdot\dfrac{\pi}{\diagdown\!\!\!\!\!\!48}\bigg)\\\sf f\bigg(\dfrac{\pi}{48}\bigg)=\dfrac{1}{2} sen\bigg(\dfrac{\pi}{12}\bigg)\\\sf f\bigg(\dfrac{\pi}{48}\bigg)=\dfrac{1}{2}\cdot\bigg(\dfrac{\sqrt{6}+\sqrt{2}}{4}\bigg)=\dfrac{\sqrt{6}+\sqrt{2}}{8}\end{array}}$