Question

considerando f e g funções de lR em lR tal que f(×) = sem ​

Answer 1

$\text {f(x)} = \text{Sen(x)} \ \ e \ \ \text{g(x)} = \text{Cos(x)}$

A)

1)

$\huge\boxed{\text{f}(\pi) = \text{Sen}(\pi) = 0}$ ;

2)

$\huge\boxed{\text{g}(\pi) = \text{Cos}(\pi) = -1}$ ;

3)

$\displaystyle \text{f}(\frac{\pi}{3})- \text g(\frac{\pi }{4}) = \text{Sen}(60^{\circ}}) - \text{Cos}(45^{\circ})$

$\displaystyle \text{f}(\frac{\pi}{3})- \text g(\frac{\pi }{4}) = \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2}$

$\huge\boxed{\displaystyle \text{f}(\frac{\pi}{3})- \text g(\frac{\pi }{4}) = \frac{(\sqrt{3}-\sqrt{2})}{2}}$

4)

$\displaystyle \frac{\displaystyle \text f(\frac{\pi}{6})}{\displaystyle \text g(\frac{\pi}{6})} = \frac{\text {Sen}(30^{\circ})}{\text{Cos}(30^{\circ})}$

$\displaystyle \frac{\displaystyle \text f(\frac{\pi}{6})}{\displaystyle \text g(\frac{\pi}{6})} = \frac{\displaystyle \frac{1}{2}}{\displaystyle \frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}$

$\huge\boxed{\displaystyle \frac{\displaystyle \text f(\frac{\pi}{6})}{\displaystyle \text g(\frac{\pi}{6})} = \frac{\sqrt{3}}{3}}$

B)

$\text x \in [0,2\pi] \ | \ \text{f(x) = g(x)}$

igualando as funções

$\text{Sen(x)=Cos(x)}$

Usando a relação fundamental da trigonometria, sabemos que :

$\text{Sen}^2(\text x)+\text{Cos}^2(\text x) = 1 \to \boxed{\text{Cos}(\text x) = \sqrt{1-\text{Sen}^2(\text x) }}$

substituindo :

$\text{Sen(x)}= \sqrt{1-\text{Sen}^2(\text x)}$

Elevando ao quadrado dos dois lados :

$\text{Sen}^2(\text x) = 1 - \text{Sen}^2(\text x)$

$\displaystyle 2.\text{Sen}^2(\text x) = 1 \to \text{Sen}^2(\text x) = \frac{1}{2}$

$\displaystyle \text{Sen}(\text x) = \pm \frac{1}{\sqrt{2}}$

$\displaystyle {\text{Sen}(\text x) =\pm \frac{\sqrt{2}}{2}}$

$\displaystyle \text {Sen(x)} = + \frac{\sqrt{2}}{2}\to \displaystyle \text x = \frac{\pi}{4}$

$\displaystyle {\text{Sen}(\text x) = - \frac{\sqrt{2}}{2}} \to \text x =\pi +\frac{\pi}{4}\to \text x = \frac{5\pi}{4}$

Portanto

$\huge\boxed{\text{x} = \frac{\pi}{4} \ \ \text{e} \ \ \text x = \frac{5.\pi}{4}}$

C) $\text{Existe x } \in \mathbb{R} \ ,\text{tal que } \displaystyle \frac{\pi}{2}<\text x < \pi \ \ \text e \ \ \text{f(x)=g(x)} \to \text{Sen(x)=Cos(x) } \ ?$

$\displaystyle \frac{\pi}{2}< \text x < \pi \to \boxed{2^{\circ} \text{Quadrante } }$

Não existe x pertencente aos reais tal que a função seno é seja a igual a função cosseno no 2º quadrante.

Justificativa : A função seno no 2º quadrante é positiva, e a função cosseno no 2º quadrante é negativa. Portanto, uma seria o simétrico da outra.