Question

(EFOMM/2012-2013) Se det $\left[\begin{array}{ccc}cosx&senx\\seny&cosy\\\end{array}\right] = \frac{-1}{3}$ então o valor de 3sen(x+y) +tg(x+y) - sec(x+y), para $\frac{ \pi }{2}$ ≤ x+y ≤ $\pi$, é igual a:

a) 0
b) 1/3
c) 2
d) 3
e) 1/2

Answer 1

$\det\!\left[ \begin{array}{cc} \cos x&\mathrm{sen\,}x\\\\ \mathrm{sen\,}y&\cos y \end{array} \right]=-\,\dfrac{1}{3}\\\\\\ \cos x\cos y-\mathrm{sen\,}y\,\mathrm{sen\,}x=-\,\dfrac{1}{3}\\\\\\ \boxed{\begin{array}{c}\cos(x+y)=-\,\dfrac{1}{3}\end{array}}$

_________

• Calculando o seno:

Elevando os dois lados ao quadrado, temos

$\cos^2(x+y)=\left(-\,\dfrac{1}{3}\right)^{\!\!2}\\\\\\ \cos^2(x+y)=\dfrac{1}{9}\\\\\\ 9\cos^2(x+y)=1~~~~~~\big(\text{mas } \cos^2(x+y)=1-\mathrm{sen^2}(x+y)\big)\\\\ 9\cdot (1-\mathrm{sen^2}(x+y))=1\\\\ 9-9\,\mathrm{sen^2}(x+y)=1\\\\ 9-1=9\,\mathrm{sen^2}(x+y)\\\\ 9\,\mathrm{sen^2}(x+y)=8\\\\ \mathrm{sen^2}(x+y)=\dfrac{8}{9}$

$\mathrm{sen}(x+y)=\pm\,\sqrt{\dfrac{8}{9}}\\\\\\ \mathrm{sen}(x+y)=\pm\,\dfrac{\sqrt{8}}{3}\\\\\\ \mathrm{sen}(x+y)=\pm\,\dfrac{2\sqrt{2}}{3}$


Como $(x+y)$ é um arco do 2º quadrante, o seno é positivo. Portanto,

$\boxed{\begin{array}{c}\mathrm{sen}(x+y)=\dfrac{2\sqrt{2}}{3} \end{array}}$

_________

• Calculando a tangente:

$\mathrm{tg}(x+y)=\dfrac{\mathrm{sen}(x+y)}{\cos(x+y)}\\\\\\ \mathrm{tg}(x+y)=\dfrac{\left(\frac{2\sqrt{2}}{3} \right )}{\left(-\,\frac{1}{3} \right )}\\\\\\ \mathrm{tg}(x+y)=\dfrac{2\sqrt{2}}{3}\cdot (-3)\\\\\\ \boxed{\begin{array}{c}\mathrm{tg}(x+y)=-2\sqrt{2} \end{array}}$

_________

• Calculando a secante:

$\sec(x+y)=\dfrac{1}{\cos(x+y)}\\\\\\ \sec(x+y)=\dfrac{1}{-\,\frac{1}{3}}\\\\\\ \boxed{\begin{array}{c}\sec(x+y)=-3 \end{array}}$

_________

Portanto, o valor da expressão pedida é

$3\,\mathrm{sen}(x+y)+\mathrm{tg}(x+y)-\sec(x+y)\\\\ =\diagup\!\!\!\! 3\cdot \dfrac{2\sqrt{2}}{\diagup\!\!\!\! 3}+(-2\sqrt{2})-(-3)\\\\\\ =2\sqrt{2}-2\sqrt{2}+3\\\\ =3$


Resposta: alternativa d) 3.


Bons estudos! :-)