Question

Dado Sen X = -3 raiz de 10 / 10 ( 3° quadrante) e cos y = 1/2 ( 4° quadrante )
Calcule ;

A) Sen ( x+y )
Cos ( x+y)


B ) Sen ( x-y )
Cos (x-y)

Answer 1

Primeiro, vamos encontrar $\cos x$ e $\mathrm{sen\,}y:$

$\bullet\;\;\mathrm{sen\,}x=-\,\dfrac{3\sqrt{10}}{10}\\\\\\ \mathrm{sen^2\,}x=\bigg(\!\!-\dfrac{3\sqrt{10}}{10}\bigg)^{\!\!2}\\\\\\ \mathrm{sen^2\,}x=\dfrac{9\cdot 10}{100}\\\\\\ 100\,\mathrm{sen^2\,}x=90~~~~~~(\text{mas }\mathrm{sen^2\,}x=1-\cos^2 x)\\\\ 100\cdot (1-\cos^2 x)=90\\\\ 100-100\cos^2 x=90\\\\ 100-90=100\cos^2 x\\\\ 100\cos^2 x=10\\\\ \cos^2 x=\dfrac{10}{100}$

$\cos x=\pm\, \sqrt{\dfrac{10}{100}}\\\\\\ \cos x=\pm\, \dfrac{\sqrt{10}}{10}$


Como $x$ é um arco do 3º quadrante, $\cos x$ é negativo. Logo,

$\boxed{\begin{array}{c}\cos x=-\, \dfrac{\sqrt{10}}{10} \end{array}}$

________

$\bullet\;\;\cos y=\dfrac{1}{2}\\\\\\ \cos^2 y=\left(\dfrac{1}{2}\right)^{\!\!2}\\\\\\ \cos^2 y=\dfrac{1}{4}\\\\\\ 4\cos^2 y=1~~~~~~(\text{mas }\cos^2 y=1-\mathrm{sen^2\,}y)\\\\ 4\cdot (1-\mathrm{sen^2\,}y)=1\\\\ 4-4\,\mathrm{sen^2\,}y=1\\\\ 4-1=4\,\mathrm{sen^2\,}y\\\\ 4\,\mathrm{sen^2\,}y=3$

$\mathrm{sen^2\,}y=\dfrac{3}{4}\\\\\\ \mathrm{sen\,}y=\pm\, \sqrt{\dfrac{3}{4}}\\\\\\ \mathrm{sen\,}y=\pm\, \dfrac{\sqrt{3}}{2}$


Como $y$ é do 4º quadrante, $\mathrm{sen\,}y$ é negativo. Logo,

$\boxed{\begin{array}{c}\mathrm{sen\,}y=-\,\dfrac{\sqrt{3}}{2}\end{array}}$

____________

A) $\mathrm{sen}(x+y)=\mathrm{sen\,}x\cos y+\mathrm{sen\,}y\cos x$

$\mathrm{sen}(x+y)=-\,\dfrac{3\sqrt{10}}{10}\cdot \dfrac{1}{2}+\bigg(\!\!-\,\dfrac{\sqrt{3}}{2}\bigg)\cdot \bigg(\!\!-\, \dfrac{\sqrt{10}}{10}\bigg)\\\\\\ \mathrm{sen}(x+y)=-\,\dfrac{3\sqrt{10}}{20}+\dfrac{\sqrt{3}\cdot \sqrt{10}}{20}\\\\\\ \mathrm{sen}(x+y)=\dfrac{-3\sqrt{10}+\sqrt{3}\cdot \sqrt{10}}{20}\\\\\\ \boxed{\begin{array}{c}\mathrm{sen}(x+y)=\dfrac{\sqrt{10}\cdot (-3+\sqrt{3})}{20} \end{array}}$

_____

$\cos(x+y)=\cos x\cos y-\mathrm{sen\,}x\,\mathrm{sen\,}y\\\\ \cos(x+y)=\bigg(\!\!-\dfrac{\sqrt{10}}{10}\bigg)\cdot \dfrac{1}{2}-\bigg(\!\!-\dfrac{3\sqrt{10}}{10}\bigg)\cdot \bigg(\!\!-\dfrac{\sqrt{3}}{2}\bigg)\\\\\\ \cos(x+y)=-\,\dfrac{\sqrt{10}}{20}-\dfrac{3\sqrt{10}\cdot \sqrt{3}}{20}\\\\\\ \cos(x+y)=\dfrac{-\sqrt{10}-3\sqrt{10}\cdot \sqrt{3}}{20}\\\\\\ \boxed{\begin{array}{c}\cos(x+y)=\dfrac{\sqrt{10}\cdot \big(\!-1-3\sqrt{3}\big)}{20} \end{array}}$

__________

B) $\mathrm{sen}(x-y)=\mathrm{sen\,}x\cos y-\mathrm{sen\,}y\cos x$

$\mathrm{sen}(x-y)=-\,\dfrac{3\sqrt{10}}{10}\cdot \dfrac{1}{2}-\bigg(\!\!-\,\dfrac{\sqrt{3}}{2}\bigg)\cdot \bigg(\!\!-\, \dfrac{\sqrt{10}}{10}\bigg)\\\\\\ \mathrm{sen}(x-y)=-\,\dfrac{3\sqrt{10}}{20}-\dfrac{\sqrt{3}\cdot \sqrt{10}}{20}\\\\\\ \mathrm{sen}(x-y)=\dfrac{-3\sqrt{10}-\sqrt{3}\cdot \sqrt{10}}{20}\\\\\\ \boxed{\begin{array}{c}\mathrm{sen}(x-y)=\dfrac{\sqrt{10}\cdot (-3-\sqrt{3})}{20} \end{array}}$

_____

$\cos(x-y)=\cos x\cos y+\mathrm{sen\,}x\,\mathrm{sen\,}y\\\\ \cos(x-y)=\bigg(\!\!-\dfrac{\sqrt{10}}{10}\bigg)\cdot \dfrac{1}{2}+\bigg(\!\!-\dfrac{3\sqrt{10}}{10}\bigg)\cdot \bigg(\!\!-\dfrac{\sqrt{3}}{2}\bigg)\\\\\\ \cos(x-y)=-\,\dfrac{\sqrt{10}}{20}+\dfrac{3\sqrt{10}\cdot \sqrt{3}}{20}\\\\\\ \cos(x-y)=\dfrac{-\sqrt{10}+3\sqrt{10}\cdot \sqrt{3}}{20}\\\\\\ \boxed{\begin{array}{c}\cos(x-y)=\dfrac{\sqrt{10}\cdot \big(\!-1+3\sqrt{3}\big)}{20} \end{array}}$


Bons estudos! :-)