Question

sendo= sen(x) = raiz 3/2 e cos (y) = 1/2, determine o valor de: a) sen( x - y) = b) cos( x + y) =

Answer 1

Explicação passo-a-passo:

a) Pela relação fundamental da trigonometria:

• $\sf sen^2~x+cos^2~x=1$

$\sf \left(\dfrac{\sqrt{3}}{2}\right)^2+cos^2~x=1$

$\sf \dfrac{3}{4}+cos^2~x=1$

$\sf cos^2~x=1-\dfrac{3}{4}$

$\sf cos^2~x=\dfrac{4-3}{4}$

$\sf cos^2~x=\dfrac{1}{4}$

$\sf cos~x=\sqrt{\dfrac{1}{4}}$

$\sf cos~x=\dfrac{1}{2}$

• $\sf sen^2~y+cos^2~y=1$

$\sf sen^2~y+\left(\dfrac{1}{2}\right)^2=1$

$\sf sen^2~y+\dfrac{1}{4}=1$

$\sf sen^2~y=1-\dfrac{1}{4}$

$\sf sen^2~y=\dfrac{4-1}{4}$

$\sf sen^2~y=\dfrac{3}{4}$

$\sf sen~y=\sqrt{\dfrac{3}{4}}$

$\sf sen~y=\dfrac{\sqrt{3}}{2}$

Temos que:

$\sf sen~(x-y)=sen~x\cdot cos~y-sen~y\cdot cos~x$

$\sf sen~(x-y)=\dfrac{\sqrt{3}}{2}\cdot\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}\cdot\dfrac{1}{2}$

$\sf sen~(x-y)=\dfrac{\sqrt{3}}{4}-\dfrac{\sqrt{3}}{4}$

$\sf sen~(x-y)=0$

b) $\sf cos~(x+y)=cos~a\cdot cos~b-sen~a\cdot sen~b$

$\sf cos~(x+y)=\dfrac{1}{2}\cdot\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{3}}{2}$

$\sf cos~(x+y)=\dfrac{1}{4}-\dfrac{3}{4}$

$\sf cos~(x+y)=\dfrac{1-3}{4}$

$\sf cos~(x+y)=\dfrac{-2}{4}$

$\sf cos~(x+y)=\dfrac{-1}{2}$