Question

usando a fórmula de soma e subtração de arcos, calcule:
a) sen 195°
b) sen 135°
c) cos 15°
d) cos 225°
alguém pode me ajudar por favor?? estou com muita dificuldade, obrigado.​​

Answer 1

Explicação passo-a-passo:

$\bullet~\text{sen}(\alpha+\beta)=\text{sen}~\alpha\cdot\text{cos}~\beta+\text{sen}~\beta\cdot\text{cos}~\alpha$

a) $\text{sen}~195^{\circ}=\text{sen}~(150^{\circ}+45^{\circ})$

$\text{sen}~195^{\circ}=\text{sen}~150^{\circ}\cdot\text{cos}~45^{\circ}+\text{sen}~45^{\circ}\cdot\text{cos}~150^{\circ}$

$\text{sen}~195^{\circ}=\dfrac{1}{2}\cdot\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{2}}{2}\cdot\left(-\dfrac{\sqrt{3}}{2}\right)$

$\boxed{\text{sen}~195^{\circ}=\dfrac{\sqrt{2}-\sqrt{6}}{4}}$

b) $\text{sen}~135^{\circ}=\text{sen}~(90^{\circ}+45^{\circ})$

$\text{sen}~135^{\circ}=\text{sen}~90^{\circ}\cdot\text{cos}~45^{\circ}+\text{sen}~45^{\circ}\cdot\text{cos}~90^{\circ}$

$\text{sen}~135^{\circ}=1\cdot\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{2}}{2}\cdot0$

$\boxed{\text{sen}~135^{\circ}=\dfrac{\sqrt{2}}{2}}$

$\bullet~\text{cos}(\alpha-\beta)=\text{cos}~\alpha\cdot\text{cos}~\beta+\text{sen}~\beta\cdot\text{sen}~\alpha$

c) $\text{cos}~15^{\circ}=\text{cos}~(60^{\circ}-45^{\circ})$

$\text{cos}~15^{\circ}=\text{cos}~60^{\circ}\cdot\text{cos}~45^{\circ}+\text{sen}~60^{\circ}\cdot\text{sen}~45^{\circ}$

$\text{cos}~15^{\circ}=\dfrac{1}{2}\cdot\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{2}}{2}$

$\boxed{\text{cos}~15^{\circ}=\dfrac{\sqrt{2}+\sqrt{6}}{4}}$

d) $\text{cos}~225^{\circ}=\text{cos}~(270^{\circ}-45^{\circ})$

$\text{cos}~225^{\circ}=\text{cos}~270^{\circ}\cdot\text{cos}~45^{\circ}+\text{sen}~270^{\circ}\cdot\text{sen}~45^{\circ}$

$\text{cos}~225^{\circ}=0\cdot\dfrac{\sqrt{2}}{2}+(-1)\cdot\dfrac{\sqrt{2}}{2}$

$\boxed{\text{cos}~225^{\circ}=-\dfrac{\sqrt{2}}{2}}$