Question

USANDO A FÓRMULA DA ADIÇÃO OU

SUBTRAÇÃO DE ARCOS, CALCULE:

a) sen 165°

b) sen 195°

c) cos 255°

d) cos 570°​

Answer 1

Fórmulas da adição

$\huge\boxed{\boxed{\mathsf{\underline{Seno~da~soma}}}}$

$\mathsf{sen(a+b) } \\ \mathsf{=sen(a)\cdot cos(b)+sen(b)\cdot cos(a)}$

Seno da diferença

$\mathsf{sen(a-b) } \\ \mathsf{=sen(a)\cdot cos(b)-sen(b)\cdot cos(a)}$

$\huge\boxed{\boxed{\mathsf{\underline{Cosseno~da~soma}}}}$

$\mathsf{cos(a+b)}\\\mathsf{=cos(a)\cdot cos(b)-sen(a)\cdot sen(b)}$

Cosseno da diferença

$\mathsf{cos(a-b)}\\\mathsf{=cos(a)\cdot cos(b)+sen(a)\cdot sen(b)}$

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a)

$\mathsf{sen(165^{\circ})=sen(120^{\circ}+45^{\circ})}\\\mathsf{sen(120^{\circ}\cdot cos(45^{\circ})+sen(45^{\circ})\cdot cos(120^{\circ})}\\\mathsf{sen(165^{\circ})=\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{2}}{2}\cdot(-\dfrac{1}{2})}\\\mathsf{sen(165^{\circ})=\dfrac{\sqrt{6}-\sqrt{2}}{4}}$

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b)

$\mathsf{sen(195^{\circ})=sen(150^{\circ}+45^{\circ})}\\\mathsf{sen(150^{\circ}\cdot cos(45^{\circ})+sen(45^{\circ})\cdot cos(150^{\circ})}\\\mathsf{sen(195^{\circ})=\dfrac{1}{2}\cdot\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{2}}{2}\cdot(-\dfrac{\sqrt{3}}{2})}\\\mathsf{sen(165^{\circ})=\dfrac{\sqrt{2}-\sqrt{6}}{4}}$

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c)

$\mathsf{cos(255^{\circ})=cos(210^{\circ}+45^{\circ})}\\\mathsf{cos(210^{\circ}\cdot cos(45^{\circ})-sen(210^{\circ})\cdot sen(45^{\circ})}\\\mathsf{cos(255^{\circ})=\dfrac{-\sqrt{3}}{2}\cdot\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}\cdot(\dfrac{\sqrt{2}}{2})}\\\mathsf{cos(255^{\circ})=-\dfrac{\sqrt{6}+\sqrt{2}}{4}}$

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d)

$\mathsf{cos(570)=cos(210-0)}\\\mathsf{cos(210^{\circ})\cdot cos(0^{\circ}) +sen(210^{\circ}).sen(0)} \\\mathsf{cos(210-0)=-\dfrac{\sqrt{3}}{2}. 1}\\\mathsf{cos(570^{\circ}) =-\dfrac{\sqrt{3}}{2}}$