Question

Se senα = 2.senβ e cosβ = 3.cosα. Ache o valor de: cos(α - β)

Answer 1

Lembrando da fórmula do cosseno da diferença:

cos(α – β) = cos α · cos β + sen α · sen β

cos(α – β) = cos α · (3 cos α) + (2 sen β) · sen β

cos(α – β) = 3 cos² α + 2 sen² β        (i)

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$\left\{ \!\begin{array}{lc} \mathrm{sen\,}\alpha=2\,\mathrm{sen\,}\beta&~~~~\mathbf{(ii)}\\\\ \cos \beta=3\cos \alpha&~~~~\mathbf{(iii)} \end{array} \right.$


Elevando as equações ao quadrado, temos

$\left\{ \!\begin{array}{lc} \mathrm{sen^2\,}\alpha=4\,\mathrm{sen^2\,}\beta&~~~~\mathbf{(iv)}\\\\ \cos^2 \beta=9\cos^2 \alpha&~~~~\mathbf{(v)} \end{array} \right.$


Aplicando convenientemente a Relação Trigonométrica Fundamental às equações (iv) e (v):

$\left\{ \!\begin{array}{lc} 1-\cos^2\alpha=4\,\mathrm{sen^2\,}\beta&~~~~\mathbf{(vi)}\\\\ 1-\mathrm{sen^2\,}\beta=9\cos^2 \alpha&~~~~\mathbf{(vii)} \end{array} \right.$


Isolando cos² α na equação (vi) e substituindo na equação (vii), temos

$\cos^2\alpha = 1-4\,\mathrm{sen^2}\beta\\\\\\ 1-\mathrm{sen^2\,}\beta=9\cdot (1-4\,\mathrm{sen^2}\beta)\\\\ 1-\mathrm{sen^2\,}\beta=9-36\,\mathrm{sen^2}\beta\\\\ -\mathrm{sen^2\,}\beta+36\,\mathrm{sen^2}\beta=9-1\\\\ 35\,\mathrm{sen^2}\beta=8\\\\ \mathrm{sen^2}\beta=\dfrac{8}{35}~~~~~~\mathbf{(viii)}$


Portanto,

$\cos^2\alpha = 1-4\cdot \dfrac{8}{35}\\\\\\ \cos^2\alpha = \dfrac{35}{35}-\dfrac{32}{35}\\\\\\ \cos^2\alpha = \dfrac{3}{35}~~~~~~\mathbf{(ix)}$

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Voltando a (i), ficamos com

cos(α – β) = 3 · (3/35) + 2 · (8/35)

cos(α – β) = 9/35 + 16/35

cos(α – β) = (9 + 16)/35

cos(α – β) = 25/35

cos(α – β) = 5/7   <——    esta é a resposta


Bons estudos! :-)