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Given that sin 0 0 a = 5/12, find (a) cos 0 0 a (b) tan 0 0 a

Mathematics

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2 parts
a

cos⁡α=11912\cos \alpha = \frac{\sqrt{119}}{12}

b

tan⁡α=5119119\tan \alpha = \frac{5\sqrt{119}}{119}

Step-by-step solution

6 steps

Part a· 3 steps

Pour trouver cos⁡α\cos \alpha, nous utilisons l'identité trigonométrique fondamentale sin⁡2α+cos⁡2α=1\sin^2 \alpha + \cos^2 \alpha = 1.

  1. Substituer la valeur de sin⁡α\sin \alpha dans l'identité.

    (512)2+cos⁡2α=1\left(\frac{5}{12}\right)^2 + \cos^2 \alpha = 1
  2. Calculer le carré de sin⁡α\sin \alpha et isoler cos⁡2α\cos^2 \alpha.

    25144+cos⁡2α=1\frac{25}{144} + \cos^2 \alpha = 1 cos⁡2α=1−25144\cos^2 \alpha = 1 - \frac{25}{144} cos⁡2α=144144−25144\cos^2 \alpha = \frac{144}{144} - \frac{25}{144} cos⁡2α=119144\cos^2 \alpha = \frac{119}{144}
  3. Prendre la racine carrée des deux côtés.

    cos⁡α=±119144\cos \alpha = \pm \sqrt{\frac{119}{144}} cos⁡α=±11912\cos \alpha = \pm \frac{\sqrt{119}}{12}

    Sans information supplémentaire sur l'angle α\alpha, nous considérons la valeur positive.

    cos⁡α=11912\boxed{\cos \alpha = \frac{\sqrt{119}}{12}}

Part b

tan⁡α=5119119\tan \alpha = \frac{5\sqrt{119}}{119}

Pour trouver tan⁡α\tan \alpha, nous utilisons la relation tan⁡α=sin⁡αcos⁡α\tan \alpha = \frac{\sin \alpha}{\cos \alpha}.

  1. Substituer les valeurs de sin⁡α\sin \alpha et cos⁡α\cos \alpha.

    tan⁡α=51211912\tan \alpha = \frac{\frac{5}{12}}{\frac{\sqrt{119}}{12}}
  2. Simplifier l'expression.

    tan⁡α=5119\tan \alpha = \frac{5}{\sqrt{119}}
  3. Rationaliser le dénominateur en multipliant le numérateur et le dénominateur par 119\sqrt{119}.

    tan⁡α=5119×119119\tan \alpha = \frac{5}{\sqrt{119}} \times \frac{\sqrt{119}}{\sqrt{119}} tan⁡α=5119119\boxed{\tan \alpha = \frac{5\sqrt{119}}{119}}
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