4. Sendo sec x = 2 e x um arco do 1º quadrante, então qual o valor do sen x ?
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A secante é o inverso do cosseno

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Relação fundamental da trigonometria:

Como cos x = 1/2:

O seno é positivo no primeiro quadrante, logo:

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Relação fundamental da trigonometria:
Como cos x = 1/2:
O seno é positivo no primeiro quadrante, logo:
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