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18 December, 06:01

Sabendo que senx=3/5 e que X € 1°quadrante, calcule tgx

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  1. 18 December, 08:50
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    sen x = 3/5 eleva ao quadrado ambos os lados

    sen² x = (3/5) ²

    Tem que partir da relação fundamental

    sen²x + cos²x = 1

    (3/5) ² + cos² x = 1

    cos² x = 1 - (3/5) ² = 16/25

    cos x = V (16/25 = 4/5

    b) tg x = sen x / cos x = 3/5 / 4/5 = 3/5 * 5/4 = 3/4

    c) cotg x = cosx / sen x = 1 / tgx = 1 / (3/4 = 4/3

    d) sec x = 1 / cos x = 1 / 4/5 = 5/4

    e) cossec x = 1 / sen x = 1 / 3/5 = 5/3

    se gostou, avalie

    edson edson · 9 anos atrás
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