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Re: [obm-l] O VALOR DE LOG
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- Subject: Re: [obm-l] O VALOR DE LOG
- From: "Nicolau C. Saldanha" <nicolau@xxxxxxxxxxxxxx>
- Date: Tue, 23 Oct 2007 18:04:50 -0200
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On 10/23/07, arkon <arkon@xxxxxxxxxx> wrote:
> (UFPB-72) Sabendo que log sen (a\2) = - 1 e log cos (a\2) = - 6 .
> O valor de log (1 – cos a)\(1 + cos a) é igual a:
Temos que spor que os logs são na base b (a ser determinado) senão a resposta
é "anulem a questão".
sen(a/2) = b^(-1)
cos(a/2) = b^(-6)
Assim sen^2(a/2) + cos^2(a/2) = b^(-2) + b^(-12) = 1
donde b ~= 1.133666191.
Suponde generosamente que seja isso o que a banca tem em mente,
1 - cos(a) = 2 sen^2(a/2) = 2 b^(-2)
1 + cos(a) = 2 cos^2(a/2) = 2 b^(-12)
(1 – cos a)\(1 + cos a) = b^10
log (1 – cos a)\(1 + cos a) = 10
> a) 8. b) 10. c) 9. d) 7. e) Nenhuma das anteriores.
Opção (b) (mas note a observação acima).
N.
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