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Re: [obm-l] Identidade trigonométrica
- To: obm-l@xxxxxxxxxxxxxx
- Subject: Re: [obm-l] Identidade trigonométrica
- From: "Renato Lira" <natolira@xxxxxxxxx>
- Date: Sat, 21 Apr 2007 09:06:20 -0300
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basta multiplicar por sen(40) em cima e em baixo e usar que [sen(x).cos(x)]/2 = [sen(2x)]/2
[sen(40º).cos(40º).cos(80º).cos(160º)]/[sen(40º)] = [sen(80º).cos(80º).cos(160º)]/[2sen(40º)]
= [sen(160º).cos(160º)]/[4sen(40º)] = [sen(320º)]/[8sen(40º)]
como sen(320º)=sen( - 40º)= - sen(40º)
Tem-se
cos(40º).cos(80º).cos(160º) = -1/8
On 4/21/07, carry_bit <carry_bit@yahoo.com.br> wrote:
Olá colegas da OBM-L,
Me deparei com um exercício que está custando resolver:
Mostre que: cos(40º).cos(80º).cos(160º) = -1/8
Agradeço desde já!!
carry_bit