Pré-Vestibular ⇒ (UFPB - 1972) Logaritmos e Trigonometria Tópico resolvido

[ALDRINMOD]

Sabendo que [tex3]\log\text{sen}\frac{a}{2}=-1[/tex3] e [tex3]\log \cos\frac{a}{2}=-6.[/tex3] O valor de [tex3]\log\frac{1-\cos a}{1+\cos a}[/tex3] é igual a:

a) [tex3]8.[/tex3]
b) [tex3]10.[/tex3]
c) [tex3]9.[/tex3]
d) [tex3]7.[/tex3]
e) nenhuma das anteriores.

[jneto]

Boa noite,

• [tex3]\log\frac{1 - \cos a }{1 + \cos a } = \log \Large\frac{\cos^{2}\left( \frac{a}{2}\right) + \text{sen}^{2}\left(\frac{a}{2}\right) - \left[\cos^{2}\left(\frac{a}{2}\right) - \text{sen}^{2}\left(\frac{a}{2}\right)\right]}{{\cos^{2}\left(\frac{a}{2}\right) + \text{sen}^{2}\left(\frac{a}{2}\right) + \cos^{2}\left(\frac{a}{2}\right) - \text{sen}^2\left(\frac{a}{2}\right)}}\large \\
  \text{ }= \log\Large \frac{2\text{sen}^2\left(\frac{a}{2}\right)}{2\cos^{2}\left(\frac{a}{2}\right)}\large \\
  \text{ }= 2\left[\log \text{sen}\left(\frac{a}{2}\right) - \log \cos\left(\frac{a}{2}\right)\right] = 2[-1 -(-6)] = 10[/tex3]

Resposta : alternativa (b).