IME / ITA ⇒ (ITA - 1975) Trigonometria: Arco Metade Tópico resolvido

[ALDRINMOD]

Sabendo-se que [tex3]\text{sen}x=\frac{m-n}{m+n},[/tex3] [tex3]n>0[/tex3] e [tex3]m>0,[/tex3] podemos afirmar que [tex3]\text{tg}(\frac{\pi}{4}-\frac{x}{2})[/tex3] é igual a:

a) [tex3]\frac{n}{m}.[/tex3]
b) [tex3]\frac{\sqrt{m}}{n}.[/tex3]
c) [tex3]1-\frac{n}{m}.[/tex3]
d) [tex3]\sqrt{\frac{n}{m}}.[/tex3]
e) nenhuma das anteriores.

[Thadeu]

Determinar o valor de [tex3]cos x :[/tex3]

• [tex3]\text{sen}^2x+ cos^2 x = 1\,\Rightarrow\, cos x = \sqrt{1-\text{sen}^2x}\,\Rightarrow\, cos x=\sqrt{1-\(\frac{m-n}{m+n}\)^2}\\
  cos x=\sqrt{(1+\frac{m-n}{m+n})(1-\frac{m-n}{m+n})}\,\Rightarrow\, cos x=\sqrt{\(\frac{2m}{m+n}\)\(\frac{2n}{m+n}\)}\,\Rightarrow\, cos x=\frac{2\sqrt{mn}}{m+n}[/tex3]

Resolvendo [tex3]\text{tg}\(\frac{\pi}{4}-\frac{x}{2}\):[/tex3]

• [tex3]\text{tg}(\frac{\pi}{4}) = 1\\
  \text{tg}\({\frac{x}{2}}\)=\pm\,\sqrt{\frac{1- cos x}{1 + cos x}} = \pm\,\sqrt{\Large \frac{1 + \frac{2 \sqrt{mn}}{m+n}}{1-\frac{2\sqrt{mn}}{m+n}}\large}=\pm\,\sqrt{\frac{m+n+2\sqrt{mn}}{m+n-2\sqrt{mn}}}=\pm\,\sqrt{\frac{(\sqrt{m}+\sqrt{n})^2}{(\sqrt{m}-\sqrt{n})^2}}=\pm\,\frac{\sqrt{m}+\sqrt{n}}{\sqrt{m}-\sqrt{n}}\\
  \text{tg}(a-b)=\frac{\text{tg}a-\text{tg}b}{1+\text{tg}a.\text{tg}b}\,\Rightarrow\,\text{tg}\(\frac{\pi}{4}-\frac{x}{2}\)=\frac{\text{tg}\(\frac{\pi}{4}\) - \text{tg}\(\frac{x}{2}\)}{1+\text{tg}\(\frac{\pi}{4}\).\text{tg}\(\frac{x}{2}\)} = \frac{1 - \text{tg}\(\frac{x}{2}\)}{1+\text{tg}\(\frac{x}{2}\)} = \frac{1\,-\(\pm\,\frac{\sqrt{m} + \sqrt{n}}{\sqrt{m} - \sqrt{n}}\)}{1\,+\(\pm\,\frac{\sqrt{m}+\sqrt{n}}{\sqrt{m}-\sqrt{n}}\)}\\
  \Rightarrow\,\frac{\sqrt{m}-\sqrt{n}-\sqrt{m}-\sqrt{n}}{\sqrt{m}-\sqrt{n}+\sqrt{m}+\sqrt{n}}=\frac{-2\sqrt{n}}{2\sqrt{m}}=-\sqrt{\frac{n}{m}}[/tex3]

Se eu não me enganei em alguma parte, a resposta é a letra (e).

[Karl Weierstrass]

Sabendo que [tex3]\text{tg} (y)=\sqrt{\frac{1-\cos(2y)}{1+\cos(2y)}}[/tex3], temos:

• [tex3]\text{tg} \left(\frac{\pi}{4} -\frac{x}{2}\right) =\sqrt{\frac{1-\cos \left(\frac{\pi}{2} -x\right)}{1+\cos \left(\frac{\pi}{2} +x\right)}}=\sqrt{\frac{1-\text{sen}x }{1-\text{sen}x}}=1[/tex3]

[tex3]\boxed{\text{E}}[/tex3]

[Karl Weierstrass]

A resposta correta é a alternativa D.

Onde se lê [tex3]1+\cos \left(\frac{\pi}{2} +x\right)[/tex3], leia-se [tex3]1+\cos \left(\frac{\pi}{2} -x\right)[/tex3].

Daí,

[tex3]\cos \left(\frac{\pi}{2} -x\right) =\text{sen}\,x=\frac{m-n}{m+n}.[/tex3]

E o resultado segue.