.Problemão: Tangente do arco metade

Problema
(Indicado a partir do 2º ano do E. M.)


Sabendo que [tex]tg \left(\frac{\theta}{2}\right) = t[/tex], determine os valores de [tex]sen\, \theta[/tex], [tex]cos\, \theta[/tex] e [tex]tg\,\theta[/tex] em função de [tex]t[/tex].

explicador_p

Lembretes

Identidades trigonométricas importantes:
[tex] \textcolor{#800000}{(*)}[/tex] [tex]sen^2(a) +cos^2(a) =1[/tex].
[tex] \textcolor{#800000}{(**)} \, sen(2a)=2sen(a)cos(a)[/tex] (seno do arco duplo).
[tex] \textcolor{#800000}{(***)} \, cos(2a)=cos^2(a) -sen^2(a)[/tex] (cosseno do arco duplo).
[tex] \textcolor{#800000}{(****)} \, sec^2(a)=1+tg^2(a)[/tex].

Solução 1


Perceba inicialmente que se [tex]tg \left(\frac{\theta}{2}\right) = t[/tex], então [tex]cos \left(\frac{\theta}{2}\right) \neq 0[/tex] e, portanto, [tex]sec\,\left(\frac{\theta}{2}\right) \in \mathbb{R} – \{0\}[/tex].

  • Da fórmula do seno do arco duplo, segue que:
    [tex]\qquad sen\,\theta = 2 \cdot sen\,\left(\frac{\theta}{2}\right)\,\cdot\, cos\left(\frac{\theta}{2}\right)[/tex]
    [tex]\qquad sen\,\theta = 2 \cdot sen\,\left(\frac{\theta}{2}\right)\,\cdot\, cos\left(\frac{\theta}{2}\right)\,\cdot \,\dfrac{sec^2\left(\frac{\theta}{2}\right)}{sec^2\left(\frac{\theta}{2}\right)}[/tex]

    [tex]\qquad sen\,\theta= \dfrac{2 \cdot sen\,\left(\frac{\theta}{2}\right)\,\cdot\, cos\left(\frac{\theta}{2}\right)\,\cdot\, sec^2\left(\frac{\theta}{2}\right)}{sec^2\left(\frac{\theta}{2}\right)}[/tex]

    [tex]\qquad sen\,\theta = \dfrac{2 \cdot sen\,\left(\frac{\theta}{2}\right)\,\cdot\, cos\left(\frac{\theta}{2}\right)\,\cdot\, \dfrac{1}{cos^2\left(\frac{\theta}{2}\right)}}{1 + tg^2\left(\frac{\theta}{2}\right)}[/tex]

    [tex]\qquad sen\,\theta = \dfrac{2 \cdot tg\,\left(\frac{\theta}{2}\right)}{1 + tg^2\left(\frac{\theta}{2}\right)}[/tex]

    [tex]\qquad \boxed{sen\,\theta = \dfrac{2t}{1 + t^2}}[/tex].

  • Da fórmula do cosseno do arco duplo, temos

    [tex]\qquad cos\,\theta = cos^2\left(\frac{\theta}{2}\right)\,-\, sen^2\left(\frac{\theta}{2}\right)[/tex]
    [tex]\qquad cos\,\theta = cos^2\left(\frac{\theta}{2}\right)\,-\, sen^2\left(\frac{\theta}{2}\right)\,\cdot \,\dfrac{sec^2\left(\frac{\theta}{2}\right)}{sec^2\left(\frac{\theta}{2}\right)}[/tex]

    [tex]\qquad cos\,\theta = \dfrac{\left(cos^2\left(\frac{\theta}{2}\right)\,-\, sen^2\left(\frac{\theta}{2}\right)\right)\,\cdot\, sec^2\left(\frac{\theta}{2}\right)}{sec^2\left(\frac{\theta}{2}\right)}[/tex]
    [tex]\qquad cos\,\theta = \dfrac{\left(cos^2\left(\frac{\theta}{2}\right)\,-\, sen^2\left(\frac{\theta}{2}\right)\right)\,\cdot\, \dfrac{1}{cos^2\left(\frac{\theta}{2}\right)}}{sec^2\left(\frac{\theta}{2}\right)}[/tex]

    [tex]\qquad cos\,\theta = \dfrac{1 – tg^2\left(\frac{\theta}{2}\right)}{1 + tg^2\left(\frac{\theta}{2}\right)}[/tex]

    [tex]\qquad \boxed{cos\,\theta = \dfrac{1 – t^2}{1 + t^2}}[/tex].

  • Finalmente,
    [tex]\qquad tg\, \theta = \dfrac{sen \theta}{cos \theta} [/tex]

    [tex]\qquad tg\, \theta = \dfrac{\dfrac{2t}{1 + t^2}}{\dfrac{1 – t^2}{1 + t^2}}[/tex]

    [tex]\qquad \boxed{ tg\, \theta = \dfrac{2t}{1 – t^2}}[/tex].


Solução elaborada pelos Moderadores do Blog.

Solução 2


Do enunciado, temos
[tex]\qquad tg\left(\frac{\theta}{2}\right)=\dfrac{sen \left(\frac{\theta}{2}\right)}{cos \left(\frac{\theta}{2}\right)}=t[/tex],
daí, segue que [tex] sen \left(\frac{\theta}{2}\right)=cos \left(\frac{\theta}{2}\right)t. \qquad \textcolor{#800000}{(i)}[/tex]
Mas [tex]sen^2\left(\frac{\theta}{2}\right)+cos^2\left(\frac{\theta}{2}\right)=1[/tex] (por [tex] \textcolor{#800000}{(*)}[/tex]), donde
[tex]\qquad (cos \left(\frac{\theta}{2}\right)t)^2+cos^2\left(\frac{\theta}{2}\right)=1[/tex]
[tex]\qquad cos^2\left(\frac{\theta}{2}\right)(1+t^2)=1[/tex]
[tex]\qquad cos^2\left(\frac{\theta}{2}\right)=\dfrac{1}{1+t^2}. \qquad \textcolor{#800000}{(ii)}[/tex]
Por [tex] \textcolor{#800000}{(i)}[/tex] e [tex] \textcolor{#800000}{(ii)}[/tex],
[tex]\qquad sen^2\left(\frac{\theta}{2}\right)=(cos \left(\frac{\theta}{2}\right)t)^2=cos^2\left(\frac{\theta}{2}\right)t^2=\dfrac{t^2}{1+t^2}. \qquad \textcolor{#800000}{(iii)}[/tex]
Utilizando [tex] \textcolor{#800000}{(***)}[/tex],
[tex]\qquad cos\, \theta =cos^2\left(\frac{\theta}{2}\right)-sen^2\left(\frac{\theta}{2}\right)[/tex],
logo, por [tex] \textcolor{#800000}{(ii)}[/tex] e [tex] \textcolor{#800000}{(iii)}[/tex], vem que:
[tex]\qquad cos\, \theta =\dfrac{1}{1+t^2}-\dfrac{t^2}{1+t^2}=\fbox{$\dfrac{1-t^2}{1+t^2}$}[/tex].
Por outro lado, usando [tex] \textcolor{#800000}{(**)}[/tex], temos
[tex]\qquad sen\,\theta=2sen \left(\frac{\theta}{2}\right)cos \left(\frac{\theta}{2}\right)[/tex],
logo, por [tex] \textcolor{#800000}{(i)}[/tex]:
[tex]\qquad sen\,\theta=2[cos \left(\frac{\theta}{2}\right)t]cos \left(\frac{\theta}{2}\right)=2tcos^2\left(\frac{\theta}{2}\right)[/tex].
Assim, por [tex] \textcolor{#800000}{(ii)}[/tex],
[tex]\qquad sen\,\theta=2t\left(\dfrac{1}{1+t^2}\right)=\fbox{$\dfrac{2t}{1+t^2}$}[/tex].
Finalmente,
[tex]\qquad tg\, \theta = \dfrac{sen \theta}{cos \theta} = \dfrac{\dfrac{2t}{1 + t^2}}{\dfrac{1 – t^2}{1 + t^2}} = \fbox{$ \dfrac{2t}{1 – t^2}$}[/tex].


Solução elaborada pelos Moderadores do Blog.

Link permanente para este artigo: http://clubes.obmep.org.br/blog/problemao-tangente-do-arco-metade/