Problema
(Indicado a partir do 1º ano do E. M.)
Determine o menor inteiro positivo [tex]n[/tex] tal que
[tex]\dfrac 1{\text{sen}45^\circ\text{sen}46^\circ}+\dfrac 1{\text{sen} 47^\circ\text{sen} 48^\circ}+\cdots+\dfrac 1{\text{sen}133^\circ\text{sen}134^\circ}=\dfrac 1{\text{sen} (n^\circ)}[/tex].
Algumas propriedades trigonométricas importantes
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[tex]\textcolor{#800000}{(i)} \text{ sen}(a\pm b)=\text{sen}\,a \cdot \cos b\pm \text{sen} b \cdot \cos a[/tex].
[tex]\textcolor{#800000}{(ii)}\,\dfrac{\cos a}{\text{sen}\, a}=\cot a[/tex].
[tex]\textcolor{#800000}{(iii)}\, \cot (180^\circ – a) = – \cot a[/tex]. |
Solução
A unidade de medida angular utilizada nesta solução é o grau, mas, para não carregar a visualização, vamos omitir nas próximas passagens a bolinha que indica a unidade de medida “grau”.
Observe, inicialmente que:
[tex]\quad \begin{align*} \dfrac{1}{\text{sen} (n) \text{ sen} (n+1)}&=\dfrac{\text{sen}1}{\text{sen}1} \cdot \dfrac{1}{\text{sen} (n) \text{ sen} (n+1)} \\
&=\dfrac{1}{\text{sen}1} \cdot \dfrac{\text{sen}1}{\text{sen} (n) \text{ sen} (n+1)} \\
&= \frac{1}{\text{sen}1} \cdot \dfrac{\text{sen} ((n+1) – n)}{\text{sen} (n) \text{sen}(n+1)}\\
& \stackrel{\textcolor{#800000}{(i)}}{=} \dfrac{1}{\text{sen} 1} \cdot \dfrac{\text{sen} (n+1) \cdot \cos (n) – \text{sen} (n) \cos (n+1)}{\text{sen} (n) \cdot \text{sen}(n+1)} \\
& =\dfrac{1}{\text{sen} 1} \left( \dfrac{\cancel{\text{sen} (n+1)} \cdot \cos (n)}{\text{sen} (n) \cdot \cancel{\text{sen}(n+1)}} -\dfrac{ \bcancel{\text{sen} (n)} \cdot \cos (n+1)}{\bcancel{\text{sen} (n)} \cdot \text{sen}(n+1)} \right)\\
&= \dfrac{1}{\text{sen} 1} \cdot \left(\frac{\cos (n)}{\text{sen} (n)} – \dfrac{\cos (n+1)}{\text{sen} (n+1)}\right)\\
&\stackrel{\textcolor{#800000}{(ii)}}{=} \dfrac{1}{\text{sen} 1} \cdot \left(\cot (n) – \cot (n+1)\right); \end{align*}[/tex]
e, portanto, [tex]\boxed{ \dfrac{1}{\text{sen} (n) \text{ sen} (n+1)}= \dfrac{1}{\text{sen} 1} \cdot \left(\cot (n) – \cot (n+1)\right)}.[/tex]
Usando esta identidade em cada parcela da soma
[tex]\qquad \dfrac 1{\text{sen}45 \text{ sen}46 }+\dfrac 1{\text{sen} 47 \text{ sen} 48 }+\cdots+\dfrac 1{\text{sen}133 \text{ sen}134 }[/tex],
obtemos:
[tex]\quad \dfrac 1{\text{sen}45 \text{ sen}46 }+\dfrac 1{\text{sen} 47 \text{ sen} 48 }+\cdots+\dfrac 1{\text{sen}133 \text{ sen}134 }=[/tex]
[tex]\quad = \dfrac{1}{\text{sen} 1} \left(\cot 45 – \cot 46\right) + \dfrac{1}{\text{sen} 1} \left(\cot 47 – \cot 48\right) +\cdots+ \dfrac{1}{\text{sen} 1} \left(\cot 133- \cot 134\right)[/tex]
[tex]\quad = \dfrac{1}{\text{sen} 1} \left(\cot 45 – \cot 46 + \cot 47 – \cot 48 + \cdots + \cot 133 – \cot 134 \right). \qquad \textcolor{#800000}{(iv)}[/tex]
Vamos observar melhor a expressão que aparece do lado direito da igualdade [tex]\textcolor{#800000}{(iv)}[/tex]:
[tex]\quad \boxed{\cot 45 – \cot 46 + \cot 47 – \cot 48 + \cdots + \cot 133 – \cot 134} =\\
\quad =\cot 45 – \cot 46 + \cot 47 – \cot 48 + \cdots +\cot 89- \cot 90 +\\
\quad \quad + \cot 91 + \cdots +\cot 133 – \cot 134\\
\quad =\cot 45 + (-\cot 46- \cot 134)+ (\cot 47+\cot 133) + (-\cot 48- \cot 132 )+\cdots\\
\quad \quad \cdots+ (\cot 89+\cot 91)- \cot 90
\quad\qquad \textcolor{#800000}{(v)}[/tex]
Como, por [tex]\textcolor{#800000}{(iii)}[/tex], [tex]\cot (180 – a) = – \cot a[/tex]; então:
- [tex]\begin{align*}-\cot46-\cot134 &= -\cot46-\cot(180-46)\stackrel{\textcolor{#800000}{(iii)}}{=}-\cot46-(-\cot46)\\
&=-\cot46+\cot46 = 0,\end{align*}[/tex] - [tex]\cot47+\cot133= \cot47+\cot(180-47)\stackrel{\textcolor{#800000}{(iii)}}{=}\cot47-\cot47= 0,[/tex]
- [tex]\cdots[/tex]
- [tex]\cot89+\cot91= \cot89+\cot(180-89)\stackrel{\textcolor{#800000}{(iii)}}{=}\cot89-\cot89= 0,[/tex]
e, dessa forma, podemos reescrever a igualdade [tex]\textcolor{#800000}{(v)}[/tex] como:
[tex]\qquad \boxed{\cot 45 – \cot 46 + \cot 47 – \cot 48 + \cdots + \cot 133 – \cot 134} =\\
\qquad =\cot 45 + 0+ 0 + 0+\cdots+ 0- \cot 90 \\
\qquad =\boxed{\cot 45 – \cot 90}.[/tex]
Portanto, de [tex]\textcolor{#800000}{(iv)}[/tex], concluímos que
[tex]\qquad \dfrac 1{\text{sen}45 \text{ sen}46 }+\dfrac 1{\text{sen} 47 \text{ sen} 48 }+\cdots+\dfrac 1{\text{sen}133 \text{ sen}134 }=[/tex]
[tex]\qquad = \dfrac{1}{\text{sen} 1} \left(\cot 45 – \cot 46 + \cot 47 – \cot 48 + \cdots + \cot 133 – \cot 134 \right)[/tex]
[tex]\qquad = \dfrac{1}{\text{sen} 1} \left(\cot 45 – \cot 90 \right)\\
\qquad = \dfrac{1}{\text{sen} 1} \left(1-0 \right)\\
\qquad = \dfrac{1}{\text{sen} 1}[/tex],
ou seja,
[tex]\qquad \boxed{ \dfrac 1{\text{sen}45 \text{ sen}46 }+\dfrac 1{\text{sen} 47 \text{ sen} 48 }+\cdots+\dfrac 1{\text{sen}133 \text{ sen}134 } = \dfrac{1}{\text{sen} 1}}[/tex].
Dessa última igualdade, finalmente, concluímos que [tex]\fcolorbox{black}{#eee0e5}{$n=1$}[/tex].
Solução elaborada pelos Moderadores do Blog.