• Matéria: Matemática
  • Autor: Lukyo
  • Perguntado 9 anos atrás

(50 PONTOS) Obtenha uma forma fechada para a soma
\displaystyle\sum\limits_{n=0}^{k}\mathrm{sen\,} n.
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Sugestão: Utilize a sequência a_{n}=\cos \left(n-\frac{1}{2}\right),

tome a diferença entre dois termos consecutivos e obtenha uma soma telescópica.

A seguinte identidade será útil para obtenção do resultado:

\cos p-\cos q=-2\,\mathrm{sen}\left(\frac{p+q}{2}\right)\mathrm{sen}\left(\frac{p-q}{2} \right ).

Respostas

respondido por: Niiya
3
cos(x+y)=cos(x)cos(y)-sen(x)sen(y)\\cos(x-y)=cos(x)cos(y)+sen(x)sen(y)

Se fizermos a diferença entre as equações, temos

cos(x+y)-cos(x-y)=-2sen(x)sen(y)

Se definirmos p = x + y e q = x - y, então

x=\dfrac{p+q}{2}~~~e~~~y=\dfrac{p-q}{2}

Daí,

\boxed{\boxed{cos(p)-cos(q)=-2sen\left(\dfrac{p+q}{2}\right)sen\left(\dfrac{p-q}{2}\right)}}
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Avaliando a diferença entre dois termos genéricos consecutivos da sequência a_{n}=cos(n-\frac{1}{2}):

a_{n+1}-a_{n}=cos(n+1-\frac{1}{2})-cos(n-\frac{1}{2})\\\\a_{n+1}-a_{n}=cos(n+\frac{1}{2})-cos(n-\frac{1}{2})

Utilizando a identidade acima:

a_{n+1}-a_{n}=-2sen\left(\dfrac{n+\frac{1}{2}+n-\frac{1}{2}}{2}\right)sen\left(\dfrac{(n+\frac{1}{2})-(n-\frac{1}{2})}{2}\right)\\\\\\a_{n+1}-a_{n}=-2sen\left(\dfrac{2n}{2}\right)sen\left(\dfrac{1}{2}\right)\\\\\\a_{n+1}-a_{n}=-2sen(n)sen(\frac{1}{2})\\\\\\~~~\therefore~~~\boxed{\boxed{sen(n)=\dfrac{a_{n+1}-a_{n}}{(-2)sen(\frac{1}{2})}}}

Substituindo sen(n) pela expressão encontrada:

\displaystyle\sum\limits_{n=0}^{k}sen(n)=\sum\limits_{n=0}^{k}\dfrac{a_{n+1}-a_{n}}{(-2)sen(\frac{1}{2})}\\\\\\\sum\limits_{n=0}^{k}sen(n)=\dfrac{1}{(-2)sen(\frac{1}{2})}\sum\limits_{n=0}^{k}(a_{n+1}-a_{n})

Como vimos,

\displaystyle\sum\limits_{n=N_{0}}^{N}(a_{n+1}-a_{n})=a_{N+1}-a_{N_{0}}

Nesse caso, temos N = k e N₀ = 0

\displaystyle\sum\limits_{n=0}^{k}(a_{n+1}-a_{n})=a_{k+1}-a_{0}\\\\\\\sum\limits_{n=0}^{k}(a_{n+1}-a_{n})=cos\left(k+1-\frac{1}{2}\right)-cos\left(0-\frac{1}{2}\right)\\\\\\\sum\limits_{n=0}^{k}(a_{n+1}-a_{n})=cos\left(k+\dfrac{1}{2}\right)-cos\left(-\dfrac{1}{2}\right)

Como cos\left(-\dfrac{1}{2}\right)=cos\left(\dfrac{1}{2}\right):

\boxed{\boxed{\sum\limits_{n=0}^{k}(a_{n+1}-a_{n})=cos\left(k+\dfrac{1}{2}\right)-cos\left(\dfrac{1}{2}\right)}}

Finalmente:

\displaystyle\sum\limits_{n=0}^{k}sen(n)=-\dfrac{1}{2}cossec\left(\dfrac{1}{2}\right)\sum\limits_{n=0}^{k}(a_{n+1}-a_{n})\\\\\\\sum\limits_{n=0}^{k}sen(n)=-\dfrac{1}{2}cossec\left(\dfrac{1}{2}\right)\cdot\left[cos\left(k+\dfrac{1}{2}\right)-cos\left(\dfrac{1}{2}\right)\right]\\\\\\\boxed{\boxed{\sum\limits_{n=0}^{k}sen(n)=\dfrac{1}{2}cossec\left(\dfrac{1}{2}\right)\cdot\left[cos\left(\dfrac{1}{2}\right)-cos\left(k+\dfrac{1}{2}\right)\right]}}

Lukyo: Perfeito!! :-)
Niiya: Demorei nessa por causa da identidade, depois notei que eram 2 senos!
respondido por: carlosmath
1
Hola. 
Por mi parte utilizaré la ley telescópica...

           \displaystyle
\boxed{\sum_{n=0}^kf(n+1)-f(n)=f(k+1)-f(0)}

Entonces centremos nuestra atención en la sumatoria

              \displaystyle
\sum_{n=0}^k\cos \left(n-\frac{1}{2}\right)

Y apliquemos la ley telescópica

            \displaystyle
\sum_{n=0}^k\cos \left(n+\frac{1}{2}\right)-\cos \left(n-\frac{1}{2}\right)=\cos \left(k+\frac{1}{2}\right)-\cos\frac{1}{2}\\ \\ \\
\sum_{n=0}^k-2\sin(n)\sin \left(\frac{1}{2}\right)=\cos \left(k+\frac{1}{2}\right)-\cos\frac{1}{2}\\ \\ \\
-2\sin \left(\frac{1}{2}\right)\sum_{n=0}^k\sin(n)=\cos \left(k+\frac{1}{2}\right)-\cos\frac{1}{2}\\ \\ \\ \\
\boxed{\sum_{n=0}^k\sin(n)=-\frac{1}{2}\csc\left(\frac{1}{2}\right)\left[\cos \left(k+\frac{1}{2}\right)-\cos\frac{1}{2}\right]}
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