Question

(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:

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

Answer 1

$cos(x+y)=cos(x)cos(y)-sen(x)sen(y)\\cos(x-y)=cos(x)cos(y)+sen(x)sen(y)$

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:

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

Answer 2

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

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