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Resposta:   E = N/2.

Algumas identidades trigonométricas

Dados [tex]x,\,y\in\mathbb{R}[/tex] valem as seguintes Identidades:

• Diferença entre senos:

     [tex]\mathrm{sen}(x)-\mathrm{sen}(y)=2\,\mathrm{sen}\!\left(\dfrac{x-y}{2}\right)\!\cos\!\left(\dfrac{x+y}{2}\right)[/tex]

• Quadrado do seno:

     [tex]\mathrm{sen}^2(x)=\dfrac{1}{2}-\dfrac{1}{2}\cos(2x)[/tex]

Algumas propriedades dos somatórios

Dados

     [tex]a,\,b[/tex] inteiros não-negativos, [tex]a\le b,[/tex]

     [tex]C,\,C_1,\,C_2[/tex] constantes reais,

     [tex]f,\,g:~~\mathbb{Z}_+\to\mathbb{R}[/tex] sequências numéricas,

valem as seguintes propriedades:

• Somatório de uma constante:

     [tex]\displaystyle\sum_{k=a}^b C=C(b-a+1)[/tex]

• Linearidade do operador somatório:

     [tex]\displaystyle\sum_{k=a}^b \big[C_1\,f(k)+C_2\,g(k)\big]=C_1\sum_{k=a}^b f(k)+C_2\sum_{k=a}^b g(k)[/tex]

• Soma telescópica:

     [tex]\displaystyle\sum_{k=a}^b \big[f(k+1)-f(k)\big]=f(b+1)-f(a)[/tex]

Uma proposição prévia

[Proposição 1] Sejam [tex]\alpha\in\mathbb{R}\setminus\{n\pi:~n\in\mathbb{Z}\}[/tex] e [tex]k\in\mathbb{N}.[/tex] Então, vale a seguinte identidade:

     [tex]\mathrm{sen}^2(\alpha k)=\dfrac{1}{2}-\dfrac{1}{4}\csc(\alpha)\cdot \Big[\mathrm{sen}\big(\alpha(2k+1)\big)-\mathrm{sen}\big(\alpha(2k-1)\big)\Big].[/tex]

     [Demonstração] De fato, pela fórmula da diferença entre senos, temos

     [tex]\mathrm{sen}\big(\alpha(2k+1)\big)-\mathrm{sen}\big(\alpha(2k-1)\big)=2\,\mathrm{sen}\!\left(\dfrac{\alpha(2k+1)-\alpha(2k-1)}{2}\right)\!\cos\!\left(\dfrac{\alpha(2k+1)+\alpha(2k-1)}{2}\right)\\\\ =2\,\mathrm{sen}\!\left(\dfrac{2\alpha k+\alpha-2\alpha k+\alpha}{2}\right)\!\cos\!\left(\dfrac{2\alpha k+\alpha+2\alpha k-\alpha}{2}\right)\\\\ =2\,\mathrm{sen}\!\left(\dfrac{2\alpha}{2}\right)\!\cos\!\left(\dfrac{4\alpha k}{2}\right)\\\\ \Longleftrightarrow\quad \mathrm{sen}\big(\alpha(2k+1)\big)-\mathrm{sen}\big(\alpha(2k-1)\big)=2\,\mathrm{sen}(\alpha)\cos(2\alpha k)[/tex]

Portanto,

     [tex]\Longleftrightarrow\quad \cos(2\alpha k)=\dfrac{\mathrm{sen}\big(\alpha(2k+1)\big)-\mathrm{sen}\big(\alpha(2k-1)\big)}{2\,\mathrm{sen}(\alpha)}\\\\ \Longleftrightarrow\quad \cos(2\alpha k)=\dfrac{1}{2}\csc(\alpha)\cdot \Big[\mathrm{sen}\big(\alpha(2k+1)\big)-\mathrm{sen}\big(\alpha(2k-1)\big)\Big][/tex]

Pela fórmula do quadrado do seno, segue que

     [tex]\mathrm{sen}^2(\alpha k)\\\\ =\dfrac{1}{2}-\dfrac{1}{2}\cos(2\alpha k)\\\\ =\dfrac{1}{2}-\dfrac{1}{2}\cdot \dfrac{1}{2}\csc(\alpha)\cdot \Big[\mathrm{sen}\big(\alpha (2k+1)\big)-\mathrm{sen}\big(\alpha (2k-1)\big)\Big]\\\\ \Longleftrightarrow\quad\mathrm{sen}^2(\alpha k)=\dfrac{1}{2}-\dfrac{1}{4}\csc(\alpha)\cdot \Big[\mathrm{sen}\big(\alpha(2k+1)\big)-\mathrm{sen}\big(\alpha(2k-1)\big)\Big][/tex]

como queríamos demonstrar.

Esta identidade nos será útil, pois ao somá-la, será possível aplicar a propriedade telescópica e efetuar as simplificações, conforme veremos a seguir.

Encontrando uma fórmula fechada para o somatório

     [tex]\displaystyle E=\mathrm{sen}^2\!\left(\frac{1\pi n}{N}\right)+\mathrm{sen}^2\!\left(\frac{2\pi n}{N}\right)+\cdots +\mathrm{sen}^2\!\left(\frac{N\pi n}{N}\right)\\\\ \Longleftrightarrow\quad E=\sum_{k=1}^N \mathrm{sen}^2\!\left(\frac{\pi n}{N}\,k\right)[/tex]

Pela proposição 1, com [tex]\alpha=\dfrac{\pi n}{N},[/tex] segue que

     [tex]\displaystyle\Longleftrightarrow\quad E=\sum_{k=1}^N \left(\dfrac{1}{2}-\dfrac{1}{4}\csc\!\left(\frac{\pi n}{N}\right)\cdot \left[\mathrm{sen}\!\left(\frac{\pi n}{N}\,(2k+1)\right)-\mathrm{sen}\!\left(\frac{\pi n}{N}\,(2k-1)\right)\right]\right)[/tex]

Pela a linearidade do operador somatório, temos

     [tex]\displaystyle\Longleftrightarrow\quad E=\sum_{k=1}^N \frac{1}{2}-\frac{1}{4}\csc\!\left(\frac{\pi n}{N}\right)\cdot \sum_{k=1}^N \left[\mathrm{sen}\!\left(\frac{\pi n}{N}\,(2k+1)\right)-\mathrm{sen}\!\left(\frac{\pi n}{N}\,(2k-1)\right)\right]\\\\ \Longleftrightarrow\quad E=\sum_{k=1}^N \frac{1}{2}-\frac{1}{4}\csc\!\left(\frac{\pi n}{N}\right)\cdot \sum_{k=1}^N \left[\mathrm{sen}\!\left(\frac{\pi n}{N}\,(2(k+1)-1)\right)-\mathrm{sen}\!\left(\frac{\pi n}{N}\,(2k-1)\right)\right][/tex]

Acima, temos o somatório de uma constante [tex]\dfrac{1}{2}[/tex] e uma soma telescópica da sequência [tex]f(k)=\mathrm{sen}\!\left(\dfrac{\pi n}{N}\,(2k-1)\right).[/tex]

Aplicando as propriedades dos somatórios, temos

     [tex]\displaystyle\Longleftrightarrow\quad E=\frac{1}{2}\,(N-1+1)-\dfrac{1}{4}\csc\!\left(\frac{\pi n}{N}\right)\cdot \left[\mathrm{sen}\!\left(\frac{\pi n}{N}\,(2(N+1)-1)\right)-\mathrm{sen}\!\left(\frac{\pi n}{N}\,(2\cdot 1-1)\right)\right]\\\\ \Longleftrightarrow\quad E=\frac{1}{2}\,N-\dfrac{1}{4}\csc\!\left(\frac{\pi n}{N}\right)\cdot \left[\mathrm{sen}\!\left(\frac{\pi n}{N}\,(2N+1)\right)-\mathrm{sen}\!\left(\frac{\pi n}{N}\right)\right][/tex]

Para simplificar, usamos novamente a identidade para a diferença entre senos:

     [tex]\Longleftrightarrow\quad E=\dfrac{1}{2}\,N-\dfrac{1}{4}\csc\!\left(\dfrac{\pi n}{N}\right)\cdot 2\,\mathrm{sen}\!\left(\dfrac{\frac{\pi n}{N}(2N+1)-\frac{\pi n}{N}}{2}\right)\cos\!\left(\dfrac{\frac{\pi n}{N}(2N+1)+\frac{\pi n}{N}}{2}\right)\\\\ \Longleftrightarrow\quad E=\dfrac{1}{2}\,N-\dfrac{1}{2}\csc\!\left(\dfrac{\pi n}{N}\right)\cdot \mathrm{sen}\!\left(\dfrac{\frac{\pi n}{N}(2N+1-1)}{2}\right)\cos\!\left(\dfrac{\frac{\pi n}{N}(2N+1+1)}{2}\right)\\\\ \Longleftrightarrow\quad E=\dfrac{1}{2}\,N-\dfrac{1}{2}\csc\!\left(\dfrac{\pi n}{N}\right)\cdot \mathrm{sen}\!\left(\dfrac{\frac{\pi n}{N}(2N)}{2}\right)\cos\!\left(\dfrac{\frac{\pi n}{N}(2N+2)}{2}\right)\\\\ E=\dfrac{1}{2}\,N-\dfrac{1}{2}\csc\!\left(\dfrac{\pi n}{N}\right)\cdot \mathrm{sen}(\pi n)\cos\!\left(\dfrac{\frac{\pi n}{N}\cdot 2(N+1)}{2}\right)\\\\ E=\dfrac{1}{2}\,N-\dfrac{1}{2}\csc\!\left(\dfrac{\pi n}{N}\right)\cdot \mathrm{sen}(\pi n)\cos\!\left(\dfrac{\pi n}{N}\,(N+1)\right)[/tex]

No entanto, [tex]\mathrm{sen}(\pi n)=0,[/tex] para todo n inteiro. Logo,

     [tex]\Longleftrightarrow\quad E=\dfrac{1}{2}\,N-\dfrac{1}{2}\csc\!\left(\dfrac{\pi n}{N}\right)\cdot 0\cdot \cos\!\left(\dfrac{\pi n}{N}\,(N+1)\right)\\\\ \Longleftrightarrow\quad E=\dfrac{1}{2}\,N-0\\\\ \Longleftrightarrow\quad E=\dfrac{N}{2}[/tex]

sendo esta a resposta.

Dúvidas? Comente.

Bons estudos! :-)