Seja um número real [tex3]r\neq 0[/tex3] e [tex3]x_{k+1}=x_k+r[/tex3] ,[tex3]k\in\mathbb N^*[/tex3]
Então,
[tex3]C\equiv\sum_{k=1}^{n}\cos x_{k}=\frac{\cos\frac{x_{1}+x_{n}}{2}\sen\frac{nr}{2}}{\sen\frac{r}{2}}[/tex3]
[tex3]S\equiv\sum_{k=1}^{n}\sen x_{k}=\frac{\sen\frac{x_{1}+x_{n}}{2}\sen\frac{nr}{2}}{\sen\frac{r}{2}}[/tex3]
Prova
[tex3]2\sen\frac r2\cdot C=\sum_{k=1}^n2\sen\frac r2\cos x_k= \sum_{k=1}^n\left[\sen \left(x_k+\frac r2\right)-\sen \left(x_k-\frac r2\right)\right]=\sen \left(x_n+\frac r2\right)-\sen \left(x_1-\frac r2\right)[/tex3]
Como [tex3]x_{k+1}=x_k+r[/tex3] então [tex3]x_{k+1}-\frac {r}{2}=x_k+\frac{r}{2}[/tex3] .
Temos que,
[tex3]2\sen\frac r2\cdot C=2\sen \frac {x_n-x_1+r}{2}\cos\frac {x_1+x_n}{2}= 2\sen \frac {(n-1)r+r}{2}\cos\frac {x_1+x_n}{2}[/tex3]
Portanto,
[tex3]\boxed{C=\frac {\cos\frac {x_1+x_n}{2}\sen\frac {nr}{2}}{\sen\frac r2}}[/tex3]
[tex3]2\sen\frac r2\cdot S=\sum_{k=1}^n2\sen\frac r2\sen x_k= \sum_{k=1}^n\left[\cos \left(x_k-\frac r2\right)-\cos\left(x_k+\frac r2\right)\right]=\cos\left(x_1-\frac r2\right)-\cos\left(x_n+\frac r2\right)[/tex3]
Como [tex3]x_{k+1}=x_k+r[/tex3] então [tex3]x_{k+1}-\frac {r}{2}=x_k+\frac{r}{2}[/tex3] .
Temos que,
[tex3]2\sen\frac r2\cdot S=2\sen \frac {x_n-x_1+r}{2}\sen\frac {x_1+x_n}{2}= 2\sen \frac {(n-1)r+r}{2}\sen\frac {x_1+x_n}{2}[/tex3]
Portanto,
[tex3]\boxed{S=\frac {\sen\left(\frac {x_1+x_n}{2}\right)\cdot\sen\left(\frac {nr}{2}\right)}{\sen\left(\frac r2\right)}}[/tex3]
Abraço.
