Donner la valeur de [math]\cos\left(\frac{\pi}{360}\right) + \cos\left(\frac{2\pi}{360}\right) + \cos\left(\frac{3\pi}{360}\right) + .... + \cos\left(\frac{359\pi}{360}\right)[/math]

On se base sur la relation [math]cos\left(\pi-x\right)=-cos\left(x\right)[/math][br][br] [math]\sum_{k=1}^{k=359}{cos\left(\frac{k \times \pi}{360}\right)} = \sum_{k=1}^{k=179}{cos\left(\frac{k \times \pi}{360}\right) + cos\left(\frac{(360 - k) \times \pi}{360}\right)} = \sum_{k=1}^{k=179}{cos\left(\frac{k \times \pi}{360}\right) + cos\left(\pi - \frac{k \times \pi}{360}\right)} [/math] [br][br][math]\sum_{k=1}^{k=359}{cos\left(\frac{k \times \pi}{360}\right)} = \sum_{k=1}^{k=179}{cos\left(\frac{k \times \pi}{360}\right) - cos\left(\frac{k \times \pi}{360}\right)} = 0 [/math]

[size=150]On se base sur la relation [math]cos\left(a\right)+cos\left(b\right)=2\times cos\left(\frac{a+b}{2}\right)\times cos\left(\frac{a-b}{2}\right)[/math][br][br] [math]\sum_{k=1}^{k=359}{cos\left(\frac{k \times \pi}{360}\right)} = \sum_{k=1}^{k=179}{cos\left(\frac{k \times \pi}{360}\right) + cos\left(\frac{(360 - k) \times \pi}{360}\right)} = 2\sum_{k=1}^{k=179}{\cos\left(\frac{\frac{k \times \pi}{360}+\frac{(360 - k)\times \pi}{360}}{2}\right) \times \cos\left(\frac{\frac{k \times \pi}{360}-\frac{(360 - k)\times \pi}{360}}{2}\right)}[/math] [br][br][math]\sum_{k=1}^{k=359}{cos\left(\frac{k \times \pi}{360}\right)} = \sum_{k=1}^{k=179}{cos\left(\frac{ \pi}{2}\right) \times \cos\left(\frac{\frac{k \times \pi}{360}-\frac{(360 - k)\times \pi}{360}}{2}\right)} = 0 [/math] car [math]cos\left(\frac{ \pi}{2}\right) = 0[/math][/size]