Сумма косинусов

[wpoms.]

Докажите, что если `n` натуральное число, то справедливо равенство `sum_{k=1}^{n} cos^4 ((kpi)/(2n+1)) = (6n-5)/(16)`.

@темы: Тригонометрия

Comments

[wpoms.]

Пишет Adjirranirr:

Пусть `S(n,\ x) = \sum_{k = 1}^{n} cos (k*x)`, тогда
`(cos (x) - 1) * S(n,\ x) = (cos(x) - 1)*\sum_{k = 1}^{n} cos(k*x) =`
`= \sum_{k = 1}^{n} (cos(x)*cos(k*x) - cos(k*x)) = \sum_{k = 1}^{n} (1/2 cos((k - 1)*x) + 1/2 cos((k + 1)*x) - 1/2 (2 * cos(x))) =`
`=1/2 [\sum_{k = 1}^{n} cos((k - 1)*x) - \sum_{k = 1}^{n} cos(k*x) + \sum_{k = 1}^{n} cos((k + 1)*x) - \sum_{k = 1}^{n} cos(k*x)] =`
`=1/2 [cos(0) + \sum_{k = 1}^{n - 1} cos (k*x) - \sum_{k = 1}^{n - 1} cos(k*x) - cos(n*x) + cos((n + 1)*x) + \sum_{k = 2}^{n} cos(k*x) - \sum_{k = 2}^{n} cos(k * x) - cos(x)] =`
`=1/2 (1 - cos(n*x) + cos((n + 1)*x) - cos(x)) = 1/2 (1 - cos(x)) - 1/2 sin(\frac{(2n + 1)*x}{2}) sin(\frac {x}{2})`
Окончательно, получим
`S(n,\ x) = -1/2 - 1/2 sin(\frac{(2n + 1)*x}{2}) \frac {sin(\frac {x}{2})}{cos(x) - 1}`

С другой стороны
`cos(a)^4 = (cos(a)^2)^2 = \frac{(1 + cos(2a))^2}{4} = \frac {1 + 2 cos(2*a) + \frac {1 + cos(4a)}{2}}{4} = 1/8 (3 + 4 cos(2*a) + cos(4*a))`.
Тогда
`\sum_{k = 1}^{n} cos(\frac {k*pi}{2n + 1})^4 = 1/8 (\sum_{k = 1}^{n} 3 + 4*\sum_{k = 1}^{n} cos(k*\frac {2*pi}{2n + 1}) + \sum_{k = 1}^{n} cos (k*\frac {4*pi}{2n + 1})) = `
`= 3/8 n + 1/2*S(n,\ \frac {2*pi}{2n + 1}) + 1/8 S(n,\ \frac {4*pi}{2n + 1}) = 3/8 n - 1/4 - 1/16 - 1/2 sin(pi)*\frac {sin(\frac {2*pi}{2n + 1})}{cos(\frac{2*pi}{2n + 1}) - 1} - 1/8 sin(2*pi)*\frac {sin(\frac {4*pi}{2n + 1})}{cos(\frac{4*pi}{2n + 1}) - 1} =`
`=\frac {6n - 5} {16}`, что и требовалось доказать.