(2n−1)π8(2 n-1) \frac{\pi}{8}(2n−1)8π
(3n−1)π4(3 n-1) \frac{\pi}{4}(3n−1)4π
(4n+1)π4(4 n+1) \frac{\pi}{4}(4n+1)4π
(2n+1)π8(2 n+1) \frac{\pi}{8}(2n+1)8π