(2n+1)π9(2 n+1) \frac{\pi}{9}(2n+1)9π
(2n−1)π9(2 n-1) \frac{\pi}{9}(2n−1)9π
(2n+1)π18(2 n+1) \frac{\pi}{18}(2n+1)18π
(2n−1)π18(2 n-1) \frac{\pi}{18}(2n−1)18π