Solution 6.8.4.6 t k = k ! n p 1; 2 : Then c(t k ) = 1; e ;! n t k p 1; 2 sin(! n q 1; 2 t k + ) = 1; e ;! n k ! n p 1; 2 p 1; 2 sin ! n p 1; 2 k ! n p 1; 2 +  ! = 1; e ;k p 1; 2 p 1; 2 sin(k + ) Then since sin(k + )=;sin;; wehave c(t k )=1+ e ;k p 1; 2 p 1; 2 sin  But, from Figure 6.2, sin  = q 1; 2 : Thus, nally c(t k )=1+e ;k p 1; 2 1