This equation can be used to assign a value to θ at every point in the ab plane at which Q − 1 exists. On the interval ((2 n − 3) π/ (2 n − 1) , π ), Q ( θ ) decreases monotonically from ∞ to 1 / 4 n $^{2}$. Thus, we may only choose a and b for which The points ( a, b ) which satisfy the inequality a (1 + b ) / ( a + b ) > Q$_{0}$ lie between the branches of a hyperbola which passes through the origin and has asymptotes at a = 1 /Q$_{0}$ and b = 1 /Q$_{0}$ − 1. Because Q decreases monotonically on the interval of interest, if θ$_{1}$ , θ$_{2}$ and θ$_{3}$ satisfy (2 n − 3) π/ (2 n − 1) ≤ θ$_{1}$ < θ$_{2}$ < θ$_{3}$ ≤ π , it follows that if θ$_{1}$ ≤ θ ( a, b$_{1}$ ) ≤ θ$_{2}$ and θ$_{2}$ ≤ θ ( a, b$_{2}$ ) ≤ θ$_{3}$ , then b$_{2}$ ≥ b$_{1}$ . In other words, the region in which θ$_{1}$ ≤ θ ≤ θ$_{2}$ lies under the region in which θ$_{2}$ ≤ θ ≤ θ$_{3}$ . We can find the allowed region for this process by looking for the curves along which the left and right hand sides of the inequality (45) are equal. Then we can test to see on which side of the curves the inequality is met. Finally we must check the inequality (47) to make sure that θ actually exists in the proposed region. We write 45 as an equation: and combine it with Eq. (28) to obtain: Also, Eq. (48) can be modified to yield: If we multiply Eqs. (49) and (50) together and solve for a , we get two solutions. The two solutions for a along with the corresponding expressions for b are as follows: and 17 a = cot( θ/ 2) cot( nθ ) (51) b = cot( nθ ) cot(( n − 1 / 2) θ ), if (52) (2 n − 2) π/ (2 n − 1) ≤ θ ≤ (2 n − 1) π/ (2 n ) a = − cot( θ/ 2) cot( nθ ) (53) b = cot( nθ ) cot(( n + 1 / 2) θ ), if (54) (2 n − 1) π/ (2 n ) ≤ θ ≤ (2 n ) π/ (2 n + 1) . sin $^{2}$( θ/ 2) = b/ (( a + b )( a + 1)) . (48) tan $^{2}$( nθ ) = 1 + a + b ab . (49) tan $^{2}$( θ/ 2) = b a (1 + a + b ) . (50) a (1 + b ) a + b > 1 / 4 n $^{2}$. (47)