$(2.3.1.3) \quad \cos h x_{k}=\frac{e^{h i x_{k}}+e^{(N-h) i x_{k}}}{2}, \quad \sin h x_{k}=\frac{e^{h i x_{k}}-e^{(N-h) i x_{k}}}{2 i}$ Making these substitutions in expressions (2.3.1.1) for $\Psi(x)$ and then collecting the powers of $e^{i x_{k}}$ produces a phase polynomial $p(x),(2.3.1 .2)$, with coefficients $\beta_{j}, j=0, \ldots, N-1$, which are related to the coefficients $A_{h}$, $B_{h}$ of $\Psi(s) $ as follows: 2 If $\sin u$ and $\cos u$ have to be both evaluated for the same argument $u$, then it may be advantageous to evaluate $t=\tan (u / 2)$ and express $\sin u$ and $\cos u$ in terms of $t$ : $$ \sin u=\frac{2 t}{1+t^{2}}, \quad \cos u=\frac{1-t^{2}}{1+t^{2}} . $$ This procedure is numerically stable for $0 \leq u \leq \pi / 4$, and the problem can always be transformed so that the argument falls into that range. SP.JG. 354 $(2.3.1.3) \quad \cos h x_{k}=\frac{e^{h i x_{k}}+e^{(N-h) i x_{k}}}{2}, \quad \sin h x_{k}=\frac{e^{h i x_{k}}-e^{(N-h) i x_{k}}}{2 i}$ Making these substitutions in expressions (2.3.1.1) for $\Psi(x)$ and then collecting the powers of $e^{i x_{k}}$ produces a phase polynomial $p(x),(2.3.1 .2)$, with coefficients $\beta_{j}, j=0, \ldots, N-1$, which are related to the coefficients $A_{h}$, $B_{h}$ of $\Psi(s) $ as follows: 2 If $\sin u$ and $\cos u$ have to be both evaluated for the same argument $u$, then it may be advantageous to evaluate $t=\tan (u / 2)$ and express $\sin u$ and $\cos u$ in terms of $t$ : $$ \sin u=\frac{2 t}{1+t^{2}}, \quad \cos u=\frac{1-t^{2}}{1+t^{2}} . $$ This procedure is numerically stable for $0 \leq u \leq \pi / 4$, and the problem can always be transformed so that the argument falls into that range. SP.JG. 354