Eufisky - The lost book

## 涉及到三角函数多倍角的行列式求解

（姚张锋问题）求行列式$\det \left( {\begin{array}{*{20}{c}}1&{\cos {\theta _2}}&{\cos {\theta _2}}& \cdots &{\cos \left( {n - 1} \right){\theta _1}}\\1&{\cos {\theta _2}}&{\cos 2{\theta _2}}& \cdots &{\cos \left( {n - 1} \right){\theta _2}}\\\vdots & \vdots & \vdots &{}& \vdots \\1&{\cos {\theta _n}}&{\cos 2{\theta _n}}& \cdots &{\cos \left( {n - 1} \right){\theta _n}}\end{array}} \right)$

\begin{align*}D &= \left| {\begin{array}{*{20}{c}}1&{\cos {\theta _1}}&{\cos 2{\theta _1}}& \cdots &{\cos \left( {n - 1} \right){\theta _1}}\\1&{\cos {\theta _2}}&{\cos 2{\theta _2}}& \cdots &{\cos \left( {n - 1} \right){\theta _2}}\\\vdots & \vdots & \vdots &{}& \vdots \\1&{\cos {\theta _n}}&{\cos 2{\theta _n}}& \cdots &{\cos \left( {n - 1} \right){\theta _n}}\end{array}} \right| = \frac{1}{{{2^{n - 1}}}}\left| {\begin{array}{*{20}{c}}1&{{\varepsilon _1} + {{\bar \varepsilon }_1}}&{\varepsilon _1^2 + \bar \varepsilon _1^2}& \cdots &{\varepsilon _1^{n - 1} + \bar \varepsilon _1^{n - 1}}\\1&{{\varepsilon _2} + {{\bar \varepsilon }_2}}&{\varepsilon _2^2 + \bar \varepsilon _2^2}& \cdots &{\varepsilon _2^{n - 1} + \bar \varepsilon _2^{n - 1}}\\\vdots & \vdots & \vdots &{}& \vdots \\1&{{\varepsilon _n} + {{\bar \varepsilon }_n}}&{\varepsilon _n^2 + \bar \varepsilon _n^2}& \cdots &{\varepsilon _n^{n - 1} + \bar \varepsilon _n^{n - 1}}\end{array}} \right|\\&= \frac{1}{{{2^{n - 1}}}}\left| {\begin{array}{*{20}{c}}1&{{\varepsilon _1} + {{\bar \varepsilon }_1}}&{{{\left( {{\varepsilon _1} + {{\bar \varepsilon }_1}} \right)}^2}}& \cdots &{{{\left( {{\varepsilon _1} + {{\bar \varepsilon }_1}} \right)}^{n - 1}}}\\1&{{\varepsilon _2} + {{\bar \varepsilon }_2}}&{{{\left( {{\varepsilon _2} + {{\bar \varepsilon }_2}} \right)}^2}}& \cdots &{{{\left( {{\varepsilon _2} + {{\bar \varepsilon }_2}} \right)}^{n - 1}}}\\\vdots & \vdots & \vdots &{}& \vdots \\1&{{\varepsilon _n} + {{\bar \varepsilon }_n}}&{{{\left( {{\varepsilon _n} + {{\bar \varepsilon }_n}} \right)}^2}}& \cdots &{{{\left( {{\varepsilon _n} + {{\bar \varepsilon }_n}} \right)}^{n - 1}}}\end{array}} \right| = \frac{1}{{{2^{n - 1}}}}\prod\limits_{1 \le j < i \le n} {\left( {{\varepsilon _i} + {{\bar \varepsilon }_i} - {\varepsilon _j} - {{\bar \varepsilon }_j}} \right)} \\&= \frac{1}{{{2^{n - 1}}}} \times {2^{\frac{{n\left( {n - 1} \right)}}{2}}}\prod\limits_{1 \le j < i \le n} {\left( {\cos {\theta _i} - \cos {\theta _j}} \right)}  = {2^{\frac{{\left( {n - 1} \right)\left( {n - 2} \right)}}{2}}}\prod\limits_{1 \le j < i \le n} {\left( {\cos {\theta _i} - \cos {\theta _j}} \right)} .\end{align*}