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由三角恒等式化来的代数题

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hbghlyj posted 2020-9-21 23:32 |Read mode
已知实数$a_{11},a_{12},a_{13},a_{21},a_{22},a_{23},c_1,c_2,c_3$满足\begin{align*}a_{11} a_{12} + a_{21} a_{22 }&= c_1,\\ a_{12 }a_{13} + a_{22} a_{23 }&= c_2, \\
a_{11} a_{13 }+ a_{21} a_{23} &= c_3,\\ a_{11}^2 + a_{21}^2 &= 1,\\ a_{12}^2 + a_{22}^2 &= 1, \\
a_{13}^2 + a_{23}^2 &=1,\end{align*}求证$c_1^2+c_2^2+c_3^2-2c_1c_2c_3=1,a_{22}^2+a_{23}^2+c_2^2-2a_{22}a_{23}c_2=1$

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