Question

13．定義運算$|{\begin{array}{l}a&b\\ c&d\end{array}}|?|{\begin{array}{l}e\\ f\end{array}}|=|{\begin{array}{l}{ae-bf}\\{ce-df}\end{array}}|$，例如$|{\begin{array}{l}1&2\\ 3&4\end{array}}|?|{\begin{array}{l}5\\ 6\end{array}}|=|{\begin{array}{l}{-7}\\{-9}\end{array}}|$．若已知$α+β=π，α-β=\frac{π}{2}$，則$|{\begin{array}{l}{sinα}&{cosα}\\{cosα}&{sinα}\end{array}}|?|{\begin{array}{l}{cosβ}\\{sinβ}\end{array}}|$=（　�。�

• A． $|{\begin{array}{l}0\\ 1\end{array}}|$
• B． $|{\begin{array}{l}1\\ 0\end{array}}|$
• C． $|{\begin{array}{l}0\\ 0\end{array}}|$
• D． $|{\begin{array}{l}1\\{-1}\end{array}}|$

Answer 1

分析 直接利用新定義結合兩角和與差的三角函數(shù)化簡得答案．

解答 解：由新定義可得，$|{\begin{array}{l}{sinα}&{cosα}\\{cosα}&{sinα}\end{array}}|?|{\begin{array}{l}{cosβ}\\{sinβ}\end{array}}|$=$|\begin{array}{l}{sinαcosβ-cosαsinβ}\\{cosαcosβ-sinαsinβ}\end{array}|$=$|\begin{array}{l}{sin（α-β）}\\{cos（α+β）}\end{array}|$=$|\begin{array}{l}{sin\frac{π}{2}}\\{cosπ}\end{array}|$=$|\begin{array}{l}{1}\\{-1}\end{array}|$．
故選：D．

點評 本題考查三角函數(shù)的化簡求值，考查了兩角和與差的三角函數(shù)，是基礎題．