Question

20．定义运算$（\begin{array}{l}{a}&{b}\\{c}&{d}\end{array}）$•$（\begin{array}{l}{e}\\{f}\end{array}）$=$（\begin{array}{l}{ae+bf}\\{ce+df}\end{array}）$，如$（\begin{array}{l}{1}&{2}\\{0}&{3}\end{array}）$•$（\begin{array}{l}{4}\\{5}\end{array}）$=$（\begin{array}{l}{14}\\{15}\end{array}）$．已知α+β=π，α-β=$\frac{π}{2}$，则$（\begin{array}{l}{sinα}&{cosα}\\{cosα}&{sinα}\end{array}）$•$（\begin{array}{l}{cosβ}\\{sinβ}\end{array}）$=（　　）

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

Answer 1

分析 利用新定义、两角和差的正弦与余弦公式即可得出．

解答 解：$（\begin{array}{l}{sinα}&{cosα}\\{cosα}&{sinα}\end{array}）$•$（\begin{array}{l}{cosβ}\\{sinβ}\end{array}）$=$（\begin{array}{l}{sinαcosβ+cosαsinβ}\\{cosαcosβ+sinαsinβ}\end{array}）$=$（\begin{array}{l}{sin（α+β）}\\{cos（α-β）}\end{array}）$=$（\begin{array}{l}{sinπ}\\{cos\frac{π}{2}}\end{array}）$=$（\begin{array}{l}{0}\\{0}\end{array}）$．
故选：A．

点评 本题考查了新定义、两角和差的正弦与余弦公式，考查了推理能力与计算能力，属于基础题．