题目内容
(2012•长春一模)类比“两角和与差的正弦公式”的形式,对于给定的两个函数:S(x)=ax-a-x,C(x)=ax+a-x,其中a>0,且a≠1,下面正确的运算公式是:( )
①S(x+y)=S(x)C(y)+C(x)S(y);②S(x-y)=S(x)C(y)-C(x)S(y);
③2S(x+y)=S(x)C(y)+C(x)S(y);④2S(x-y)=S(x)C(y)-C(x)S(y).
①S(x+y)=S(x)C(y)+C(x)S(y);②S(x-y)=S(x)C(y)-C(x)S(y);
③2S(x+y)=S(x)C(y)+C(x)S(y);④2S(x-y)=S(x)C(y)-C(x)S(y).
分析:根据S(x)=ax-a-x,C(x)=ax+a-x,求出S(x+y)与S(x)C(y)+C(x)S(y),从而得到它们的关系,以及求出S(x-y)与S(x)C(y)-C(x)S(y),从而得到它们的关系,即可得到答案.
解答:解:∵S(x)=ax-a-x,C(x)=ax+a-x,
∴S(x+y)=ax+y-a-x-y,S(x)C(y)+C(x)S(y)=(ax-a-x)(ay+a-y)+(ax+a-x)(ay-a-y)=2ax+y-2a-x-y,
∴2S(x+y)=S(x)C(y)+C(x)S(y),故③正确;
∵S(x-y)=ax-y-a-x+y,S(x)C(y)-C(x)S(y)=(ax-a-x)(ay+a-y)-(ax+a-x)(ay-a-y)=2ax-y-2a-x+y,
∴2S(x-y)=S(x)C(y)-C(x)S(y),故④正确.
故选B.
∴S(x+y)=ax+y-a-x-y,S(x)C(y)+C(x)S(y)=(ax-a-x)(ay+a-y)+(ax+a-x)(ay-a-y)=2ax+y-2a-x-y,
∴2S(x+y)=S(x)C(y)+C(x)S(y),故③正确;
∵S(x-y)=ax-y-a-x+y,S(x)C(y)-C(x)S(y)=(ax-a-x)(ay+a-y)-(ax+a-x)(ay-a-y)=2ax-y-2a-x+y,
∴2S(x-y)=S(x)C(y)-C(x)S(y),故④正确.
故选B.
点评:本题主要考查了类比推理,写类比结论时:先找类比对象,再找类比元素,本题只需逐一验证即可,属于基础题.
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