题目内容
设A、B、C为△ABC的三内角,则不管△ABC的形状如何变化,表达式:①sin(A+B)+sinC;②cos(A+B)+cosC;③tan
②③④
解析:cos(A+B)+cosC=cos(π-C)+cosC=0;
tan
·tan
=tan
·tan
=cot
·tan
=1;
cos
∶sin
=cos
∶sin
=sin
∶sin
=1.
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题目内容
设A、B、C为△ABC的三内角,则不管△ABC的形状如何变化,表达式:①sin(A+B)+sinC;②cos(A+B)+cosC;③tan
②③④
解析:cos(A+B)+cosC=cos(π-C)+cosC=0;
tan
·tan
=tan
·tan
=cot
·tan
=1;
cos
∶sin
=cos
∶sin
=sin
∶sin
=1.