论文部分内容阅读
题目:判断函数y=(1-cosx+sinx)/(1+cosx+sinx)的奇偶性。解:y=(1-cosx+sinx)/(1+cosx+sinx)=(2sin~2(x/2)+2sin(x/2)cos(x/2))/(2sin~2(x/2)+2sin(x/2)-cos(x/2))=2sin(x/2)(sin(x/2)+cos(x/2))/2cos(x/2)(cos(x/2)+sin(x/2))=tg(x/2)。∵ y=tg(x/2)是奇函数。∴ y=(1-cosx+sinx)/(1+cosx+sinx)是奇函数。表面看来,以上解法无懈可击。但如果注意到当
Title: Determine the parity of the function y=(1-cosx+sinx)/(1+cosx+sinx). Solution: y=(1-cosx+sinx)/(1+cosx+sinx)=(2sin~2(x/2)+2sin(x/2)cos(x/2))/(2sin~2(x /2) +2sin(x/2)-cos(x/2))=2sin(x/2)(sin(x/2)+cos(x/2))/2cos(x/2)(cos( x/2)+sin(x/2))=tg(x/2). ∵ y=tg(x/2) is an odd function. ∴ y=(1-cosx+sinx)/(1+cosx+sinx) is an odd function. On the surface, the above solution is flawless. But if you notice