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How is the following function an odd function? $S(x) = \frac{\sin x}{x}$, $x \neq 0$

I get $$\frac{\sin(-x)}{-x} = \frac{\sin x}{x}$$

which is even right? because $S(-x) = S(x)$?

So unless the book is wrong, what am I doing wrong?

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    $\begingroup$ It is indeed even. $\endgroup$
    – mrf
    Feb 8, 2013 at 7:00
  • $\begingroup$ Ha, okay, sorry to bother!! Thanks for the confirmation! $\endgroup$
    – KKendall
    Feb 8, 2013 at 7:04

2 Answers 2

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You are correct to say that $\frac{\sin x}{x}$ is an even function.

In fact, odd and even functions sort of behave like odd and even numbers. When you add two even numbers or two odd numbers, you get an even number. If you add an even and and odd, you get an odd number. The same is true about subtraction.

Similarly, if you multiply two even functions or two odd functions, you get an even function. If you multiply an even and an odd, you get odd. The same is true about division, where it's defined.

Here, both $\sin x$ and $x$ are odd, so their quotient is even.

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From odd function definition we have $f(-x)=-f(x)$.

Now $g(x)=\frac{\sin x}x$ consists of $2$ odd functions. It can be represented as $g(x)=h(x)v(x)$, where $h(x)=\sin x$ and $v(x)=\frac1x$.

Since $h,v$ are both odd, we get $$g(-x)=h(-x)v(-x)=(-h(x))(-v(x))=h(x)v(x)=g(x).$$ Which means that $g$ is an even function. This is true for any odd functions $h,v$.

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