Classify the function y = sin x / x and the function y = cos x / x as even, odd, or neither.

Step-by-step solution

1. 1

   State the test. A function is even if $f(-x)=f(x)$ and odd if $f(-x)=-f(x)$, for every $x$ in the domain. Both functions here have domain $x\neq 0$, which is symmetric about the origin, so the test is meaningful.

2. 2

   Recall the parity of the building blocks.

   $\sin(-x)=-\sin x\ (\text{odd}),\qquad \cos(-x)=\cos x\ (\text{even}),\qquad (-x)=-x\ (\text{odd}).$

3. 3

   Test the first function.

   $f(-x)=\frac{\sin(-x)}{-x}=\frac{-\sin x}{-x}=\frac{\sin x}{x}=f(x),$

   so $\dfrac{\sin x}{x}$ is even. The two minus signs cancel — odd divided by odd is even.

4. 4

   Test the second function.

   $g(-x)=\frac{\cos(-x)}{-x}=\frac{\cos x}{-x}=-\frac{\cos x}{x}=-g(x),$

   so $\dfrac{\cos x}{x}$ is odd. Here only the denominator flips — even divided by odd is odd.

5. 5

   Confirm numerically. At $x=\pm 2$: $\dfrac{\sin 2}{2}=0.454649$ and $\dfrac{\sin(-2)}{-2}=0.454649$, equal as required for an even function. Meanwhile $\dfrac{\cos 2}{2}=-0.208073$ while $\dfrac{\cos(-2)}{-2}=+0.208073$, opposite as required for an odd function.

6. 6

   Note the general parity rules. For quotients: even/even = even, odd/odd = even, even/odd = odd, odd/even = odd. They behave exactly like multiplying signs, which makes them easy to reconstruct rather than memorise.