Question

Which function is odd? pic attached

Answer 1

An odd function is where: \(f(-x) = -f(x)\) for every \(x\).

Answer 2

right

Answer 3

b?

Answer 4

So, test letting \(x=\pi\) and \(f(x)\) being each function listed.

Answer 5

There are multiple correct solutions. I wanna see you do some tests.

Answer 6

d is not one

Answer 7

i did them all and none came out to a negative answer

Answer 8

What? \[
\sin(\pi) = 1 \\
\sin(-\pi) = -1 = -\sin(\pi)
\]

Answer 9

i was suppose to do -pi?

Answer 10

i put pi into the place of x

Answer 11

We know \(\sin(x)\) is odd not because of that test (it can only tell us if something is NOT odd when the test fails), but because of the unit circle.

Answer 12

B C D

Answer 13

Suppose \(f(x)\) is an odd function, and \(g(x)\) is the reciprocal: \(g(x) =1/f(x)\)\[
g(-x) = \frac{1}{f(-x)} = \frac{1}{-f(x)} = -\frac{1}{f(x)} = -g(x)
\]So the inverse of an odd function is an odd function.
\[
\csc(x) = \frac{1}{\sin(x) }
\]Thus \(\csc(x)\) is an odd function.

Answer 14

Since \(\cos(-x) = \cos(x)\) it is an even function.
The proof that an even functions inverse is even is similar to the one I did for odd functions.
This means \(\sec(x) = 1/\cos(x)\) is also even.

Answer 15

So all that's left to test is \(\cot(x)\). \[
\cot(-x) = \frac{\cos(-x)}{\sin(-x)} = \frac{\cos(x)}{-\sin(x)} = -\frac{\cos(x)}{\sin(x)}
=- \cot(x)
\]So \(\cot(x)\) is odd.

Answer 16

Recap:
Even functions: \(\cos(x),\ \sec(x) \)
Odd functions: \(\sin(x),\ \csc(x),\ \cot(x) \)

Answer 17

For completeness, \(\tan(x)\) is just the inverse of \(\cot(x)\) so it is odd.