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I understand the rules of if $f(-x)=-f(x)$ then odd and if $f(-x) = f(x)$ then even. I also know it's possible for the function to be neither even nor odd.

For simple polynomials these rules are easy to apply. For trig functions with phase shifts not so easy.

Without using a visual of symmetry about the origin or y-axis, how would I determine if something like $f(x)=sin(x-\pi/8)$ is even, odd or neither? When I compute $f(-x)$ I get $f(-x)=sin(-x-\pi/8)$ and it's not easy to see if this is the same as $=-sin(x-\pi/8)$.

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    Do you know $\sin(-x)=-\sin(x)$? – 2017-01-16
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    Try a couple of _specific_ well-chosen values of $x$. Here, for example, $f(0) = -\sin(\pi/8) \neq 0$, so $f$ is not odd, while $f(-\pi/8) = \sin(-\pi/4) \neq 0 = f(\pi/8)$, so $f$ is not even. – 2017-01-16
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    $\sin$ is odd, $\cos$ is even. Other phases are neither. – 2017-01-16
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    Yves, I think some phases, specifically $\pi/2)$ could be still even / odd depending on sin or cos. – 2017-01-16

2 Answers 2

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$$f(x)=\sin(x+a)= \sin(x)\cos (a)+\sin(a)\cos(x)$$

So,

$$f(-x)=\sin(-x+a)=\sin(-x)\cos (a)+\sin(a)\cos(-x)=-\sin(x)\cos (a)+\sin(a)\cos(x)$$

$f$ is even if:

$$\sin(x)\cos (a)=0 \rightarrow \cos (a)=0$$

$f$ is odd if:

$$\sin(a)\cos (x)=0 \rightarrow \sin (a)=0$$

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    Couldn't we say that $f$ is even if $-sin(x)cos(a) = sin(x) cos(a)$ implying if $cos(a) <0$ then even? – 2017-01-16
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    @user163862: $-\sin(x)\cos(a) = \sin(x) \cos(a) \Leftrightarrow 2\sin(x)\cos(a) =0 \Leftrightarrow \sin(x)\cos(a) =0$. What you claim is not right because the last equality must hold for any value of $x$ even when $\sin x \ne 0$ – 2017-01-16
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    could we talk in chat room for a moment? – 2017-01-16
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    Do I just select Mathematics room and you will be there or ? – 2017-01-16
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/51900/discussion-between-user163862-and-arnaldo). – 2017-01-16
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A simple example, you just have to apply the definition. Given $k$ real, define $$ f_k(x):=\sin(x+k\pi). $$ Then $f_k$ is odd if and only if $f_k(-x)=-f_k(x)$, i.e. $$ \sin(-x+k\pi)=-\sin(x+k\pi). $$ iff $$ -\sin(-x+k\pi)=\sin(x+k\pi) $$ But $-\sin(-y)=\sin(y)$, hence the above is equivalent to $$ \sin(x-k\pi)=\sin(x+k\pi)=\sin((x-k\pi)+2k\pi). $$ Hence $f_k$ is an odd function if and only if $k$ is an integer.

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    I followed all of that except for the part where you say --"is equivalent to $sin(x-k\pi)=sin(x+k\pi)$. It would seem to me that it should be: $sin(-x+k\pi)=sin(x+k\pi)$ – 2017-01-16
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    is $sin(-x+k\pi)$ the same as $sin(x-k\pi)$? – 2017-01-16
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    Nope, I edit it, hoping it is clearer ;) – 2017-01-16