Even or Odd function
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 $=-sin(x-\pi/8)$.
$\endgroup$ 42 Answers
$\begingroup$$$f(x)=\sin(x+a)= \sin(x)\cos (a)+\sin(a)\cos(x)$$
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:
$$\sin(a)\cos (x)=0 \rightarrow \sin (a)=0$$
$\endgroup$ 5 $\begingroup$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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