So I know this is probably a really obvious question, but I'm a bit unclear on this. Why is $\sin(x^3)$ odd?
Is it because you can substitute $u = x^3$ and then $\sin(x^3)=\sin(u)$
Then you can say $\sin(-u) = -\sin(u)$?
The part about this that seems weird to me is that I don't understand why rules for $\sin(x)$ should work for $\sin(f(x))$. If someone can please explain to me what is going on that would be really helpful. Thanks.
$\endgroup$ 16 Answers
$\begingroup$Pure definition:
A function $f$ is odd if $f(-x) = - f(x)$ for all $x$.
So $\sin((-x)^3) = \sin(-x^3) = - \sin (x^3)$.
So it is odd.
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One thing to note, is that if $f$ is odd, and $g$ is also odd then $f\circ g$ is odd because:
$f(g(-x) = f(-g(x)) = -f(g(x))$.
And $x^3$ is odd because $(-x)^3 = (-1*x)^3 = (-1)^3x^3 = (-1)*x^3 = -x^3$
An $\sin x$ is odd because .... well, it's a basic trigonometric identity that $\sin (-x) = -\sin (x)$. Just draw a picture. The $y$ value of a point of a circle at an angle will be the negative of the $y$ value of the angle in the opposite direction.
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Is it because you can substitute $u=x^3$ and then $sin(x^3)=sin(u)$ Then you can say $sin(−u)=−sin(u)$?
Only if you can also say that $-u = (-x)^3$.
$\endgroup$ 0 $\begingroup$Sometimes looking at a general case is useful.
A composition of odd functions $f$ and $g$ will be odd, since $$f \circ g (-x)=f(g(-x))=f(-g(x))=-f(g(x))=-f\circ g(x).$$
In your case, writing $\sin(x^3)$ as the composition $f \circ g$ of $f(x)=\sin(x)$ and $g(x)=x^3$ shows that $\sin(x^3)$ is odd.
However, if you were to compose two functions, one odd and one even (let $f$ be odd and $g$ even),
$$f\circ g(-x)=f(g(-x))=f(g(x))=f\circ g(x)$$ $$g \circ f (-x)=g(f(-x))=g(-f(x))=g(f(x))=g \circ f(x) $$
you'd get an even function.
$\endgroup$ $\begingroup$Both $\sin$ and $x\mapsto x^3$ are odd functions -- and it is easy to show that the composition of two odd functions is odd: $$ (f\circ g)(-x) = f(g(-x)) = f(-g(x)) = -f(g(x)) = -(f\circ g)(x) $$
$\endgroup$ $\begingroup$we have $$(-x)^3=-x^3$$ and $$\sin(-x^3)=-\sin(x^3)$$
$\endgroup$ $\begingroup$In more generality (which you suggest you want): if $f$ is odd then $$ f(-x) = -f(x). $$ When you use that fact inside $\sin(f(x)$ then the fact that $\sin$ is odd tells you the composition is odd.
What could you say about the composition if $f$ were even?
$\endgroup$ $\begingroup$Because $$\sin((-x)^3)=\sin(-x^3)=-\sin{x^3}$$
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