I watched a video where a problem involved recognizing that $\sin x$ is an odd function and $\sin^3 x$ is also odd. But the presenter didn't explain why $\sin^3 x$ is also odd. Why does the fact that the function is odd not change when it is cubed? Is there a rule where for every even power the odd function is even and for every odd power the odd function remains odd? What about for even functions?

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    $\begingroup$ If $f$ is odd and $g = f^3$, then $g(-x) = f(-x)^3 = (-f(x))^3 = (-1)^3f(x)^3 = -f(x)^3 = -g(x)$. You can play that game now with all variants. $\endgroup$ – amsmath Oct 6 '19 at 2:14
  • $\begingroup$ @amsmath Why are you allowed to put the the negative sign from inside the parentheses to outside? $\endgroup$ – user532874 Oct 6 '19 at 2:16
  • $\begingroup$ Because $(ab)^n = a^n\cdot b^n$. $\endgroup$ – amsmath Oct 6 '19 at 2:16
  • $\begingroup$ 532874, that's the definition of odd function, no? $\endgroup$ – Gerry Myerson Oct 6 '19 at 2:17
  • $\begingroup$ @user532874 A function $f$ is odd if $f(-x) = -f(x)$ for all $x$. $\endgroup$ – amsmath Oct 6 '19 at 2:18

Note that $$(-1)^{2k}=(+1)$$ and $$(-1)^{2k+1}=(-1)$$ $$(+1)^k = (+1)$$

Thus if a function is odd we have $$f^{2k}(-x) = (-1)^{2k}f^{2k}(x)=f^{2k}(x)$$ and $$f^{2k+1}(-x) = (-1)^{2k+1}f ^{2k+1}(x)=-f^{2k+1}(x)$$

Thus odd functions to the odd powers are odd and to the even powers are even.

Even functions to any power stay even.

  • $\begingroup$ I think you are missing a $2k$ exponent after "Thus if a function is odd we have" $\endgroup$ – user532874 Oct 6 '19 at 2:35
  • $\begingroup$ @user532874 is it fixed now? $\endgroup$ – Mohammad Riazi-Kermani Oct 6 '19 at 2:43
  • $\begingroup$ @MohammadRiazi-Kermani Check your calculations. You are missing minuses. $\endgroup$ – amsmath Oct 6 '19 at 2:59
  • $\begingroup$ @amsmath Thanks for informative comment. I should have been more careful. $\endgroup$ – Mohammad Riazi-Kermani Oct 6 '19 at 3:12

It is not only the cubing, it works for anyother odd function, it is always true that the composite of odd functions is an odd function.

In your case, $f(x)=\sin x$ is an odd function and $g(x)=x^3$ is also an odd function so $g\circ f(x)= g(f(x))=g(\sin x)=\sin^3x$

Another example is $\sin x$ with any odd power for example $x^{12345}$, since both are odd functions then $\sin^{12345}x$ is also an odd function.

Or $\frac{1}{\sin x}$ is also an odd function since it is the composite of $\sin x$ and $\frac{1}{x}$,and they are both odd functions.

Anyway you can just check if $f(-x)=-f(x)$ without worrying about compositions, $f(-x)=\sin^3(-x)=(\sin (-x))^3=(-\sin x)^3 =-(\sin^3x)=-f(x)$


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