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This question already has an answer here:

Prove that $\{ \sin(x), \sin^2(x), \sin^3(x)\}$ is linearly independent in $F(\Bbb R)$.

I tried plugging in $\left\{ 0, \frac {\pi} {2}, \pi, \frac {3\pi}{2}\right\}$ but that doesn't work. How should I prove this?

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marked as duplicate by Potato, Marc van Leeuwen, Marso, Giuseppe Negro, nbubis Oct 17 '13 at 15:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ what is $F(R)$ by the way? $\endgroup$ – Marso Oct 17 '13 at 15:20
  • $\begingroup$ I guess it is $\Bbb R^\Bbb R=\{\, f:\Bbb R\to\Bbb R\mid \,\}$ $\endgroup$ – Marc van Leeuwen Oct 17 '13 at 15:22
  • $\begingroup$ Try evaluating in $k\frac\pi6$ for $k=0,1,\ldots$, you are bound to find $3$ independent $3$-tuples soon. $\endgroup$ – Marc van Leeuwen Oct 17 '13 at 15:24
  • $\begingroup$ But for a more insightful answer, see the "duplicate" (actually much more general) question. $\endgroup$ – Marc van Leeuwen Oct 17 '13 at 15:30
  • $\begingroup$ Thanks, the other post makes sense. $\endgroup$ – user95087 Oct 17 '13 at 15:32
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Assume: $$\sin(x)^3=A\sin(x)^2+B\sin(x)$$ then from $x=\dfrac{\pi}{2}$ and $x=\dfrac{3\pi}{2}$: $$1=A+B,\quad-1=A-B\quad\Rightarrow\quad A=0,\quad B=1$$ which implies: $$\sin(x)^3=\sin(x)\quad\Rightarrow\quad \sin(x)^2=1\quad \forall x \ne n\pi$$

which is false.

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