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To give some background, I am trying to solve this Determine if this is a subring. problem. I have determined that, if I can show that the statement in yellow is false, i.e. contradictory, my proof should be complete.

$\frac12 sin(2t) \equiv b_1cos(t)+...+b_kcos(kt)+c_1sin(t)+...c_lsin(lt)$

where $b_1,...,b_k,c_1,...,c_l\in \mathbb Z$

My thought is that a contradiction might be obtainaned by either taking integrals or derivatives of both sides, but so far I have not been very successful. Please help me!

Please keep the following in mind when answering:

  1. I do not know Fourier Analysis

  2. I do not know Complex Analysis

  3. I am still a beginner at Ring Theory

  4. I am looking for a solution that is easy (for me) to understand

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    $\begingroup$ The set $\{1, \cos(t), \cos(2t), \ldots, \cos(kt), \sin(t), \sin(2t), \ldots, \sin(lt)\}$ is linearly independent so the only way for this equation to hold is if $c_2 = 1/2$, a contradiction since $c_2 \in \mathbb{Z}$. To show linear independence, you can show orthogonality (this would be the first insight of the Fourier analysis idea). $\endgroup$ – Tob Ernack Feb 2 '18 at 15:22
  • $\begingroup$ If I can prove that the set is linearly independent, I should be done. But it is something that needs to be proven rather than assumed. $\endgroup$ – Pascal's Wager Feb 2 '18 at 15:28
  • $\begingroup$ To prove linear independence, you can check that the set is orthogonal, where we use the inner product $\langle f, g \rangle = \int\limits_{-\pi}^{\pi}f(t)g(t)dt$. See for example this. Once you've shown orthogonality, linear independence is easy because if $0 = c_1f_1(t) + \ldots + c_nf_n(t)$ then $0 = \langle f_i, c_1f_1 + \ldots + c_nf_n \rangle = c_i\langle f_i, f_i \rangle$. But $\langle f_i, f_i \rangle \neq 0$ so $c_i = 0$ for all $i$. $\endgroup$ – Tob Ernack Feb 2 '18 at 15:37
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Hint: Suppose this is possible. Set $t=\frac\pi4$ and show this implies$\frac12$ is an integer combination of $1, \frac1{\sqrt2}$, which leads to a contradiction.

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    $\begingroup$ Thank you very much! The simplest answer is the best answer, as far as I'm concerned, and yours was about as simple as it gets! I solved the problem. $\endgroup$ – Pascal's Wager Feb 2 '18 at 15:38
  • $\begingroup$ Thanks- you may need the orthogonality argument mentioned in comments if the sums need not be finite though! In that case, just take the dot product on both sides with the LHS to get the contradiction. $\endgroup$ – Macavity Feb 2 '18 at 15:45

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