# Show this is contradictory: $\frac12 sin(2t) \equiv b_1cos(t)+…+b_kcos(kt)+c_1sin(t)+…c_lsin(lt)$ where $b_1,…,b_k,c_1,…,c_l\in \mathbb Z$

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!

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

• 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). – Tob Ernack Feb 2 '18 at 15:22
• 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. – Pascal's Wager Feb 2 '18 at 15:28
• 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 $i$. – Tob Ernack Feb 2 '18 at 15:37

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.