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Let a trigonometric polynomial be a function defined as $$a_0 + \sum_{k=0}^n \left[a_k \cos(kx) + b_k \sin(kx)\right].$$

Clearly the set of all such functions forms a vector space. Is it true that they form also an algebra?

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  • $\begingroup$ Yes. You can rewrite products $\cos(nx)\sin(mx)$ using trigonometric identities obtaining expressions containing only $cos$ or $\sin$ but not products of these. $\endgroup$ – b00n heT Mar 9 '17 at 13:29
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    $\begingroup$ I'll give you a hint. Look instead at the space of sums $\sum_{k=0}^n e^{i k x}$., i.e. $\mathbb{C}[e^{i x}]$. This is clearly an algebra. $\endgroup$ – Daniel Miller Mar 9 '17 at 13:29
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By linearity, it suffices to check that the products $$\cos(kx) \cos(lx), \cos(kx) \sin(lx), \sin(kx) \sin(lx)$$ can all be written in this form, but this follows immediately from the usual product-to-sum identities and the symmetry properties of $\sin$ and $\cos$. For example, $$\cos(kx) \cos(lx) = \tfrac{1}{2}\left(\cos[(k + l) x] + \cos [(k - l) x]\right) .$$

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Yes. Use

$\sin (x+y)=\sin x\cos y+\cos x\sin y$,

$\cos (x+y)=\cos x\cos y-\sin x\sin y$

for a proof.

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