Is Pythagoras the only relation to hold between cos\cos and sin\sin?

Question 1

Pythagoras says that $\cos^2 \theta + \mathrm{sin}^2\theta = 1$ for all real $\theta$ .

(Vague) Question. Is this the only relationship between the functions $\cos$ and $\sin$ ?

More precisely:

Let $\langle \cos,\sin\rangle$ denote the intersection of all subalgebras of the $\mathbb{R}$ -algebra of all functions $\mathbb{R} \rightarrow \mathbb{R}$ containing $\{\cos,\sin\}$ . (By default, all my algebras are unital, associative and commutative.) Let $A$ denote the $\mathbb{R}$ -algebra presented by the generators $\{c,s\}$ and the relation $c^2+s^2=1$ . There is a unique $\mathbb{R}$ -algebra homomorphism $\varphi : A \rightarrow \langle \cos,\sin\rangle$ given as follows. $\varphi(c) = \cos, \,\,\varphi(s) = \sin$

We know that $\varphi$ is surjective.

Question. Is $\varphi$ injective?

So consider $f \in A$ . Then $f = \sum_{i,j : \mathbb{N}}a_{ij}s^ic^i$ for certain choices of $a_{ij} : \mathbb{R}$ . Now suppose $\varphi(f)=0$ . We want to show that $f=0$ . Ideas, anyone?

Question 2

If I understand well, you are asking if the $\mathbb R$ -algebra generated by $\sin$ and $\cos$ , that is, $\mathbb R[\sin,\, \cos]$ is isomorphic to $\mathbb R[X,Y]/(X^2+Y^2-1)$ .

Consider the surjective morphism $\varphi:\mathbb R[X,Y]\to\mathbb R[\sin,\,\cos]$ defined by $\varphi(X)=\sin$ , $\varphi(Y)=\cos$ . Then $(X^2+Y^2-1)\subseteq\ker\varphi$ . Conversely, let $f\in\ker\varphi$ . We can write $f=(X^2+Y^2-1)g+r$ where $\deg_Xr\le 1$ , so $r=a(Y)+b(Y)X$ . Moreover, $a(\cos)+b(\cos)\sin=0$ . This means that $a(\cos x)+b(\cos x)\sin x=0$ for all $x\in\mathbb R$ . Changing $x$ by $-x$ we get $a(\cos x)-b(\cos x)\sin x=0$ for all $x\in\mathbb R$ , so $a(\cos x)=0$ for all $x\in\mathbb R$ , and $b(\cos x)=0$ for all $x\in\mathbb R$ , $x\ne k\pi$ . Since $a,b$ are polynomials this is enough to conclude $a=b=0$ , and therefore $r=0$ . We thus proved that $\ker\varphi=(X^2+Y^2-1)$ .

Is Pythagoras the only relation to hold between cos\cos and sin\sin?

Answer

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