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?

  • 4
    $\begingroup$ If there were another relation, you'd have 2 equations in 2 unknowns, and you could probably (in theory) solve for the two functions. $\endgroup$ Sep 19, 2014 at 7:20
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    $\begingroup$ This is the first time I've heard $\sin^2+\cos^2=1$ being called "Pythagoras". But it makes sense, yes. By the way, the question of relations between $\sin$ and $\cos$ becomes more interesting if you consider composition, not just pointwise addition and pointwise multiplication. $\endgroup$ Sep 19, 2014 at 14:08
  • $\begingroup$ Most pairs of functions do not satisfy $\sin(x+y) = \sin x \cos y + \cos x \sin y;$ consider $\sin x^2 $ and $\cos x^2.$ $\endgroup$
    – Will Jagy
    Sep 19, 2014 at 15:48

2 Answers 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 $\ker\varphi=(X^2+Y^2-1)$.


It suffices to show that all functions of the form $x \mapsto P_0(\cos x)+\sin x P_1(\cos x)$ are nonzero, whenever $P_0$ and $P_1$ are polynomials not both equal to zero.

Suppose we have some relation of the form

$$P_0(\cos x)+\sin x P_1(\cos x)\equiv0 \, .$$

Then $$ \sin x= -\frac{P_0(\cos x)}{P_1(\cos x)} $$ whenever $P_1(\cos x) \neq 0$. If $P_1$ is a nonzero polynomial, this implies that $\sin x$ is even, which is absurd; thus $P_1 \equiv 0$.

But that means our relation reduces to $P_0(\cos x) \equiv 0$; as $\cos$ assumes infinitely many values, it follows that $P_0 \equiv 0$.


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