Pythagoras says that cos2θ+sin2θ=1 for all real θ.

(Vague) Question.Is this theonlyrelationship between the functions cos and sin?More precisely:

Let ⟨cos,sin⟩ denote the intersection of all subalgebras of the R-algebra of all functions R→R containing {cos,sin}. (By default, all my algebras are unital, associative and commutative.) Let A denote the R-algebra presented by the generators {c,s} and the relation c2+s2=1. There is a unique R-algebra homomorphism φ:A→⟨cos,sin⟩ given as follows. φ(c)=cos,φ(s)=sin

We know that φ is surjective.

Question.Is φ injective?So consider f∈A. Then f=∑i,j:Naijsici for certain choices of aij:R. Now suppose φ(f)=0. We want to show that f=0. Ideas, anyone?

**Answer**

If I understand well, you are asking if the R-algebra generated by sin and cos, that is, R[sin,cos] is isomorphic to R[X,Y]/(X2+Y2−1).

Consider the surjective morphism φ:R[X,Y]→R[sin,cos] defined by φ(X)=sin, φ(Y)=cos. Then (X2+Y2−1)⊆kerφ. Conversely, let f∈kerφ. We can write f=(X2+Y2−1)g+r where degXr≤1, so r=a(Y)+b(Y)X. Moreover, a(cos)+b(cos)sin=0. This means that a(cosx)+b(cosx)sinx=0 for all x∈R. Changing x by −x we get a(cosx)−b(cosx)sinx=0 for all x∈R, so a(cosx)=0 for all x∈R, and b(cosx)=0 for all x∈R, x≠kπ. Since a,b are polynomials this is enough to conclude a=b=0, and therefore r=0. We thus proved that kerφ=(X2+Y2−1).

**Attribution***Source : Link , Question Author : goblin GONE , Answer Author :
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