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

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

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

More precisely:

Let cos,sin denote the intersection of all subalgebras of the R-algebra of all functions RR 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:Acos,sin given as follows. φ(c)=cos,φ(s)=sin

We know that φ is surjective.

Question. Is φ injective?

So consider fA. 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+Y21).

Consider the surjective morphism φ:R[X,Y]R[sin,cos] defined by φ(X)=sin, φ(Y)=cos. Then (X2+Y21)kerφ. Conversely, let fkerφ. We can write f=(X2+Y21)g+r where degXr1, 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 xR. Changing x by x we get a(cosx)b(cosx)sinx=0 for all xR, so a(cosx)=0 for all xR, and b(cosx)=0 for all xR, xkπ. Since a,b are polynomials this is enough to conclude a=b=0, and therefore r=0. We thus proved that kerφ=(X2+Y21).

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