Is it the right way to determine the coefficients? I am having trouble with understanding the logical completeness of a solution of exercise on my textbook (Linear Algebra Done Right).
 we have $Tp=(bp(1)p(2),c\sin p(0))$ &$p\in P(R)$. Prove that if T is a linear map, then $b=c=0$. It answers:
consider $f(x)=\pi/2$ and $g(x)=\pi/2$, then they both are polynomials. So examining the property of the linear map: $T(x+y) = T(x)+T(y)$. we have $T(f+g) = (b\pi^2,c\sin(\pi))$ and $T(f)+T(g) = (b\pi^2/4,csin(\pi/2))+(b\pi^2/4,csin(\pi/2)) = (b\pi^2/2,2csin(\pi/2))$
Thus b = c = 0.
My logic is that since the $f$ and $g$ are only two instances of the polynomials, using which to determine the value of $b$ and $c$ can be dangerous. Because we did not try polynomials with degree 1,2,3,4..... I think the more appropriate way to prove is using the property of polynomials but it would make this way more complex.
I am curious about your opinion. I am really confused about whether I am being critical or I am just lack of experience in mathematic proof. Thanks for your time

A proof is either correct or not.  There's no issue of danger. 
I dare say mathematics is not dangerous.   Nobody ever cut himself on an integral (though metaphorically, yes, to be sure). 
One might come up with a simpler, more elegant proof,  say;   but I don't think it's very likely in this case.  
Mathematicians tend to prefer simpler proofs, for fairly obvious reasons.    Btw, the polynomials used are degree zero, that is,constant polynomials.   They are in a sense the simplest types of polynomials. 
So, in sum, I think it's clearly inexperience at this point.  But don't worry.  Practice makes perfect.  
