Prove that sinx is not a rational function using the fact that it is not of the form p(x)/q(x) where p and q are polynomials. Then, by using the above proof, prove that sinx does not satisfy a "quadratic equation" of the form: ( sin ⁡ x ) 2 f 2 ( x ) + ( sin ⁡ x ) f 1 ( x ) + f 0 ( x ) = 0 ,where f 0 , f 1 , f 2 are rational functions.I know that a rational function cannot be zero at infinitely many points unless it is 0 everywhere, but how does one use this information to formulate p(x)/q(x) argument? If anybody could please help.

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Prove that sinx is not a rational function using the fact that it is not of the form p(x)/q(x) where p and q are polynomials. Then, by using the above proof, prove that sinx does not satisfy a &quot;quadratic equation&quot; of the form:  (  sin  ⁡  x      )    2        f    2    (  x  )  +  (  sin  ⁡  x  )      f    1    (  x  )  +      f    0    (  x  )  =  0  ,where       f    0    ,      f    1    ,      f    2   are rational functions.I know that a rational function cannot be zero at infinitely many points unless it is 0 everywhere, but how does one use this information to formulate p(x)/q(x) argument? If anybody could please help.

Bridget Acevedo · Accepted Answer

Step 1In fact, the curve   y  =  sin  ⁡  x is not equal to (or contained in) any algebraic curve   f  (  x  ,  y  )  =  0, where f(x,y) is any polynomial in two variables with real coefficients.Step 2Indeed, the algebraic curve   f  (  x  ,  y  )  =  0 meets the x-axis in only finitely many points, but   y  =  sin  ⁡  x meets the x-axis in infinitely many points.

Gisselle Hodges · Accepted Answer

Step 1For the second part, by multiplying the equation with a suitable polynomial, we may assume that       f    0  ,       f    1  ,       f    2   are polynomials. We have:  ∀  n  ∈      Z    :      sin    2    ⁡  (  π  n  )      f    2    (  π  n  )  +  sin  ⁡  (  π  n  )      f    1    (  π  n  )  +      f    0    (  π  n  )  =  0Therefore:      f    0    (  π  n  )  =  0Step 2Since a non-constant polynomial function can only have a finite number of zeros,       f    0   must be 0 everywhere. The equation reduces to:      sin    2    ⁡  (  x  )      f    2    (  x  )  +  sin  ⁡  (  x  )      f    1    (  x  )  =  0Assume   x  ≠  π  n and factor out sin(x) to get:  ∀  x  ≠  π  n  :  sin  ⁡  (  x  )      f    2    (  x  )  +      f    1    (  x  )  =  0Which means that:  sin  ⁡  (  x  )  =  −                    f        1            (      x      )                      f        2            (      x      )      But   −      f    1        /        f    2   is also a rational function. By the continuity of sin and   −      f    1        /        f    2  , equality must hold everywhere, which contradicts the fact that sin is not a rational function.

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