Posted by: Dave Richeson | August 27, 2014

Tangent lines to the sine function with rational slope

Today I was wondering the following thing (I won’t bore you with how I ended up with this question):

Are there any rational values of {m} for which the line {y=mx} is tangent to the graph of {y=\sin x?}

Clearly the answer is yes: {m=1.} But my gut feeling was that this was the only such {m.} After some head scratching, I obtained the following proof.

Suppose they are tangent at {x=x_{0}}. Then the point of tangency is {(x_{0},mx_{0})=(x_{0},\sin(x_{0})).} Moreover, the slope of the tangent line is {\cos(x_{0})}. Thus, {x_{0}} must satisfy the two equations:

{\sin(x_{0})=mx_{0},} and

{\cos(x_{0})=m}.

Using a trig identity,

{1=\sin^{2}(x_{0})+\cos^{2}(x_{0})=m^{2}x_{0}^{2}+m^{2}}.

So,

{x_{0}=\sqrt{\frac{1-m^{2}}{m^{2}}}.}

Note that {x_{0}} is an algebraic number—it is the root of a polynomial with integer coefficients.

In 1882 Ferdinand von Lindemann famously proved that {e^{a}} is transcendental (that is, non-algebraic) for every non-zero algebraic number {a,} and a similar proof holds for the sine and cosine functions.

Thus, because {\cos(x_{0})=m,} and {m} is algebraic, it must be the case that {x_{0}=0}. In particular, {m=1.}

This concludes the proof.

 

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