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.