Justify that $\tan 1^\circ$ is an irrational number.
I don't know how to go about it
Justify that $\tan 1^\circ$ is an irrational number
irrational-numbersnumber theorytrigonometry
Related Solutions
First, the way many scientific calculators work is to calculate more digits than they show, and then they round the value before displaying it. This happens even if you calculate something like $1/7 \times 7$. The calculator may believe the result is slightly lower than $1$, but the rounded number is $1$. You can test how many digits of precision your calculator uses by multiplying by $10^n$ and then subtracting the integer part. This will often reveal a few more digits.
Second, those are irrational numbers. Proving that takes some number theory. Let $\xi = \cos 1^\circ + i \sin 1^\circ$, a $360$th root of unity. $\xi$ is conjugate to $\xi^n$ for each $n$ coprime to $360$ including $\xi^{49} = \cos 49^\circ + i \sin 49^\circ$. The minimal polynomial of $\xi$ and $\xi^{49}$ has degree $\phi(360)=96$. If $\cos^2 49^\circ$ were rational, then $(\xi^{49} + \xi^{-49})^2$ would be rational, which would mean that $\xi^{49}$ satisfies a polynomial with rational coefficients of degree $4 \lt 96$. Similarly, $\sin^2 49^\circ = (\xi^{49} - \xi^{-49})^2/4$ is not rational.
A proof without words (but it uses some geometry). Is that OK?
Best Answer
Here is a more algebraic approach not given in Minus One-Twelfth's link. Let $c$ and $s$ denote $\cos 1^\circ$ and $\sin 1^\circ$, respectively. We have $$(c+is)^{180}=e^{\pi i}=-1,$$ so $$\Im(c+is)^{180}=0.$$ That is $$\sum_{k=1}^{90}(-1)^{k-1}\binom{180}{2k-1}s^{2k-1}c^{180-2k+1}=0.$$ Note that $s,c\ne 0$. If $t$ is $\tan 1^\circ$, we have $$\sum_{k=1}^{90}(-1)^{k-1}\binom{180}{2k-1}t^{2(k-1)}=\frac{\sum_{k=1}^{90}(-1)^{k-1}\binom{180}{2k-1}s^{2k-1}c^{180-2k+1}}{sc^{179}}=0.$$
If $t$ is a rational number, then $t=\frac{p}{q}$ for some positive integers $p,q$ such that $\gcd(p,q)=1$. By the rational root theorem, $p,q\mid180$, so we can write $t=\frac{n}{180}$ for some integer $n$. Because $$\frac{\pi}{180}<\tan\frac{\pi}{180}<\frac{\tan(\pi/4)}{45}=\frac{1}{45},$$ where the first inequality is due to the inequality $\tan x>x$ for all $x\in(0,\pi/2)$, and the second inequality is true by convexity of $\tan$ on $(0,\pi/2)$. However, this means $$\pi<n<4,$$ but this is a contradiction (no integers lie strictly between $\pi$ and $4$). So $t=\tan 1^\circ$ cannot be rational.