Justify that tan ⁡ 1 ∘ is an irrational number.

Question

Justify that   tan  ⁡      1    ∘   is an irrational number.

trajeronls · Accepted 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 + i s)^{180} = e^{π i} = - 1,

so

ℑ (c + i s)^{180} = 0.

That is

\sum_{k = 1}^{90} (- 1)^{k - 1} (\binom{180}{2 k - 1}) s^{2 k - 1} c^{180 - 2 k + 1} = 0.

Note that

s, c \neq 0

. If

t

is

\tan 1^{\circ}

, we have

\sum_{k = 1}^{90} (- 1)^{k - 1} (\binom{180}{2 k - 1}) t^{2 (k - 1)} = \frac{\sum_{k = 1}^{90} (- 1)^{k - 1} (\binom{180}{2 k - 1}) s^{2 k - 1} c^{180 - 2 k + 1}}{s c^{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 ∣ 180

, so we can write

t = \frac{n}{180}

for some integer

n

. Because

\frac{π}{180} < \tan \frac{π}{180} < \frac{\tan (π / 4)}{45} = \frac{1}{45},

where the first inequality is due to the inequality

\tan x > x

for all

x \in (0, π / 2)

, and the second inequality is true by convexity of

\tan

on

(0, π / 2)

. However, this means

π < n < 4,

but this is a contradiction (no integers lie strictly between

π

and

4

). So

t = \tan 1^{\circ}

cannot be rational.

Justify that tan ⁡