We have an answer to "Testing with my calculator in degree mode, I..."

Lesnaoq73

Answered question

2022-04-02

Testing with my calculator in degree mode, I have found the following to be true:

\tan (90 - \frac{1}{10^{n}}) \approx \frac{180}{π} \times 10^{n}, n \in N

enriuadaziaa

Beginner2022-04-03Added 7 answers

First, note that by standard trigonometric identities,

\tan (90^{\circ} - \frac{1^{\circ}}{10^{n}}) = \frac{\sin (90^{\circ} - \frac{1^{\circ}}{10^{n}})}{\cos (90^{\circ} - \frac{1^{\circ}}{10^{n}})} = \frac{\cos (\frac{1^{\circ}}{10^{n}})}{\sin (\frac{1^{\circ}}{10^{n}})}

= \frac{\cos (\frac{1}{10^{n}} \times \frac{π}{180})}{\sin (\frac{1}{10^{n}} \times \frac{π}{180})}

Also, the power series for

\sin

and

\cos

are

\sin (x) = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \dots

\cos (x) = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \dots

so that, for small x, we have

\sin (x) \approx x

\cos (x) \approx 1

and therefore

\tan (90^{\circ} - \frac{1^{\circ}}{10^{n}}) \approx \frac{1}{\frac{1}{10^{n}} \times \frac{π}{180}} = \frac{180}{π} \times 10^{n}

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