Question for Trig Gurus

[Thomas Klemm]

(07-29-2022 10:19 PM)ttw Wrote:  There are also some logarithmic forms and a nice continued fraction for arctan(x).

From Inverse trigonometric functions:Logarithmic_forms:

\(
\begin{align}
\arctan(z)=-{\frac {i}{2}}\ln \left({\frac {i-z}{i+z}}\right)
\end{align}
\)

These formulas can't be used as the calculator lacks complex numbers.

From Inverse trigonometric functions:Continued fractions for arctangent:

\(
\begin{align}
\arctan(z)={\frac {z}{1+{\cfrac {(1z)^{2}}{3+{\cfrac {(2z)^{2}}{5+{\cfrac {(3z)^{2}}{7+{\cfrac {(4z)^{2}}{9+\ddots }}}}}}}}}}
\end{align}
\)

To get five correct figures we have to use 4 terms (i.e. the formula above without the \(\cdots\)).

Example

For x = 0.6 we get:

16 × 0.36 ÷ 9
+ 7 ÷ 9 ÷ 0.36 = 1/x
+ 5 ÷ 4 ÷ 0.36 = 1/x
+ 3 ÷ 0.36 = 1/x
+ 1 ÷ 0.6 = 1/x

0.5404217 (0.5404195)

And if we want the result in degrees we add:

× 180 ÷ \(\pi\) =

30.963884 (30.963757)

It can get a bit more tedious to enter \(x^2\) four times if \(x\) is an arbitrary number.

Example

Let's assume we want to calculate \(\cos^{-1}(x)\) for \(x=0.7\).
As before we can use:

\(
\begin{align}
\tan \frac{x}{2} &= \sqrt{\frac{1-\cos x}{1+\cos x}} \\
\end{align}
\)


0.3 ÷ 1.7 =
0.1764706

16 × 0.1764706 ÷ 9
+ 7 ÷ 9 ÷ 0.1764706 = 1/x
+ 5 ÷ 4 ÷ 0.1764706 = 1/x
+ 3 ÷ 0.1764706 = 1/x
+ 1 ÷ 0.1764706 √ = 1/x
× 2 =

0.7953990 (0.7953988)

And again if we want the result in degrees we add:

× 180 ÷ \(\pi\) =

45.573005 (45.572996)

---

As I don't have this calculator the results were just rounded to 7 places.
Therefore the numbers may vary slightly.