Question for Trig Gurus

[Thomas Klemm]

(12-01-2014 07:49 PM)Namir Wrote:  is there a secret formula to calculate or estimate the inverse trig functions from one of the trig functions??

You could use these trigonometric identities to calculate \(\tan \frac{x}{4}\) from \(\cos x\):
\[
\begin{align}
\cos \frac{x}{2} &= \sqrt{\frac{1+\cos x}{2}} \\
\tan \frac{x}{2} &= \sqrt{\frac{1-\cos x}{1+\cos x}} \\
\end{align}
\]

From this you could use the Taylor series of \(\arctan x \approx x-\frac{x^3}{3}+\frac{x^5}{5}+\dots\) to estimate \(x\). Thus we can calculate \(\arccos x\).

Example:
\(\cos x = 0.5\)
\(\cos \frac{x}{2} = \sqrt{\frac{1.5}{2}} = \sqrt{0.75} \approx 0.866025\)
\(\tan \frac{x}{4} = \sqrt{\frac{1-0.866025}{1+0.866025}} = \sqrt{\frac{0.133975}{1.866025}} \approx 0.267949\)

We slightly rewrite the Taylor series using Horner's method: \(((\frac{x^2}{5}-\frac{1}{3}) x^2+1) x\).
Just keep in mind that we already calculated \(x^2 = \frac{0.133975}{1.866025} \approx 0.071797\).

This is how you could calculate it:
0.071797
÷ 5
− 3 [1/x]
× 0.071797
+ 1
× 0.267949
=
0.261813

The correct value is:
0.261799


We can use these two identities to calculate \(\arcsin x\) or \(\arctan x\):
\[
\begin{align}
\cos x &= \sqrt{1-\sin^2 x} \\
\cos x &= \frac{1}{\sqrt{1+\tan^2 x}} \\
\end{align}
\]

Example:
\(\tan x = \frac{1}{\sqrt{3}}\)
\(\cos x = \frac{1}{\sqrt{1+\frac{1}{3}}} \approx 0.866025\)
\(\cos \frac{x}{2} = \sqrt{\frac{1.866025}{2}} \approx 0.965926\)
\(\tan \frac{x}{4} = \sqrt{\frac{1-0.965926}{1+0.965926}} = \sqrt{\frac{0.034074}{1.965926}} \approx 0.131652\)

Here we end up with:
0.523598

The correct value is:
0.523599

Kind regards
Thomas


PS: From John Wolff's Web Museum:

Quote:

• Trig functions operate in degrees, but there are no inverses.
• Other keys give reciprocals, square roots, pi, and degree-minute-second conversions.
• There are no memory functions.
• The calculator does not use scientific notation, but operates with a floating decimal point within the 8-digit range of the display.
• The trig functions return only five figures, and take about two seconds.
  

Thus we're not too bad with our results.