# Some peculiar Trigonometric Identities

Have you ever wondered what $\sin( \arccos(x) )$ is expressed just in x? Probably not, but it might actually sometimes be useful to know. The expression might for example turn up when you’ve done a trigonometric substitution in an integral. So I decided how to go about to show the trigonometric identities $\cos( \arcsin(x) ) = \sin(\arccos (x)) = \sqrt{1-x^2}$, $\cos(\arctan(x)) = \frac{1}{\sqrt{1+x^2}}$, $\sin(\arctan(x)) = \frac{x}{\sqrt{1+x^2}}$.

The idea is the use the well known identity $\sin ^2 x + \cos ^2 x = 1$ and try to get a relation between the basic trig functions and the arcus functions trough that. It goes like this $x = \sin( \arcsin (x)) = \sqrt{1-\cos ^2 (\arcsin (x))}$

using the well known identity. We see that we already have the wanted relation, so that the rest is algebra. By squaring both sides we get $x^2 = 1 - \cos^2 (\arcsin (x)) \Leftrightarrow \cos( \arcsin (x)) = \sqrt{1-x^2}$.

And there we have the first one. Similarly to obtain the next one we go $x = \cos (\arccos (x)) = \sqrt{1- \sin ^2 (\arccos (x) )}$, $\Leftrightarrow x^2 = 1-\sin^2 (\arccos (x)) \Leftrightarrow \sin( \arccos (x) ) = \sqrt{1-x^2}$

Obtaining the symmetric identity between sine and cosine. For tangent we go $x = \tan(\arctan(x)) = \frac{\sin(\arctan (x))}{\cos(\arctan(x))} = \frac{\sqrt{1-\cos ^2 (\arctan (x))}}{\cos (\arctan(x))} = \frac{\sin(\arctan(x))}{\sqrt{1 - \sin ^2 (\arctan(x))}}$

where we separately use expression 3 and 4 in the equal-chain above to obtain $x^2 = \frac{1 - \cos ^2 (\arctan (x) )}{\cos ^2 (\arctan (x))} = \frac{1}{\cos ^2 (\arctan (x))} - 1 \Leftrightarrow \cos (\arctan (x)) = \frac{1}{\sqrt{1 + x^2}}$

and then for nr.4 $x^2 = \frac{ \sin ^2 (\arctan (x))}{1 - \sin^2 (\arctan (x))} \Leftrightarrow x^2 = (x^2 + 1)\sin^2 (\arctan (x)) \Leftrightarrow \sin ( \arctan(x)) = \frac{x}{\sqrt{1 + x^2}}$.

And that was that. Later i will do $\arccos(\sin(x))$ etc.. but that requires complex numbers.

## 2 thoughts on “Some peculiar Trigonometric Identities”

1. Hehe takk for det Mathias 😉 Fortsett med det.
Du må bare si ifra hvis du har noen forslag til hva jeg kan skrive om.

2. Mathias on said:

Flott artikkel, Mikael! Jeg trenger alltid en liten trigonometri-repetisjon. Nå trenger jeg bare en liten gjennomgang av enhetssirkelen.

Det virker kanskje som om jeg leter etter oppdateringer på bloggen deres hele tiden, men jeg har faktisk en RSS-feed til e-postklienten min.

🙂