Arccosine in terms of Arctangent

Theorem

$\arccos x = 2 \map \arctan {\sqrt {\dfrac {1 - x} {1 + x} } }$

where $x$ is a real number with $-1 < x \le 1$.

Proof

Let:

$\theta = \arccos x$

Then by the definition of arccosine:

$x = \cos \theta$

and:

$0 \le \theta < \pi$

Then:

\(\ds 2 \map \arctan {\sqrt {\frac {1 - x} {1 + x} } }\)

\(=\)

\(\ds 2 \map \arctan {\sqrt {\frac {1 - \cos \theta} {1 + \cos \theta} } }\)

\(\ds \)

\(=\)

\(\ds 2 \map \arctan {\sqrt {\frac {2 \sin^2 \frac \theta 2} {2 \cos^2 \frac \theta 2} } }\)

Double Angle Formula for Cosine: Corollary $1$, Double Angle Formula for Cosine: Corollary $2$

\(\ds \)

\(=\)

\(\ds 2 \map \arctan {\tan \frac \theta 2}\)

for $0 \le \theta < \dfrac \pi 2$ we have $\sin \theta \ge 0$ and $\cos \theta > 0$

\(\ds \)

\(=\)

\(\ds \theta\)

Definition of Real Arctangent

\(\ds \)

\(=\)

\(\ds \arccos x\)

$\blacksquare$

Also see

• Arcsine in terms of Arctangent