Editing Inverse trigonometric functions (section)

===Relationships among the inverse trigonometric functions===
[[Image:Arcsine Arccosine.svg|168px|right|thumb|The usual principal values of the arcsin(''x'') (red) and arccos(''x'') (blue) functions graphed on the cartesian plane.]]
[[Image:Arctangent Arccotangent.svg|294px|right|thumb|The usual principal values of the arctan(''x'') and arccot(''x'') functions graphed on the cartesian plane.]]
[[Image:Arcsecant Arccosecant.svg|294px|right|thumb|Principal values of the arcsec(''x'') and arccsc(''x'') functions graphed on the cartesian plane.]]

Complementary angles:
:<math>\begin{align}
\arccos(x) &= \frac{\pi}{2} - \arcsin(x) \\[0.5em]
\arccot(x) &= \frac{\pi}{2} - \arctan(x) \\[0.5em]
\arccsc(x) &= \frac{\pi}{2} - \arcsec(x)
\end{align}</math>

Negative arguments:
:<math>\begin{align}
\arcsin(-x) &= -\arcsin(x) \\
\arccsc(-x) &= -\arccsc(x) \\
\arccos(-x) &= \pi -\arccos(x) \\
\arcsec(-x) &= \pi -\arcsec(x) \\
\arctan(-x) &= -\arctan(x) \\
\arccot(-x) &= \pi -\arccot(x)
\end{align}</math>

Reciprocal arguments:
:<math>\begin{align}
\arcsin\left(\frac{1}{x}\right) &= \arccsc(x) & \\[0.3em]
\arccsc\left(\frac{1}{x}\right) &= \arcsin(x) & \\[0.3em]
\arccos\left(\frac{1}{x}\right) &= \arcsec(x) & \\[0.3em]
\arcsec\left(\frac{1}{x}\right) &= \arccos(x) & \\[0.3em]
\arctan\left(\frac{1}{x}\right) &= \arccot(x)       &= \frac{\pi}{2} - \arctan(x) \, , \text{ if } x > 0 \\[0.3em]
\arctan\left(\frac{1}{x}\right) &= \arccot(x) - \pi &= -\frac{\pi}{2} - \arctan(x) \, , \text{ if } x < 0 \\[0.3em]
\arccot\left(\frac{1}{x}\right) &= \arctan(x)       &= \frac{\pi}{2} - \arccot(x) \, , \text{ if } x > 0 \\[0.3em]
\arccot\left(\frac{1}{x}\right) &= \arctan(x) + \pi &= \frac{3\pi}{2} - \arccot(x) \, , \text{ if } x < 0 
\end{align}</math>
The identities above can be used with (and derived from) the fact that <math>\sin</math> and <math>\csc</math> are [[Multiplicative inverse|reciprocals]] (i.e. <math>\csc = \tfrac1{\sin}</math>), as are <math>\cos</math> and <math>\sec,</math> and <math>\tan</math> and <math>\cot.</math>

Useful identities if one only has a fragment of a sine table:
:<math>\begin{align}
\arcsin(x) &= \frac{1}{2}\arccos\left(1-2x^2\right) \, , \text{ if } 0 \leq x \leq 1 \\
\arcsin(x) &= \arctan\left(\frac{x}{\sqrt{1 - x^2}}\right) \\
\arccos(x) &= \frac{1}{2}\arccos\left(2x^2-1\right) \, , \text{ if } 0 \leq x \leq 1 \\
\arccos(x) &= \arctan\left(\frac{\sqrt{1 - x^2}}{x}\right) \\
\arccos(x) &= \arcsin\left(\sqrt{1 - x^2}\right) \, , \text{ if } 0 \leq x \leq 1 \text{ , from which you get } \\
\arccos    &\left(\frac{1-x^2}{1 + x^2}\right) = \arcsin \left (\frac{2x}{1 + x^2}\right) \, , \text{ if } 0 \leq x \leq 1 \\
\arcsin    &\left(\sqrt{1 - x^2}\right) =\frac{\pi}{2}-\sgn(x)\arcsin(x) \\
\arctan(x) &= \arcsin\left(\frac{x}{\sqrt{1 + x^2}}\right) \\
\arccot(x) &= \arccos\left(\frac{x}{\sqrt{1 + x^2}}\right)
\end{align}</math>

Whenever the square root of a complex number is used here, we choose the root with the positive real part (or positive imaginary part if the square was negative real).

A useful form that follows directly from the table above is

:<math>\arctan(x) = \arccos\left(\sqrt{\frac{1}{1+x^2}}\right)\, , \text{ if } x\geq 0 </math>.

It is obtained by recognizing that <math>\cos\left(\arctan\left(x\right)\right) = \sqrt{\frac{1}{1+x^2}} = \cos\left(\arccos\left(\sqrt{\frac{1}{1+x^2}}\right)\right)</math>.

From the [[tangent half-angle formula|half-angle formula]], <math>\tan\left(\tfrac{\theta}{2}\right) = \tfrac{\sin(\theta)}{1 + \cos(\theta)}</math>, we get:
:<math>\begin{align}
\arcsin(x) &= 2 \arctan\left(\frac{x}{1 + \sqrt{1 - x^2}}\right) \\[0.5em]
\arccos(x) &= 2 \arctan\left(\frac{\sqrt{1 - x^2}}{1 + x}\right) \, , \text{ if } -1 < x \leq  1 \\[0.5em]
\arctan(x) &= 2 \arctan\left(\frac{x}{1 + \sqrt{1 + x^2}}\right)
\end{align}</math>