Inverse trigonometric functions

The inverse trigonometric functions invert trigonometric functions on branches where they are one-to-one. Standard real ranges are $[-\pi /2,\pi /2]$ for $\arcsin$, $[0,\pi ]$ for $\arccos$, and $(-\pi /2,\pi /2)$ for $\arctan$. Thus $\arcsin (1/2)=\pi /6$ but $\arcsin (\sin (5\pi /6))=\pi /6$: the inverse returns the selected branch. The notation ${\sin }^{-1}x$ usually means $\arcsin x$, not $1/\sin x$. Implicit differentiation gives $(\arcsin x{)}^{\prime }=(1-x^2{)}^{-1/2}$ for $|x|<1$.

For real $-1\le x\le 1$, the principal arccosine is the unique angle $\theta \in [0,\pi ]$ satisfying $\cos \theta =x$. Restricting to this interval makes the cosine one-to-one. Differentiating the identity on $-1<x<1$ gives $d(\arccos x)/dx=-1/\sqrt{1-x^2}$. Other solutions of $\cos \theta =x$ are $\theta =\pm \arccos x+2\pi n$, with integer $n$; an initial condition can select one of these branches.

Set $u=e^{iw}$ in $\cos w=z$. The resulting quadratic equation is $u^2-2zu+1=0$, with roots $z\pm i\sqrt{1-z^2}$. Their product is one, so taking a complex logarithm never encounters zero. Solving the exponential equation gives the displayed full inverse relation, with integer $n$. The principal complex logarithm and principal square root do not make both signs identical to the principal inverse cosine: at $z=0$ they yield $\pm \pi /2$, while principal $\arccos 0=\pi /2$. A single analytic branch requires compatible choices of sign and branches.

The inverse of the tangent restricted to $(-\pi /2,\pi /2)$, taking real values in that interval. Its derivative is $(1+x^2{)}^{-1}$. A formula involving an inverse tangent of a ratio may need a continuous branch of a multivalued function to describe a smooth field through a pole of that ratio.

For real $A,B$ with $AB<1$, the principal inverse tangent satisfiesTo prove it, multiply the complex numbers $1+iA$ and $1+iB$. Their product is $(1-AB)+i(A+B)$, whose positive real part places its complex argument in $(-\pi /2,\pi /2)$. The individual complex arguments are $\arctan A$ and $\arctan B$, and their sum is in $(-\pi ,\pi )$. Multiplication adds these complex arguments, so the positive real part forces the sum into the same principal range and gives the displayed formula.

If $AB>1$, both $A,B$ have the same sign. Add $\pi$ to the right-hand side when $A,B>0$, and subtract $\pi$ when $A,B<0$. If $AB=1$, the sum is $\pi /2$ or $-\pi /2$ according to the sign of $A+B$. These cases record the branch information that can be lost by applying the tangent addition formula alone.

The principal inverse sine is the inverse of the sine restricted to $[-\pi /2,\pi /2]$:The range restriction is essential because the sine is not injective on all of the real numbers. In the interior $-1<x<1$, implicit differentiation givesThe positive square root appears because the cosine is positive in $(-\pi /2,\pi /2)$. For example, $\arcsin (\sin (3\pi /4))=\pi /4$, not $3\pi /4$. All real solutions of $\sin \theta =x$ are $\theta =\arcsin x+2k\pi$ or $\theta =\pi -\arcsin x+2k\pi$, with $k$ an integer.