Inverse trigonometric functions

In order to use inverse trigonometric functions, we need to understand that an inverse trigonometric function “undoes” what the original trigonometric function “does,” as is the case with any other function and its inverse. In other words, the domain of the inverse function is the range of the original function, and vice versa, as summarized in .

For example, if \(f(x)=\sin \ x,\) then we would write \({f}^{-1}(x)={\sin }^{-1}x.\) Be aware that \({\sin }^{-1}x\) does not mean \(\frac{1}{\sin x}.\) The following examples illustrate the inverse trigonometric functions:

• Since \(\text{sin}(\frac{\pi }{6})=\frac{1}{2},\) then \(\frac{\pi }{6}={\text{sin}}^{-1}(\frac{1}{2}).\)
• Since \(\cos (\pi )=-1,\) then \(\pi ={\cos }^{-1}(-1).\)
• Since \(\tan (\frac{\pi }{4})=1,\) then \(\frac{\pi }{4}={\tan }^{-1}(1).\)

In previous sections, we evaluated the trigonometric functions at various angles, but at times we need to know what angle would yield a specific sine, cosine, or tangent value. For this, we need inverse functions. Recall that, for a one-to-one function, if \(f(a)=b,\) then an inverse function would satisfy \({f}^{-1}(b)=a.\)

Bear in mind that the sine, cosine, and tangent functions are not one-to-one functions. The graph of each function would fail the horizontal line test. In fact, no periodic function can be one-to-one because each output in its range corresponds to at least one input in every period, and there are an infinite number of periods. As with other functions that are not one-to-one, we will need to restrict the domain of each function to yield a new function that is one-to-one. We choose a domain for each function that includes the number 0. shows the graph of the sine function limited to \([-\frac{\pi }{2},\frac{\pi }{2}]\) and the graph of the cosine function limited to \([0,\pi ].\)

shows the graph of the tangent function limited to \((-\frac{\pi }{2},\frac{\pi }{2}).\)

These conventional choices for the restricted domain are somewhat arbitrary, but they have important, helpful characteristics. Each domain includes the origin and some positive values, and most importantly, each results in a one-to-one function that is invertible. The conventional choice for the restricted domain of the tangent function also has the useful property that it extends from one vertical asymptote to the next instead of being divided into two parts by an asymptote.

On these restricted domains, we can define the inverse trigonometric functions.

Condensed: the full section is in OpenStax Algebra and Trigonometry 2e.

Now that we can identify inverse functions, we will learn to evaluate them. For most values in their domains, we must evaluate the inverse trigonometric functions by using a calculator, interpolating from a table, or using some other numerical technique. Just as we did with the original trigonometric functions, we can give exact values for the inverse functions when we are using the special angles, specifically \(\frac{\pi }{6}\) (30°), \(\frac{\pi }{4}\) (45°), and \(\frac{\pi }{3}\) (60°), and their reflections into other quadrants.

Condensed: the full section is in OpenStax Algebra and Trigonometry 2e.

To evaluate inverse trigonometric functions that do not involve the special angles discussed previously, we will need to use a calculator or other type of technology. Most scientific calculators and calculator-emulating applications have specific keys or buttons for the inverse sine, cosine, and tangent functions. These may be labeled, for example, SIN \({\ }^{-1}\), ARCSIN, or ASIN.

In the previous chapter, we worked with trigonometry on a right triangle to solve for the sides of a triangle given one side and an additional angle. Using the inverse trigonometric functions, we can solve for the angles of a right triangle given two sides, and we can use a calculator to find the values to several decimal places.

In these examples and exercises, the answers will be interpreted as angles and we will use \(\theta\) as the independent variable. The value displayed on the calculator may be in degrees or radians, so be sure to set the mode appropriate to the application.

Condensed: the full section is in OpenStax Algebra and Trigonometry 2e.

There are times when we need to compose a trigonometric function with an inverse trigonometric function. In these cases, we can usually find exact values for the resulting expressions without resorting to a calculator. Even when the input to the composite function is a variable or an expression, we can often find an expression for the output. To help sort out different cases, let \(f(x)\) and \(g(x)\) be two different trigonometric functions belonging to the set \(\{\sin (x),\cos (x),\tan (x)\}\) and let \({f}^{-1}(y)\) and \({g}^{-1}(y)\) be their inverses.

Condensed: the full section is in OpenStax Algebra and Trigonometry 2e.

• An inverse function is one that “undoes” another function. The domain of an inverse function is the range of the original function and the range of an inverse function is the domain of the original function.
• Because the trigonometric functions are not one-to-one on their natural domains, inverse trigonometric functions are defined for restricted domains.
• For any trigonometric function \(f(x),\) if \(x={f}^{-1}(y),\) then \(f(x)=y.\) However, \(f(x)=y\) only implies \(x={f}^{-1}(y)\) if \(x\) is in the restricted domain of \(f.\)
• Special angles are the outputs of inverse trigonometric functions for special input values; for example, \(\frac{\pi }{4}={\tan }^{-1}(1)\ \text{and}\ \frac{\pi }{6}={\sin }^{-1}(\frac{1}{2}).\)
• A calculator will return an angle within the restricted domain of the original trigonometric function.
• Inverse functions allow us to find an angle when given two sides of a right triangle.
• In function composition, if the inside function is an inverse trigonometric function, then there are exact expressions; for example, \(\sin ({\cos }^{-1}(x))=\sqrt{1-{x}^{2}}.\)
• If the inside function is a trigonometric function, then the only possible combinations are \({\sin }^{-1}(\cos \ x)=\frac{\pi }{2}-x\) if \(0\le x\le \pi\) and \({\cos }^{-1}(\sin \ x)=\frac{\pi }{2}-x\) if \(-\frac{\pi }{2}\le x\le \frac{\pi }{2}.\)
• When evaluating the composition of a trigonometric function with an inverse trigonometric function, draw a reference triangle to assist in determining the ratio of sides that represents the output of the trigonometric function.
• When evaluating the composition of a trigonometric function with an inverse trigonometric function, you may use trig identities to assist in determining the ratio of sides.

For the following exercises, sketch the graph of each function for two full periods. Determine the amplitude, the period, and the equation for the midline.

For the following exercises, determine the amplitude, period, and midline of the graph, and then find a formula for the function.

For the following exercises, find the amplitude, period, phase shift, and midline.

For the following exercises, find the period and horizontal shift of each function.

For the following exercises, graph the functions on the specified window and answer the questions.

For the following exercises, let \(f(x)=\frac{3}{5}\cos (6x).\)

For the following exercises, find and graph one period of the periodic function with the given amplitude, period, and phase shift.

Condensed: the full section is in OpenStax Algebra and Trigonometry 2e.

In order to use inverse trigonometric functions, we need to understand that an inverse trigonometric function “undoes” what the original trigonometric function “does,” as is the case with any other function and its inverse. In other words, the domain of the inverse function is the range of the original function, and vice versa, as summarized in .

For example, if \(f(x)=\sin \ x,\) then we would write \({f}^{-1}(x)={\sin }^{-1}x.\) Be aware that \({\sin }^{-1}x\) does not mean \(\frac{1}{\sin x}.\) The following examples illustrate the inverse trigonometric functions:

• Since \(\text{sin}(\frac{\pi }{6})=\frac{1}{2},\) then \(\frac{\pi }{6}={\text{sin}}^{-1}(\frac{1}{2}).\)
• Since \(\cos (\pi )=-1,\) then \(\pi ={\cos }^{-1}(-1).\)
• Since \(\tan (\frac{\pi }{4})=1,\) then \(\frac{\pi }{4}={\tan }^{-1}(1).\)

In previous sections, we evaluated the trigonometric functions at various angles, but at times we need to know what angle would yield a specific sine, cosine, or tangent value. For this, we need inverse functions. Recall that, for a one-to-one function, if \(f(a)=b,\) then an inverse function would satisfy \({f}^{-1}(b)=a.\)

Bear in mind that the sine, cosine, and tangent functions are not one-to-one functions. The graph of each function would fail the horizontal line test. In fact, no periodic function can be one-to-one because each output in its range corresponds to at least one input in every period, and there are an infinite number of periods. As with other functions that are not one-to-one, we will need to restrict the domain of each function to yield a new function that is one-to-one. We choose a domain for each function that includes the number 0. shows the graph of the sine function limited to \([-\frac{\pi }{2},\frac{\pi }{2}]\) and the graph of the cosine function limited to \([0,\pi ].\)

shows the graph of the tangent function limited to \((-\frac{\pi }{2},\frac{\pi }{2}).\)

These conventional choices for the restricted domain are somewhat arbitrary, but they have important, helpful characteristics. Each domain includes the origin and some positive values, and most importantly, each results in a one-to-one function that is invertible. The conventional choice for the restricted domain of the tangent function also has the useful property that it extends from one vertical asymptote to the next instead of being divided into two parts by an asymptote.

On these restricted domains, we can define the inverse trigonometric functions.

Condensed: the full section is in OpenStax Precalculus 2e.

Now that we can identify inverse functions, we will learn to evaluate them. For most values in their domains, we must evaluate the inverse trigonometric functions by using a calculator, interpolating from a table, or using some other numerical technique. Just as we did with the original trigonometric functions, we can give exact values for the inverse functions when we are using the special angles, specifically \(\frac{\pi }{6}\) (30°), \(\frac{\pi }{4}\) (45°), and \(\frac{\pi }{3}\) (60°), and their reflections into other quadrants.

Condensed: the full section is in OpenStax Precalculus 2e.

To evaluate inverse trigonometric functions that do not involve the special angles discussed previously, we will need to use a calculator or other type of technology. Most scientific calculators and calculator-emulating applications have specific keys or buttons for the inverse sine, cosine, and tangent functions. These may be labeled, for example, SIN \({\ }^{-1}\), ARCSIN, or ASIN.

In the previous chapter, we worked with trigonometry on a right triangle to solve for the sides of a triangle given one side and an additional angle. Using the inverse trigonometric functions, we can solve for the angles of a right triangle given two sides, and we can use a calculator to find the values to several decimal places.

In these examples and exercises, the answers will be interpreted as angles and we will use \(\theta\) as the independent variable. The value displayed on the calculator may be in degrees or radians, so be sure to set the mode appropriate to the application.

Condensed: the full section is in OpenStax Precalculus 2e.

There are times when we need to compose a trigonometric function with an inverse trigonometric function. In these cases, we can usually find exact values for the resulting expressions without resorting to a calculator. Even when the input to the composite function is a variable or an expression, we can often find an expression for the output. To help sort out different cases, let \(f(x)\) and \(g(x)\) be two different trigonometric functions belonging to the set \(\{\sin (x),\cos (x),\tan (x)\}\) and let \({f}^{-1}(y)\) and \({g}^{-1}(y)\) be their inverses.

Condensed: the full section is in OpenStax Precalculus 2e.

• An inverse function is one that “undoes” another function. The domain of an inverse function is the range of the original function and the range of an inverse function is the domain of the original function.
• Because the trigonometric functions are not one-to-one on their natural domains, inverse trigonometric functions are defined for restricted domains.
• For any trigonometric function \(f(x),\) if \(x={f}^{-1}(y),\) then \(f(x)=y.\) However, \(f(x)=y\) only implies \(x={f}^{-1}(y)\) if \(x\) is in the restricted domain of \(f.\)
• Special angles are the outputs of inverse trigonometric functions for special input values; for example, \(\frac{\pi }{4}={\tan }^{-1}(1)\ \text{and}\ \frac{\pi }{6}={\sin }^{-1}(\frac{1}{2}).\)
• A calculator will return an angle within the restricted domain of the original trigonometric function.
• Inverse functions allow us to find an angle when given two sides of a right triangle.
• In function composition, if the inside function is an inverse trigonometric function, then there are exact expressions; for example, \(\sin ({\cos }^{-1}(x))=\sqrt{1-{x}^{2}}.\)
• If the inside function is a trigonometric function, then the only possible combinations are \({\sin }^{-1}(\cos \ x)=\frac{\pi }{2}-x\) if \(0\le x\le \pi\) and \({\cos }^{-1}(\sin \ x)=\frac{\pi }{2}-x\) if \(-\frac{\pi }{2}\le x\le \frac{\pi }{2}.\)
• When evaluating the composition of a trigonometric function with an inverse trigonometric function, draw a reference triangle to assist in determining the ratio of sides that represents the output of the trigonometric function.
• When evaluating the composition of a trigonometric function with an inverse trigonometric function, you may use trig identities to assist in determining the ratio of sides.

For the following exercises, sketch the graph of each function for two full periods. Determine the amplitude, the period, and the equation for the midline.

For the following exercises, determine the amplitude, period, and midline of the graph, and then find a formula for the function.

For the following exercises, find the amplitude, period, phase shift, and midline.

For the following exercises, find the period and horizontal shift of each function.

For the following exercises, graph the functions on the specified window and answer the questions.

For the following exercises, let \(f(x)=\frac{3}{5}\cos (6x).\)

For the following exercises, find and graph one period of the periodic function with the given amplitude, period, and phase shift.

Condensed: the full section is in OpenStax Precalculus 2e.

In mathematics, the inverse trigonometric functions (occasionally also called antitrigonometric, cyclometric, or arcus functions) are the inverse functions of the trigonometric functions, under suitably restricted domains. Specifically, they are the inverses of the sine, cosine, tangent, cotangent, secant, and cosecant functions, and are used to obtain an angle from any of the angle's trigonometric ratios. Inverse trigonometric functions are widely used in engineering, navigation, physics, and geometry.

There are several notations for the inverse trigonometric functions. The most common uses the prefix arc-: arcsin(x), arccos(x), arctan(x), etc. (This convention is used throughout this article.) This notation arises from the following geometric relationships: when measuring in radians, an angle of θ radians will correspond to an arc whose length is rθ, where r is the radius of the circle. Thus in the unit circle, the cosine of x function is both the arc and the angle, because the arc of a circle of radius 1 is the same as the angle. Or, "the arc whose cosine is x" is the same as "the angle whose cosine is x", because the length of the arc of the circle in radii is the same as the measurement of the angle in radians. In computer programming languages, the inverse trigonometric functions are often called by the abbreviated forms asin, acos, atan.

Another common notation in English sources uses a superscript f: sin(x), cos(x), tan(x), etc., as introduced by John Herschel in 1813. This conflicts with the common syntax sin(x), which denote exponentiation rather than function composition, and therefore may result in confusion between notation for the reciprocal (multiplicative inverse) and inverse function. The inverse function convention using a superscript in brackets is unambiguous but not commonly used: sin(x), cos(x), tan(x).

The ambiguity is mitigated by the use of special names for reciprocal trigonometric functions, for example, (cos(x)) = sec(x). Nevertheless, certain authors advise against using it, since it is ambiguous. Another problematic convention used by a small number of authors is an uppercase first letter, along with a “−1” superscript: Sin(x), Cos(x), Tan(x), etc. Although it is intended to avoid confusion with the reciprocal, which should be represented by sin(x), cos(x), etc., or, better, by sin x, cos x, etc., it in turn creates yet another major source of ambiguity, especially since many popular high-level programming languages (e.g. Mathematica and MAGMA) use capitalised names for the standard trig functions.

Since 2019, the ISO 80000-2 standard has specified solely the "arc" prefix for the inverse functions.

Since none of the six trigonometric functions are one-to-one, they must be restricted in order to have inverse functions. Therefore, the result ranges of the inverse functions are proper (i.e. strict) subsets of the domains of the original functions.

For example, using function in the sense of multivalued functions, just as the square root function \(y = \sqrt{x}\) could be defined from \(y^2 = x,\) the function \(y = \arcsin(x)\) is defined so that \(\sin(y) = x.\) For a given real number \(x,\) with \(-1 \leq x \leq 1,\) there are multiple (in fact, countably infinitely many) numbers \(y\) such that \(\sin(y) = x\); for example, \(\sin(0) = 0,\) but also \(\sin(\pi) = 0,\) \(\sin(2 \pi) = 0,\) etc. When only one value is desired, the function may be restricted to its principal branch. With this restriction, for each \(x\) in the domain, the expression \(\arcsin(x)\) will evaluate only to a single value, called its principal value. These properties apply to all the inverse trigonometric functions.

The principal inverses are listed in the following table.

Note: Some authors define the range of arcsecant to be (\(0 \leq y < \frac{\pi}{2}\) or \(\pi \leq y < \frac{3 \pi}{2}\) ), because the tangent function is nonnegative on this domain. This makes some computations more consistent. For example, using this range, \(\tan(\arcsec(x)) = \sqrt{x^2 - 1},\) whereas with the range (\(0 \leq y < \frac{\pi}{2}\) or \frac{\pi}{2} < y \leq \pi\)), we would have to write \(\tan(\arcsec(x)) = \pm \sqrt{x^2 - 1},\) since tangent is nonnegative on \(0 \leq y < \frac{\pi}{2},\) but nonpositive on \frac{\pi}{2} < y \leq \pi.\) For a similar reason, the same authors define the range of arccosecant to be ( - \pi < y \leq - \frac{\pi}{2}\) or 0 < y \leq \frac{\pi}{2} ) .\)

Each of the trigonometric functions is periodic in the real part of its argument, running through all its values twice in each interval of \(2 \pi:\)

• Sine and cosecant begin their period at \(2 \pi k-\frac{\pi}{2}\) (where \(k\) is an integer), finish it at \(2 \pi k+\frac{\pi}{2},\) and then reverse themselves over \(2 \pi k+\frac{\pi}{2}\) to \(2 \pi k+\frac{3\pi}{2}.\)
• Cosine and secant begin their period at \(2 \pi k,\) finish it at \(2 \pi k+\pi.\) and then reverse themselves over \(2 \pi k+\pi\) to \(2 \pi k+2 \pi.\)
• Tangent begins its period at \(2 \pi k-\frac{\pi}{2},\) finishes it at \(2 \pi k+\frac{\pi}{2},\) and then repeats it (forward) over \(2 \pi k+\frac{\pi}{2}\) to \(2 \pi k+\frac{3 \pi}{2}.\)
• Cotangent begins its period at \(2 \pi k,\) finishes it at \(2 \pi k+\pi,\) and then repeats it (forward) over \(2 \pi k+\pi\) to \(2 \pi k+2 \pi.\)

This periodicity is reflected in the general inverses, where \(k\) is some integer.

The following table shows how inverse trigonometric functions may be used to solve equalities involving the six standard trigonometric functions. It is assumed that the given values \(\theta,\) \(r,\) \(s,\) \(x,\) and \(y\) all lie within appropriate ranges so that the relevant expressions below are well-defined. Note that "for some \(k \in \Z\)" is just another way of saying "for some integer \(k.\)"

The symbol \(\,\iff\,\) is logical equality and indicates that if the left hand side is true then so is the right hand side and, conversely, if the right hand side is true then so is the left hand side (see this footnote for more details and an example illustrating this concept).

where the first four solutions can be written in expanded form as:

For example, if \(\cos \theta = -1\) then \(\theta = \pi+2 \pi k = -\pi+2 \pi (1+k)\) for some \(k \in \Z.\) While if \(\sin \theta = \pm 1\) then \(\theta = \frac{\pi}{2}+\pi k =-\frac{\pi}{2}+\pi (k+1)\) for some \(k \in \Z,\) where \(k\) will be even if \(\sin \theta = 1\) and it will be odd if \(\sin \theta = -1.\) The equations \(\sec \theta = -1\) and \(\csc \theta = \pm 1\) have the same solutions as \(\cos \theta = -1\) and \(\sin \theta = \pm 1,\) respectively. In all equations above except for those just solved (i.e. except for \(\sin\)/\(\csc \theta = \pm 1\) and \(\cos\)/\(\sec \theta =-1\)), the integer \(k\) in the solution's formula is uniquely determined by \(\theta\) (for fixed \(r, s, x,\) and \(y\)).

With the help of integer parity \[\operatorname{Parity}(h) = \begin{cases} 0 & \text{if } h \text{ is even } \\ 1 & \text{if } h \text{ is odd } \\ \end{cases}\] it is possible to write a solution to \(\cos \theta = x\) that doesn't involve the "plus or minus" \(\,\pm\,\) symbol:

\(cos \; \theta = x \quad\) if and only if \(\quad \theta = (-1)^h \arccos(x) + \pi h + \pi \operatorname{Parity}(h) \quad\) for some \(h \in \Z.\)

\(sec \; \theta = r \quad\) if and only if \(\quad \theta = (-1)^h \arcsec(r) + \pi h + \pi \operatorname{Parity}(h) \quad\) for some \(h \in \Z,\)

Condensed: the full section is in Wikipedia.

The equations above can be transformed by using the reflection and shift identities:

These formulas imply, in particular, that the following hold:

\[\begin{aligned} \sin \theta &= -\sin(-\theta) &&= -\sin(\pi+\theta) &&= \phantom{-}\sin(\pi-\theta) \\ &= -\cos\left(\frac{\pi}{2}+\theta\right) &&= \phantom{-}\cos\left(\frac{\pi}{2}-\theta\right) &&= -\cos\left(-\frac{\pi}{2}-\theta\right) \\ &= \phantom{-}\cos\left(-\frac{\pi}{2}+\theta\right) &&= -\cos\left(\frac{3\pi}{2}-\theta\right) &&= -\cos\left(-\frac{3\pi}{2}+\theta\right) \\[0.3ex] \cos \theta &= \phantom{-}\cos(-\theta) &&= -\cos(\pi+\theta) &&= -\cos(\pi-\theta) \\ &= \phantom{-}\sin\left(\frac{\pi}{2}+\theta\right) &&= \phantom{-}\sin\left(\frac{\pi}{2}-\theta\right) &&= -\sin\left(-\frac{\pi}{2}-\theta\right) \\ &= -\sin\left(-\frac{\pi}{2}+\theta\right) &&= -\sin\left(\frac{3\pi}{2}-\theta\right) &&= \phantom{-}\sin\left(-\frac{3\pi}{2}+\theta\right) \\[0.3ex] \tan \theta &= -\tan(-\theta) &&= \phantom{-}\tan(\pi+\theta) &&= -\tan(\pi-\theta) \\ &= -\cot\left(\frac{\pi}{2}+\theta\right) &&= \phantom{-}\cot\left(\frac{\pi}{2}-\theta\right) &&= \phantom{-}\cot\left(-\frac{\pi}{2}-\theta\right) \\ &= -\cot\left(-\frac{\pi}{2}+\theta\right) &&= \phantom{-}\cot\left(\frac{3\pi}{2}-\theta\right) &&= -\cot\left(-\frac{3\pi}{2}+\theta\right) \\[0.3ex] \end{aligned}\]

where swapping \(\sin \leftrightarrow \csc,\) swapping \(\cos \leftrightarrow \sec,\) and swapping \(\tan \leftrightarrow \cot\) gives the analogous equations for \(\csc, \sec, \text{ and } \cot,\) respectively.

So for example, by using the equality \(\sin \left(\frac{\pi}{2}-\theta\right) = \cos \theta,\) the equation \(\cos \theta = x\) can be transformed into \(\sin \left(\frac{\pi}{2}-\theta\right) = x,\) which allows for the solution to the equation \(\;\sin \varphi = x\;\) (where \(\varphi := \frac{\pi}{2}-\theta\)) to be used; that solution being: \(\varphi = (-1)^k \arcsin (x)+\pi k \; \text{ for some } k \in \Z,\) which becomes: \[\frac{\pi}{2}-\theta ~=~ (-1)^k \arcsin (x)+\pi k \quad \text{ for some } k \in \Z\] where using the fact that \((-1)^{k} = (-1)^{-k}\) and substituting \(h :=-k\) proves that another solution to \(\;\cos \theta = x\;\) is: \[\theta ~=~ (-1)^{h+1} \arcsin (x)+\pi h+\frac{\pi}{2} \quad \text{ for some } h \in \Z.\] The substitution \(\;\arcsin x = \frac{\pi}{2}-\arccos x\;\) may be used express the right hand side of the above formula in terms of \(\;\arccos x\;\) instead of \(\;\arcsin x.\;\)

Trigonometric functions of inverse trigonometric functions are tabulated below. They may be derived from the Pythagorean identities. Another way is by considering the geometry of a right-angled triangle, with one side of length 1 and another side of length \(x,\) then applying the Pythagorean theorem and definitions of the trigonometric ratios. It is worth noting that for arcsecant and arccosecant, the diagram assumes that \(x\) is positive, and thus the result has to be corrected through the use of absolute values and the signum (sgn) operation.

Complementary angles:

\(\begin{aligned} \arccos(x) &= \frac{\pi}{2} - \arcsin(x) \\[0.5em] \arccot(x) &= \frac{\pi}{2} - \arctan(x) \\[0.5em] \arccsc(x) &= \frac{\pi}{2} - \arcsec(x) \end{aligned}\)

Negative arguments:

\(\begin{aligned} \arcsin(-x) &= -\arcsin(x) \\ \arccsc(-x) &= -\arccsc(x) \\ \arccos(-x) &= \pi -\arccos(x) \\ \arcsec(-x) &= \pi -\arcsec(x) \\ \arctan(-x) &= -\arctan(x) \\ \arccot(-x) &= \pi -\arccot(x) \end{aligned}\)

Reciprocal arguments:

\(\begin{aligned} \arcsin\left(\frac{1}{x}\right) &= \arccsc(x) & \\[0.3em] \arccsc\left(\frac{1}{x}\right) &= \arcsin(x) & \\[0.3em] \arccos\left(\frac{1}{x}\right) &= \arcsec(x) & \\[0.3em] \arcsec\left(\frac{1}{x}\right) &= \arccos(x) & \\[0.3em] \arctan\left(\frac{1}{x}\right) &= \arccot(x) &= \frac{\pi}{2} - \arctan(x) \, , \text{ if } x > 0 \\[0.3em] \arctan\left(\frac{1}{x}\right) &= \arccot(x) - \pi &= -\frac{\pi}{2} - \arctan(x) \, , \text{ if } x < 0 \\[0.3em] \arccot\left(\frac{1}{x}\right) &= \arctan(x) &= \frac{\pi}{2} - \arccot(x) \, , \text{ if } x > 0 \\[0.3em] \arccot\left(\frac{1}{x}\right) &= \arctan(x) + \pi &= \frac{3\pi}{2} - \arccot(x) \, , \text{ if } x < 0 \end{aligned}\)

The identities above can be used with (and derived from) the fact that \(\sin\) and \(\csc\) are reciprocals (i.e. \(\csc = \tfrac1{\sin}\)), as are \(\cos\) and \(\sec,\) and \(\tan\) and \(\cot.\)

Useful identities if one only has a fragment of a sine table:

\(\begin{aligned} \arcsin(x) &= \frac{1}{2}\arccos\left(1-2x^2\right) \, , \text{ if } 0 \leq x \leq 1 \\ \arcsin(x) &= \arctan\left(\frac{x}{\sqrt{1 - x^2}}\right) \\ \arccos(x) &= \frac{1}{2}\arccos\left(2x^2-1\right) \, , \text{ if } 0 \leq x \leq 1 \\ \arccos(x) &= \arctan\left(\frac{\sqrt{1 - x^2}}{x}\right) \\ \arccos(x) &= \arcsin\left(\sqrt{1 - x^2}\right) \, , \text{ if } 0 \leq x \leq 1 \text{ , from which you get } \\ \arccos &\left(\frac{1-x^2}{1 + x^2}\right) = \arcsin \left (\frac{2x}{1 + x^2}\right) \, , \text{ if } 0 \leq x \leq 1 \\ \arcsin &\left(\sqrt{1 - x^2}\right) =\frac{\pi}{2}-\sgn(x)\arcsin(x) \\ \arctan(x) &= \arcsin\left(\frac{x}{\sqrt{1 + x^2}}\right) \\ \arccot(x) &= \arccos\left(\frac{x}{\sqrt{1 + x^2}}\right) \end{aligned}\)

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

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

\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}\)

Condensed: the full section is in Wikipedia.

\(\arctan(u) \pm \arctan(v) = \arctan\left(\frac{u \pm v}{1 \mp uv}\right) \pmod \pi \, , \quad u v \ne 1 \, .\)

This is derived from the tangent addition formula

\(\tan(\alpha \pm \beta) = \frac{\tan(\alpha) \pm \tan(\beta)}{1 \mp \tan(\alpha)\tan(\beta)} \, ,\)

by letting

\(\alpha = \arctan(u) \, , \quad \beta = \arctan(v) \, .\)

The derivatives for complex values of z are as follows:

\(\begin{aligned} \frac{d}{dz} \arcsin(z) &{} = \frac{1}{\sqrt{1-z^2}} \; ; &z &{}\neq -1, +1 \\ \frac{d}{dz} \arccos(z) &{} = -\frac{1}{\sqrt{1-z^2}} \; ; &z &{}\neq -1, +1 \\ \frac{d}{dz} \arctan(z) &{} = \frac{1}{1+z^2} \; ; &z &{}\neq -i, +i\\ \frac{d}{dz} \arccot(z) &{} = -\frac{1}{1+z^2} \; ; &z &{}\neq -i, +i \\ \frac{d}{dz} \arcsec(z) &{} = \frac{1}{z^2 \sqrt{1 - \frac{1}{z^{2}}}} \; ; &z &{}\neq -1, 0, +1 \\ \frac{d}{dz} \arccsc(z) &{} = -\frac{1}{z^2 \sqrt{1 - \frac{1}{z^{2}}}} \; ; &z &{}\neq -1, 0, +1 \end{aligned}\)

Only for real values of x:

\(\begin{aligned} \frac{d}{dx} \arcsec(x) &{} = \frac{1}{|x| \sqrt{x^2-1}} \; ; & |x| > 1\\ \frac{d}{dx} \arccsc(x) &{} = -\frac{1}{|x| \sqrt{x^2-1}} \; ; & |x| > 1 \end{aligned}\)

These formulas can be derived in terms of the derivatives of trigonometric functions. For example, if \(x = \sin \theta\), then \(dx/d\theta = \cos \theta = \sqrt{1-x^2},\) so

\(\frac{d}{dx}\arcsin(x) = \frac{d \theta}{dx} = \frac{1}{dx/d\theta} = \frac{1}{\sqrt{1-x^2}}.\)

Integrating the derivative and fixing the value at one point gives an expression for the inverse trigonometric function as a definite integral:

\(\begin{aligned} \arcsin(x) &{}= \int_0^x \frac{1}{\sqrt{1 - z^2}} \, dz \; , & |x| &{} \leq 1\\ \arccos(x) &{}= \int_x^1 \frac{1}{\sqrt{1 - z^2}} \, dz \; , & |x| &{} \leq 1\\ \arctan(x) &{}= \int_0^x \frac{1}{z^2 + 1} \, dz \; ,\\ \arccot(x) &{}= \int_x^\infty \frac{1}{z^2 + 1} \, dz \; ,\\ \arcsec(x) &{}= \int_1^x \frac{1}{z \sqrt{z^2 - 1}} \, dz = \pi + \int_{-x}^{-1} \frac{1}{z \sqrt{z^2 - 1}} \, dz\; , & x &{} \geq 1\\ \arccsc(x) &{}= \int_x^\infty \frac{1}{z \sqrt{z^2 - 1}} \, dz = \int_{-\infty}^{-x} \frac{1}{z \sqrt{z^2 - 1}} \, dz \; , & x &{} \geq 1\\ \end{aligned}\)

When x equals 1, the integrals with limited domains are improper integrals, but still well-defined.

Similar to the sine and cosine functions, the inverse trigonometric functions can also be calculated using power series, as follows. For arcsine, the series can be derived by expanding its derivative, \(\tfrac{1}{\sqrt{1-z^2}}\), as a binomial series, and integrating term by term (using the integral definition as above). The series for arctangent can similarly be derived by expanding its derivative \(\frac{1}{1+z^2}\) in a geometric series, and applying the integral definition above (see Leibniz series).

\[\begin{aligned} \arcsin(z) & = z + \left( \frac{1}{2} \right) \frac{z^3}{3} + \left( \frac{1 \cdot 3}{2 \cdot 4} \right) \frac{z^5}{5} + \left( \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \right) \frac{z^7}{7} + \cdots \\[5pt] & = \sum_{n=0}^\infty \frac{(2n-1)!!}{(2n)!!}\frac{z^{2n+1}}{2n+1} \\[5pt] & = \sum_{n=0}^\infty \frac{(2n)!}{(2^n n!)^2} \frac{z^{2n+1}}{2n+1} \, ; \qquad |z| \le 1 \end{aligned}\] The Taylor series for arctangent is also known as the arctangent series or Gregory's series. \[\arctan(z) = z - \frac{z^3}{3} +\frac{z^5}{5} - \frac{z^7}{7} + \cdots = \sum_{n=0}^\infty \frac{(-1)^n z^{2n+1}}{2n+1} \, ; \qquad |z| \le 1 \qquad z \neq i,-i\]

Series for the other inverse trigonometric functions can be given in terms of these according to the relationships given above. For example, \(\arccos(x) = \pi/2 - \arcsin(x)\), \(\arccsc(x) = \arcsin(1/x)\), and so on. Another series is given by:

\[2\left(\arcsin\left(\frac{x}{2}\right) \right)^2 = \sum_{n=1}^\infty \frac{x^{2n}}{n^2\binom {2n} n}.\]

Leonhard Euler found a series for the arctangent that converges more quickly than its Taylor series:

\[\arctan(z) = \frac z {1 + z^2} \sum_{n=0}^\infty \prod_{k=1}^n \frac{2k z^2}{(2k + 1)(1 + z^2)}.\] (The term in the sum for n = 0 is the empty product, so is 1.)

Alternatively, this can be expressed as

Condensed: the full section is in Wikipedia.

For real and complex values of z:

\(\begin{aligned} \int \arcsin(z) \, dz &{}= z \, \arcsin(z) + \sqrt{1 - z^2} + C\\ \int \arccos(z) \, dz &{}= z \, \arccos(z) - \sqrt{1 - z^2} + C\\ \int \arctan(z) \, dz &{}= z \, \arctan(z) - \frac{1}{2} \ln \left( 1 + z^2 \right) + C\\ \int \arccot(z) \, dz &{}= z \, \arccot(z) + \frac{1}{2} \ln \left( 1 + z^2 \right) + C\\ \int \arcsec(z) \, dz &{}= z \, \arcsec(z) - \ln \left[ z \left( 1 + \sqrt{ \frac{z^2-1}{z^2} } \right) \right] + C\\ \int \arccsc(z) \, dz &{}= z \, \arccsc(z) + \ln \left[ z \left( 1 + \sqrt{ \frac{z^2-1}{z^2} } \right) \right] + C \end{aligned}\)

For real x ≥ 1:

\(\begin{aligned} \int \arcsec(x) \, dx &{}= x \, \arcsec(x) - \ln \left( x + \sqrt{x^2-1} \right) + C\\ \int \arccsc(x) \, dx &{}= x \, \arccsc(x) + \ln \left( x + \sqrt{x^2-1} \right) + C \end{aligned}\)

For all real x not between -1 and 1:

\(\begin{aligned} \int \arcsec(x) \, dx &{}= x \, \arcsec(x) - \sgn(x) \ln\left| x + \sqrt{x^2-1}\right| + C\\ \int \arccsc(x) \, dx &{}= x \, \arccsc(x) + \sgn(x) \ln\left| x + \sqrt{x^2-1}\right| + C \end{aligned}\)

The absolute value is necessary to compensate for both negative and positive values of the arcsecant and arccosecant functions. The signum function is also necessary due to the absolute values in the derivatives of the two functions, which create two different solutions for positive and negative values of x. These can be further simplified using the logarithmic definitions of the inverse hyperbolic functions:

\(\begin{aligned} \int \arcsec(x) \, dx &{}= x \, \arcsec(x) - \operatorname{arcosh}(|x|) + C\\ \int \arccsc(x) \, dx &{}= x \, \arccsc(x) + \operatorname{arcosh}(|x|) + C\\ \end{aligned}\)

The absolute value in the argument of the arcosh function creates a negative half of its graph, making it identical to the signum logarithmic function shown above.

All of these antiderivatives can be derived using integration by parts and the simple derivative forms shown above.

Since the inverse trigonometric functions are analytic functions, they can be extended from the real line to the complex plane. This results in functions with multiple sheets and branch points. One possible way of defining the extension is:

\(\arctan(z) = \int_0^z \frac{dx}{1 + x^2} \quad z \neq -i, +i\)

where the part of the imaginary axis which does not lie strictly between the branch points (−i and +i) is the branch cut between the principal sheet and other sheets. The path of the integral must not cross a branch cut. For z not on a branch cut, a straight line path from 0 to z is such a path. For z on a branch cut, the path must approach from Re[x] > 0 for the upper branch cut and from Re[x] < 0 for the lower branch cut.

The arcsine function may then be defined as:

\(\arcsin(z) = \arctan\left(\frac{z}{\sqrt{1 - z^2}}\right) \quad z \neq -1, +1\)

where (the square-root function has its cut along the negative real axis and) the part of the real axis which does not lie strictly between −1 and +1 is the branch cut between the principal sheet of arcsin and other sheets;

\(\arccos(z) = \frac{\pi}{2} - \arcsin(z) \quad z \neq -1, +1\)

which has the same cut as arcsin;

\(\arccot(z) = \frac{\pi}{2} - \arctan(z) \quad z \neq -i, i\)

which has the same cut as arctan;

\(\arcsec(z) = \arccos\left(\frac{1}{z}\right) \quad z \neq -1, 0, +1\)

where the part of the real axis between −1 and +1 inclusive is the cut between the principal sheet of arcsec and other sheets;

\(\arccsc(z) = \arcsin\left(\frac{1}{z}\right) \quad z \neq -1, 0, +1\)

Condensed: the full section is in Wikipedia.

These functions may also be expressed using complex logarithms. This extends their domains to the complex plane in a natural fashion. The following identities for principal values of the functions hold everywhere that they are defined, even on their branch cuts.

\(\begin{aligned} \arcsin(z) &{}= -i \ln \left( \sqrt{1-z^2} + iz \right) = i \ln \left( \sqrt{1-z^2} - iz \right) &{}= \arccsc\left(\frac{1}{z}\right) \\[10pt] \arccos(z) &{}= -i \ln \left( z + i \sqrt {1-z^2} \right) = i \ln \left( z - i \sqrt {1-z^2} \right) = \frac{\pi}{2} - \arcsin(z) &{}= \arcsec\left(\frac{1}{z}\right) \\[10pt] \arctan(z) &{}= -\frac{i}{2}\ln \left(\frac{i - z}{i + z}\right) = -\frac{i}{2}\ln \left(\frac{1 + iz}{1 - iz}\right) &{}= \arccot\left(\frac{1}{z}\right) \\[10pt] \arccot(z) &{}= -\frac{i}{2}\ln\left( \frac{z + i}{z - i} \right) = -\frac{i}{2}\ln\left( \frac{iz - 1}{iz + 1} \right) &{}= \arctan\left(\frac{1}{z}\right) \\[10pt] \arcsec(z) &{}= -i \ln \left( \frac{1}{z} + i \sqrt{1 - \frac{1}{z^2}} \right) = i \ln \left( \frac{1}{z} - i \sqrt{1 - \frac{1}{z^2}} \right) = \frac{\pi}{2} - \arccsc(z) &{}= \arccos\left(\frac{1}{z}\right) \\[10pt] \arccsc(z) &{}= -i \ln \left( \sqrt{1 - \frac{1}{z^2}} + \frac{i}{z} \right) = i \ln \left( \sqrt{1 - \frac{1}{z^2}} - \frac{i}{z} \right) &{}= \arcsin\left(\frac{1}{z}\right) \end{aligned}\)