# List of Inverse Trig Function Identities, Integrals, and Derivatives

TR Smith is a product designer and former teacher who uses math in her work every day.

## Inverse Trig Function Graphs

Graphs of inverse trig functions.

Inverse trigonometric functions output an angle measure when you input a trigonometric ratio. For instance, the input for the inverse tangent function, arctan, is the tangent (opposite side divided by adjacent side) of some angle. The output is the corresponding angle of that ratio. For example, the inverse tangent of sqrt(3) is π/3 radians = 60°. This makes sense because the tangent of 60° or π/3 radians is sqrt(3).

The list below gives various algebraic identities of arcsin, arccos, and arctan along with the most common derivatives and integrals (antiderivatives). You can use these identities and expressions to simplify algebra, geometry, and calculus problems involving trigonometric and inverse trig functions.

tan-1(x) + tan-1(1/x) = π/2
(x > 0)

tan-1(x) + tan-1(1/x) = -π/2
(x < 0)

sin-1(x) + sin-1(y) =
sin-1(x*sqrt(1-y2) + y*sqrt(1-x2))

cos-1(x) + cos-1(y) =
cos-1[xy - sqrt((1-y2)(1-x2))]

tan-1(x) + tan-1(y) =
tan-1[(x+y)/(1-xy)]

sin-1(x) + cos-1(x) = π/2

sin-1(x) = cos-1(sqrt(1-x2))

cos-1(x) = sin-1(sqrt(1-x2))

cos(2*cos-1(x)) = 2x2 - 1

cos(2*sin-1(x)) = 1 - 2x2

## Algebraic Identities

sin(arcsin(x)) = x

cos(arccos(x)) = x

tan(arctan(x)) = x

arcsin(sin(x)) = x

arccos(cos(x)) = x

arctan(tan(x)) = x

arcsin(cos(x)) = π/2 - x

arccos(sin(x)) = π/2 - x

sin(arccos(x)) = sqrt(1 - x^2)

cos(arcsin(x)) = sqrt(1 - x^2)

tan(arcsin(x)) = x/sqrt(1 - x^2)

tan(arccos(x)) = sqrt(1 - x^2)/x

cos(arctan(x)) = 1/sqrt(1 + x^2)

sin(arctan(x)) = x/sqrt(1 + x^2)

## Derivatives

[arctan(x)]' = 1/(1 + x^2)

[arcsin(x)]' = 1/sqrt(1 - x^2)

[arccos(x)]' = -1/sqrt(1 - x^2)

[arccot(x)]' = -1/(1 + x^2)

[arccsc(x)]' = -1/sqrt(x^4 - x^2)

[arcsec(x)]' = 1/sqrt(x^4 - x^2)

## Inverse Trigonometric Integrals/Antiderivatives

tan-1(x) dx = x*tan-1(x) - 0.5*Ln(x2 + 1)

∫ sin-1(x) dx = x*sin-1(x) + sqrt(1 - x2)

∫ cos-1(x) dx = x*cos-1(x) - sqrt(1 - x2)

∫ cot-1(x) dx = x*cot-1(x) + 0.5*Ln(x2 + 1)

∫ csc-1(x) dx = x*csc-1(x) + Ln(x + sqrt(x2 - 1))

∫ sec-1(x) dx = x*sec-1(x) - Ln(x + sqrt(x2 - 1))

x*tan-1(x) dx = 0.5x2tan-1(x) + 0.5*tan-1(x) - 0.5x

x*sin-1(x) dx = 0.5x2sin-1(x) - 0.25*sin-1(x) + 0.25x*sqrt(1 - x2)

sin-1(x)2 dx = -2x + x*sin-1(x)2 + 2*sin-1(x)*sqrt(1 - x2)

x*cos-1(x) dx = 0.5x2cos-1(x) - 0.25*cos-1(x) - 0.25x*sqrt(1 - x2)

cos-1(x)2 dx = -2x + x*cos-1(x)2 - 2*cos-1(x)*sqrt(1 - x2)

∫ sin-1(sqrt(x)) dx = (x-0.5)*sin-1(x) + 0.5*sqrt(x - x2)

## Example Calculus Problem

What is the area under the curve f(x) = x*arctan(1/x) from x = 1 to x = 2? To solve this problem we first note that "arctan" is alternative notation for tan-1. Next, we use the identity

arctan(1/x) = π/2 - arctan(x)

to simplify the function f(x) into

f(x) = x*[π/2 - arctan(x)]
= xπ/2 - x*arctan(x)

Now we can integrate it using one of the basic inverse trig integral formulas given above.

∫ xπ/2 - x*arctan(x) dx

= 0.25πx^2 - 0.5(x^2)arctan(x) - 0.5arctan(x) + 0.5x

Plugging in the limits of integration for x and subtracting gives you 0.873721 as the area under the curve between x = 1 and x = 2.

## Example Algebra Problem

Solve the equation arcsin(x) = arccos(sqrt(x)) for x in the interval [0, 1]. This problem can be approached two ways, either by taking the sine of both sides or by taking the cosine of both sides. If w take the sine of both sides, we cancel out the arcsin function to get

x = sin(arccos(sqrt(x)))

Using the fact that sin(arccos(z)) = sqrt(1 - z^2), the right-hand side of the equation can be simplified to give us

x = sqrt(1 - sqrt(x)^2)
x = sqrt(1 - x)
x^2 = 1 - x
x^2 + x - 1 = 0
x = [-1 +/- sqrt(1 + 4)]/2
x = -1.618034 and x = 0.618034

Since we only want solutions in the interval [0, 1], the solution is x = 0.618034, also known as the golden mean or golden section.

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• Tomás de España 3 years ago

is very helpful and with more equations than my book, but is printable version available?

• Author

TR Smith 3 years ago from Eastern Europe

Glad you found it useful, Tomás. Unfortunately there isn't a separate printable version with simpler formatting.