Inverse Trig Functions: Arcsin, Arccos and Arctan

Inverse trig functions return an angle from a ratio: ( arcsin x ) (answers in ( [-frac{pi}{2}, frac{pi}{2}] )), ( arccos x ) (in ( [0, pi] )) and ( arctan x ) (in ( (-frac{pi}{2}, frac{pi}{2}) )). For example ( arcsinfrac12 = frac{pi}{6} ) and ( arctan 1 = frac{pi}{4} ). Their derivatives are ( frac{1}{sqrt{1 – x^2}} ), ( -frac{1}{sqrt{1 – x^2}} ) and ( frac{1}{1 + x^2} ), which is why ( intfrac{dx}{1 + x^2} = arctan x + C ).

The inverse trig functions undo sine, cosine and tangent. If ( sintheta = x ), then ( theta = arcsin x ): you start with a ratio and get back an angle. They are written two ways:

Function	Also written	Read as
( arcsin x )	( sin^{-1}x )	“the angle whose sine is x”
( arccos x )	( cos^{-1}x )	“the angle whose cosine is x”
( arctan x )	( tan^{-1}x )	“the angle whose tangent is x”

The ( -1 ) means inverse function, not a reciprocal: ( sin^{-1}x neq frac{1}{sin x} ). Conversely, the reciprocal of sine is ( csc x ).

Why the ranges are restricted

A function only has an inverse if it’s one-to-one: each output comes from exactly one input. Sine isn’t. ( sinfrac{pi}{6} = sinfrac{5pi}{6} = sinfrac{13pi}{6} = frac12 ), and infinitely many more. So we keep one piece of each graph where the function takes every value exactly once, and invert that piece. The angles that come out are called principal values.

Function	Domain	Range	Quadrants of the answer
( arcsin x )	( [-1, 1] )	( left[-frac{pi}{2}, frac{pi}{2}right] )	I and IV
( arccos x )	( [-1, 1] )	( [0, pi] )	I and II
( arctan x )	( (-infty, infty) )	( left(-frac{pi}{2}, frac{pi}{2}right) )	I and IV

The other three, ( text{arccsc} ), ( text{arcsec} ) and ( text{arccot} ), are defined similarly but are used far less often.

How to evaluate inverse trig functions

1. Ask which angle has the given sine, cosine or tangent. The unit circle and special right triangles give the standard ones.
2. Make sure the angle is inside the principal range. If not, find the angle in the range with the same value.

Example 1: exact values

• ( arcsinfrac{sqrt3}{2} = frac{pi}{3} ), since ( sinfrac{pi}{3} = frac{sqrt3}{2} ) and ( frac{pi}{3} ) is in ( left[-frac{pi}{2}, frac{pi}{2}right] ).
• ( arccos 0 = frac{pi}{2} ).
• ( arctansqrt3 = frac{pi}{3} ).
• ( arcsin(-1) = -frac{pi}{2} ).

Example 2: negative inputs

( arccosleft(-frac{sqrt2}{2}right) ). Arccos returns angles in ( [0, pi] ), so the answer is in Quadrant II with reference angle ( frac{pi}{4} ): ( pi – frac{pi}{4} = frac{3pi}{4} ).

( arctan(-1) ). Arctan returns angles between ( -frac{pi}{2} ) and ( frac{pi}{2} ), so the answer is ( -frac{pi}{4} ), not ( frac{3pi}{4} ).

Example 3: compositions

( sin(arcsin 0.7) = 0.7 ), because ( sin(arcsin x) = x ) for every x in ( [-1, 1] ).

But ( arcsinleft(sinfrac{5pi}{6}right) = arcsinfrac12 = frac{pi}{6} ), not ( frac{5pi}{6} ). The inverse only undoes the function on the principal range.

Example 4: a triangle trick

Find ( cos(arctanfrac34) ). Let ( theta = arctanfrac34 ), so ( tantheta = frac34 ): draw a right triangle with opposite side 3 and adjacent side 4. The hypotenuse is 5, so ( costheta = frac45 ). The same method gives the identity ( cos(arcsin x) = sqrt{1 – x^2} ).

Inverse trig calculator

Pick a function and enter x. You get the principal angle in radians (exact when possible) and degrees.

The full-page inverse trig calculator has more examples. For the forward direction, use the trig calculator.

Graphs

• ( y = arcsin x ) rises from ( left(-1, -frac{pi}{2}right) ) to ( left(1, frac{pi}{2}right) ), steepest at the ends.
• ( y = arccos x ) falls from ( (-1, pi) ) to ( (1, 0) ). In fact ( arccos x = frac{pi}{2} – arcsin x ).
• ( y = arctan x ) is defined for every x and has horizontal asymptotes ( y = pmfrac{pi}{2} ).

Each graph is the reflection of the restricted trig graph across the line ( y = x ), as for any inverse function (see domain and range).

Derivatives of inverse trig functions

$$frac{d}{dx}arcsin x = frac{1}{sqrt{1 - x^2}}, qquad frac{d}{dx}arccos x = -frac{1}{sqrt{1 - x^2}}, qquad frac{d}{dx}arctan x = frac{1}{1 + x^2}$$

Derivation for arcsin. Let ( y = arcsin x ), so ( sin y = x ). Differentiate implicitly: ( cos y,frac{dy}{dx} = 1 ), so ( frac{dy}{dx} = frac{1}{cos y} ). Because y is in ( left[-frac{pi}{2}, frac{pi}{2}right] ), ( cos y ge 0 ), so ( cos y = sqrt{1 – sin^2 y} = sqrt{1 – x^2} ). The implicit differentiation guide explains the method, and the derivative of arcsin x and derivative of arctan x guides go further.

With the chain rule, ( frac{d}{dx}arctan(x^2) = frac{2x}{1 + x^4} ).

Integration of inverse trigonometric functions

Reading the derivative table backwards gives three integrals that appear constantly in Calculus 2:

$$int frac{dx}{sqrt{a^2 - x^2}} = arcsinfrac{x}{a} + C$$

$$int frac{dx}{a^2 + x^2} = frac{1}{a}arctanfrac{x}{a} + C$$

$$int frac{dx}{xsqrt{x^2 - a^2}} = frac{1}{a},text{arcsec}frac{|x|}{a} + C$$

Example 5

$$int_0^1 frac{dx}{1 + x^2} = arctan 1 - arctan 0 = frac{pi}{4}$$

This is one way to compute ( pi ): it equals 4 times this integral.

Example 6

( int frac{dx}{sqrt{9 – x^2}} = arcsinfrac{x}{3} + C ), using ( a = 3 ).

Example 7: completing the square

( int frac{dx}{x^2 + 4x + 13} ). First, complete the square: ( x^2 + 4x + 13 = (x + 2)^2 + 9 ). Then, with ( u = x + 2 ) and ( a = 3 ):

$$int frac{du}{u^2 + 9} = frac13arctanfrac{x + 2}{3} + C$$

That’s a u-substitution followed by the arctan formula, and it is how partial fractions finish when a quadratic factor doesn’t factor further.

Common mistakes

• Treating ( sin^{-1}x ) as ( frac{1}{sin x} ). Different functions entirely.
• Answers outside the range. ( arccos ) never returns a negative angle, and ( arcsin ) never returns more than ( frac{pi}{2} ).
• Domain errors. ( arcsin 1.2 ) is undefined in the real numbers.
• Forgetting the ( frac1a ) in the arctan integral. ( int frac{dx}{4 + x^2} = frac12arctanfrac{x}{2} + C ), not ( arctanfrac{x}{2} ).
• Degrees in calculus. The derivative formulas assume radians.

Practice problems

1. Evaluate ( arcsinleft(-frac12right) ).
2. Evaluate ( arccosfrac{sqrt3}{2} ).
3. Evaluate ( tan(arcsinfrac{5}{13}) ).
4. Differentiate ( y = arcsin(3x) ).
5. Evaluate ( int frac{dx}{16 + x^2} ).
6. Evaluate ( int_0^{1/2} frac{dx}{sqrt{1 – x^2}} ).

Answers: (1) ( -frac{pi}{6} ); (2) ( frac{pi}{6} ); (3) ( frac{5}{12} ); (4) ( frac{3}{sqrt{1 – 9x^2}} ); (5) ( frac14arctanfrac{x}{4} + C ); (6) ( frac{pi}{6} ).

FAQ

What are inverse trig functions used for?

Finding an angle from a ratio: the angle of a ramp from its rise and run, the direction of a vector, or the angles of a triangle. In calculus they also appear as antiderivatives.

Is arcsin the same as sin⁻¹?

Yes. Both mean the inverse sine function. However, neither means ( frac{1}{sin x} ).

Why is the range of arccos different from arcsin?

Cosine isn’t one-to-one on ( left[-frac{pi}{2}, frac{pi}{2}right] ) (it’s symmetric there), so a different interval, ( [0, pi] ), is used.

What is the derivative of arctan x?

( frac{1}{1 + x^2} ). It’s defined for every x. As a result, arctan is one of the friendliest functions to integrate against.

How do I integrate 1/(1 + x²)?

( int frac{dx}{1 + x^2} = arctan x + C ).

Further reading

• Inverse trigonometric functions (Wikipedia) — notation, principal values, identities, derivatives and integrals.
• Derivatives of Inverse Trig Functions (Paul’s Online Math Notes) — derivations of all six derivatives with examples.