# Inverse Trigonometric Functions: Value, Derivatives, Examples, Exercises

The **inverse trigonometric functions**, as the name implies, are the corresponding inverse functions of the sine, cosine, tangent, cotangent, secant, and cosecant functions.

Inverse trigonometric functions are denoted by the same name as their corresponding direct trigonometric function plus the prefix *arc*. Thus:

one.- *arcsen (x)* is the inverse trigonometric function of the function *sen (x)*

two.- *arccos (x)* is the inverse trigonometric function of the function *cos (x)*

3.- *arctan (x)* is the inverse trigonometric function of the function *so (x)*

4.-* arccot (x)* is the inverse trigonometric function of the function *cot (x)*

5.- *arcsec (x)* is the inverse trigonometric function of the function *sec (x)*

6.- *arccsc (x)* is the inverse trigonometric function of the function *csc (x)*

The function *θ = arcsen (x)* results in a unit arc *θ* (or angle in radians *θ*) such that *sin (θ) = x*.

So for example, arcsen (√3 / 2) = π / 3 since, as is known, the sine of π / 3 radians is equal to √3 / 2.

**Principal value of inverse trigonometric functions**

For a mathematical function f (x) to have an inverse g (x) = f^{-one}(x) it is necessary that this function be *injective*, which means that each y value of the arrival set of the function f (x) comes from one and only one x value.

It is clear that this requirement is not fulfilled by any trigonometric function. To clarify the point, let us note that the value y = 0.5 pIt can be obtained from the sine function in the following ways:

- sin (π / 6) = 0.5
- sin (5π / 6) = 0.5
- sin (7π / 6) = 0.5

And many more, since the sine function is periodic with period 2π.

In order to define inverse trigonometric functions, it is necessary to restrict the domain of their corresponding direct trigonometric functions, such that they fulfill the requirement of injectivity.

That restricted domain of the direct function will be the range or main branch of its corresponding inverse function.

**Table of domains and ranges of inverse trigonometric functions**

**Derivatives of inverse trigonometric functions**

To obtain the derivatives of the inverse trigonometric functions, the properties of the derivatives are applied, in particular the derivative of an inverse function.

If we denote by f (y) the function and by f^{-one}(x) to its inverse function, then the derivative of the inverse function is related to the derivative of the direct function by the following relationship:

[f ^{-1}(x)] ‘= 1 / f’ [f ^{-1}(x)]

For example: if x = f (y) = √y is the direct function, its inverse will be

y = f ^{-1}(x) = x ^{2} . Let’s apply the rule of the derivative of the inverse to this simple case to see that this rule is actually fulfilled:

[x ^{2} ] ‘= 1 / [√y]’ = 1 / (½ y ^{-½} = 2 and ^{½} = 2 (x ^{2} ) ^{½} = 2x

Well, we can use this trick to find the derivatives of the inverse trigonometric functions.

For example, we take *θ = arcsen (x)* as the direct function, then its inverse function will be *sin (θ) = x*.

[arcsen (x)] ‘= 1 / [sin (θ)]’ = 1 / cos (θ) = 1 / √ (1 – sin (θ) ^{2} ) =…

… = 1 / √ (1 – x ^{2} ).

In this way, all the derivatives of the inverse trigonometric functions can be obtained, which are shown below:

These derivatives are valid for any argument z belonging to the complex numbers and therefore they are also valid for any real argument x, since z = x + 0i.

**Examples**

**– Example 1**

Find arctan (1).

**Solution**

The arctan (1) is the unit arc (angle in radians) ፀ such that tan (ፀ) = 1. That angle is ፀ = π / 4 because tan (π / 4) = 1. So arctan (1) = π /4.

**– Example 2**

Calculate arcsen (cos (π / 3)).

**Solution**

The angle π / 3 radians is a remarkable angle whose cosine is ½, so the problem boils down to finding arcsen (½).

Then it is about finding which is the angle whose sine gives ½. That angle is π / 6, since sin (π / 6) = sin (30º) = ½. Therefore arcsen (cos (π / 3)) = π / 6.

**Exercises**

**– Exercise 1**

Find the result of the following expression:

sec (arctan (3)) + csc (arccot (4))

**Solution**

We start by naming α = arctan (3) and β = arccot (4). Then the expression that we have to calculate looks like this:

sec (α) + csc (β)

The expression α = arctan (3) is equivalent to saying tan (α) = 3.

Since the tangent is the opposite leg over the adjacent one, a right triangle with leg opposite α of 3 units and an adjacent leg of 1 unit are constructed, so that tan (α) = 3/1 = 3.

In a right triangle the hypotenuse is determined by the Pythagorean theorem. With these values the result is √10, so that:

sec (α) = hypotenuse / adjacent leg = √10 / 1 = √10.

Similarly, β = arccot (4) is equivalent to affirming that cot (β) = 4.

We construct a right leg triangle adjacent to β of 4 units and an opposite leg of 1 unit, so that cot (β) = 4/1.

The triangle is immediately completed by finding its hypotenuse thanks to the Pythagorean theorem. In this case it turned out to have √17 units. Then the csc (β) = hypotenuse / opposite leg = √17 / 1 = √17 is calculated.

Remembering that the expression that we must calculate is:

sec (arctan (3)) + csc (arccot (4)) = sec (α) + csc (β) =…

… = √10 + √17 = 3.16 + 4.12 = 7.28.

**– Exercise 2**

Find the solutions of:

Cos (2x) = 1 – Sen (x)

**Solution**

It is necessary that all trigonometric functions are expressed in the same argument or angle. We will use the identity of the double angle:

Cos (2x) = 1 – 2 Sin ^{2} (x)

Then the original expression is reduced to:

1 – 2 Sin ^{2} (x) = 1 – Sin x

Once simplified and factored, it is expressed as:

sin (x) (2 sin (x) – 1) = 0

Which gives rise to two possible equations: Sen (x) = 0 with solution x = 0 and another equation sin (x) = ½ with x = π / 6 as solution.

The solutions to the proposed equation are: x = 0 or x = π / 6.

**– Exercise 3**

Find the solutions of the following trigonometric equation:

cos (x) = sin^{two}(x)

**Solution**

To solve this equation, it is convenient to place a single type of trigonometric function, so we will use the fundamental trigonometric identity so that the original equation is rewritten as follows:

cos (x) = 1 – cos ^{2}(x)

If we name y = cos (x), the expression can be rewritten as:

and ^{2} + and – 1 = 0

It is an equation of the second degree in y, whose solutions are:

y = (-1 ± √5) / 2

Then the values of x that satisfy the original equation are:

x = arccos ((-1 ± √5) / 2)

The real solution being the one with a positive sign x = 0.9046 rad = 51.83º.

The other solution is complex: x = (π – 1.06 i) rad.

**References**

- Hazewinkel, M. 1994. Encyclopaedia of Mathematics. Kluwer Academic Publishers / Springer Science & Business Media.
- Mate Mobile. Inverse trigonometric functions. Recovered from: matemovil.com
- Universe formulas. Inverse trigonometric functions. Recovered from: universoformulas.com
- Weisstein, Eric W. Inverse Trigonometric Functions. Recovered from: mathworld.wolfram.com
- Wikipedia. Inverse trigonometric functions. Recovered from: en.wikipedia.com