 # Graphical Representation of Inverse Trigonometric Functions

Inverse trigonometric functions can also be named as arc functions. They are mainly known as arcsine, arccosine, arctangent, arcsecant, arccotangent, and arccosecant. Similarly, like we presented a graph of trigonometric functions, we can represent graphs of inverse trigonometric functions. In trigonometric functions of sin, cos, and tan we had to find the length of the side of the right triangle when the length of one side was given and the measure of one of the acute angles was given. In inverse trigonometric function, we will have to find the measure of the angle when we are given length of the two sides.

Let us know what the formula for these functions is.

## Domain and Range of Inverse Trigonometric Formulas

 Function Domain Range x [-1,1] [-$\frac{\pi}{2}$, $\frac{\pi}{2}$] x [-1, 1] [0, $\pi$ ] x R [-$\frac{\pi}{2}$, $\frac{\pi}{2}$] x R [0, $\pi$] x R- [-1, 1] [0, $\pi$] – {$\frac{\pi}{2}$} x R- [-1, 1] [-$\frac{\pi}{2}$, $\frac{\pi}{2}$] – {0}

It is important to note the following formulas considering the domain and range of inverse function

• sin(sin-1x) = x, if -1 ≤ x ≤ 1 and sin-1(sin y) = y if -$\frac{\pi}{2}$ ≤ y ≤ $\frac{\pi}{2}$.

• cos(cos-1x) = x, if -1 ≤ x ≤ 1 and cos-1(cos y) = y if 0 ≤ y(arccos) ≤ π.

• tan(tan-1x) = x, if -∞ < x < ∞ and cos-1(cos y) = y if -$\frac{\pi}{2}$ ≤ y(arctan) ≤ $\frac{\pi}{2}$.

• cot(cot-1x) = x, if -∞ < x < ∞ and cot-1(cot y) = y if 0 < y <π.

• sec(sec-1x) = x, if -∞ ≤ x ≤ -1 or 1 ≤ x ≤ ∞ and sec-1(sec y) = y if -0 ≤ y ≤ π, y ≠ $\frac{\pi}{2}$.

• cosec(cosec-1x) = x, if -∞ ≤ x ≤- 1 or 1 ≤ x ≤ ∞ and cosec-1(cosec y) = y if -$\frac{\pi}{2}$ ≤ y ≤ $\frac{\pi}{2}$, y ≠ 0.

Inverse trigonometric functions are also known as ‘arc functions’ because, for a given value of the trigonometric function, they produce the length of arc needed to get the particular value.

### 1 - Arcsine Function

inverse sine function is defined as

y = arcsin x for – $\frac{\pi}{2}$ ≤ y ≤ $\frac{\pi}{2}$

y is the angle with sine x which means x = sin y

the graph of y = arcsin x

### 2 - Arccosine Function

The graph of cosine does not extend beyond the point you see in the graph (if it extended, there would be multiple values of y for each x value and we would no longer have a function). The start and endpoints are indicated with dots (-1,) and (1,0)

### 3 - Arctangent Function

This graph can extend beyond what you see in the positive and negative direction of x and it does not cross the dashed line.

The domain of arctan x is all values of x

The range for arctan x is  - $\frac{\pi}{2}$ < arctan x < $\frac{\pi}{2}$

### 4 - Arccotangent Function

The graph of arccotangent extends in the positive and negative x directions. As shown in the graph it does not stop at -8, 8

The domain of arccot x is all values of x

The range of arccot x is −2π ​< arccot x ≤ 2π​ (arccot x ≠ 0)

### 5 - Arcsecant Function

Here, in the graph of sec inverse x, the curve is defined outside of the portion between -1 and 1. The starting points (-1, π) and (1,0) with dots.

The domain of arcsec x is all values of x except −1 < x < 1

The range of arcsec x is 0 ≤ arcsec x ≤ π, arcsec x $\neq \frac{\pi}{2}$

### 6 - Arccosecant Function

The graph extends from positive and negative x direction and is not defined between – 1 and 1

The domain of arccsc x is all values of x except – 1 < x < 1

The range

The range of arccsc x is  - $\frac{\pi}{2}$ ≤ arc csc x ≤ $\frac{\pi}{2}$ , arccsc x $\neq 0$

### Solved Examples of Inverse Trigonometric Functions

1. Find the accurate value of each of the expression in  [0, 2$\pi$].

1. sin-1(−3$\sqrt{2}$)

2. cos-1(−2$\sqrt{2}$)

3. tan-1$\sqrt{3}$

solution:

a. We get – 3 $\sqrt{2}$ from 30 - 60 - 90 triangle. Therefore, the reference angle for 3 $\sqrt{2}$ would be 60°. As it is sine it is negative and must be in the third or fourth quadrant. Here the answer is either 4 $\frac{\pi}{3}$ or 5$\frac{\pi}{3}$

b. From the isosceles right triangle we get -2$\sqrt{2}$. The reference angle will be 45° as it is cosine and negative. The angle is either on the second or third quadrant. The answer is 3 $\frac{\pi}{4}$ or 5$\frac{\pi}{4}$

c. From the 30 - 60 - 90 triangle we get $\sqrt{3}$. For the reference angle 60°, a tangent is $\sqrt{3}$. In the first and third quadrant, the tangent is positive, therefore, the answer is $\frac{\pi}{3}$ or 4 $\frac{\pi}{3}$

Note: Every example here has two answers which can be a problem when finding a single inverse for each trigonometric function. So, we have to restrict the domain in which inverse is found.

2. Get the value of (1.1106).

Solution: let B = (1.1106)

Then tan B = 1.1106

B = 48°

Tan 48 = 1.1106

Therefore, (1.1106 ) = 48°.