We will now define those inverse functions, \(\boxed{sin^{−1}}\), \(\boxed{cos^{−1}}\), and \(\boxed{tan^{−1}}\) , and determine their graphs.

Recall that a **function** is a rule that assigns a single object *y* from one set (the **range**) to each object *x* from another set (the **domain**). We can write that rule as *y* = *f*(*x*), where *f* is the function (see Figure 1). There is a simple *vertical rule* for determining whether a rule *y* = *f*(*x*) is a function: *f* is a function if and only if every vertical line intersects the graph of *y* = *f*(*x*) in the *xy*-coordinate plane at most once (see Figure 2).

Recall that a function *f* is **one-to-one** (often written as 1−1) if it assigns distinct values of *y* to distinct values of *x*. In other words, if *x*_{1} \(\ne\) *x*_{2} then *f*(*x*_{1}) \(\ne\) *f*(*x*_{2}). Equivalently, *f* is one-to-one if *f*(*x*_{1}) = *f*(*x*_{2}) implies *x*_{1} = *x*_{2}. There is a simple *horizontal rule* for determining whether a function *y* = *f*(*x*) is one-to-one: *f* is one-to-one if and only if every horizontal line intersects the graph of *y* = *f*(*x*) in the *xy*-coordinate plane at most once (see Figure 3).

If a function *f* is one-to-one on its domain, then *f* has an **inverse function**, denoted by f^{−1}, such that *y* = *f*(*x*) if and only if *f*^{−1}(*y*) = *x*. The domain of *f*^{−1} is the range of *f*.

The basic idea is that *f*^{−1} “undoes” what f does, and vice versa. In other words,
*f*^{−1}(*f*(*x*)) = *x* for all *x* in the domain of *f*, and
*f*(*f*^{−1}(*y*)) = *y* for all *y* in the range of *f*.

We know from their graphs that none of the trigonometric functions are one-to-one over their entire domains. However, we can restrict those functions to *subsets* of their domains where they *are* one-to-one. For example, *y* = sin *x* is one-to-one over the interval \(\left[−\frac{π}{2}, \frac{π}{2} \right]\), as we see in the graph below:

For −\(\frac{π}{2}\) ≤ *x* ≤ \(\frac{π}{2}\) we have −1 ≤ sin *x* ≤ 1, so we can define the **inverse sine** function *y* = sin^{−1} *x* (sometimes called the **arc sine** and denoted by *y* = arcsin *x*) whose domain is the interval [−1,1] and whose range is the interval \(\left[−\frac{π}{2}, \frac{π}{2} \right]\). In other words:

$$\text{sin}^{-1}(\text{sin } y) = y \text{ for } −\frac{π}{2} ≤ y ≤ \frac{π}{2}\; \; \; (1)$$
$$\text{sin}(\text{sin}^{-1} x) = x \text{ for } −1 ≤ x ≤ 1 \; \; \; \; (2)$$

Find sin^{−1} \(\left(sin \frac{π}{4}\right)\).

**Solution**: Since \(−\frac{π}{2} ≤ \frac{π}{4} ≤ \frac{π}{2}\), we know that sin^{−1} \(\left(sin \frac{π}{4}\right)\) = \(\boxed{\frac{π}{4}}\), by formula (1).

**Example 2**

Find sin^{−1} \(\left(sin \frac{5π}{4}\right)\).

**Solution**: Since \(\frac{5π}{4} > \frac{π}{2}\), we cannot use formula (1). But we know that sin \(\frac{5π}{4} = −\frac{1}{\sqrt{2}}\). Thus, sin^{−1} sin \(\left(sin \frac{5π}{4}\right)\) = sin^{−1} \(\left(\frac{1}{\sqrt{2}} \right)\) is, by definition, the angle *y* such that \(−\frac{π}{2} ≤ y ≤ \frac{π}{2}\) and sin *y* = −\(\frac{1}{\sqrt{2}}\). That angle is *y* = \(−\frac{π}{4}\), since

$$sin \left(−\frac{π}{2}\right) = −sin \left(\frac{π}{2}\right) = \frac{1}{\sqrt{2}}$$

Thus, sin^{−1 } \(\left(sin \frac{5π}{4}\right) = −\boxed{\frac{π}{4}}\).

Example 2 illustrates an important point: sin^{−1} *x* should *always* be a number between \(−\frac{π}{2}\) and \(\frac{π}{2}\) . If you get a number outside that range, then you made a mistake somewhere. This why we get sin^{−1}(−0.682) = −43º when using the \(\boxed{sin^{−1}}\) button on a calculator. Instead of an angle between 0º and 360º (i.e. 0 to 2π radians) we got an angle between −90º and 90º (i.e. \(−\frac{π}{2}\) to \(\frac{π}{2}\) radians).

In general, the graph of an inverse function *f*^{−1} is the reflection of the graph of *f* around the line *y* = *x*. The graph of *y* = sin^{−1} *x* is shown in Figure 5. Notice the symmetry about the line *y* = *x* with the graph of *y* = sin *x*.

The **inverse cosine** function *y* = cos^{−1} *x* (sometimes called the **arc cosine** and denoted by *y* = arccos *x*) can be determined in a similar fashion. The function *y* = cos *x* is one-to-one over the interval [0,*π*], as we see in the graph below:

Thus, y = cos^{−1} *x* is a function whose domain is the interval [−1,1] and whose range is the interval [0,*π*]. In other words:

$$\text{cos}^{-1}(\text{cos } y) = y \text{ for } 0 ≤ y ≤ π \; \; \; \; (3)$$
$$\text{cos}(\text{cos}^{-1} x) = x \text{ for } −1 ≤ x ≤ 1 \; \; \; \; (4)$$

The graph of *y* = cos^{−1} *x* is shown below in Figure 9. Notice the symmetry about the line *y* = *x* with the graph of *y* = cos *x*.

Find cos^{−1} \(\left(cos \frac{π}{3}\right)\).

**Solution**: Since 0 ≤ \(\frac{π}{3}\) ≤ *π*, we know that cos^{−1} \(\left(cos \frac{π}{3}\right)\) = \(\boxed{\frac{π}{3}}\), by formula (3).

**Example 4**

Find cos^{−1} \(\left(cos \frac{4π}{3}\right)\).

**Solution**: Since \(\frac{4π}{3}\) > *π*, we cannot use formula (3). But we know that cos \(\frac{4π}{3}\) = −\(\frac{1}{2}\). Thus, cos^{−1} \(\left(cos \frac{4π}{3}\right)\) = cos^{−1} \(\left(−cos \frac{1}{2}\right)\) is, by definition, the angle *y* such that 0 ≤ *y* ≤ *π* and cos *y* = −\(\frac{1}{2}\). That angle is *y* = \(\frac{2π}{3}\) (i.e. 120º). Thus, \(\left(cos \frac{4π}{3}\right)\) = \(\frac{2π}{3}\)

Examples 2 and 4 may be confusing, since they seem to violate the general rule for inverse functions that *f*^{−1}(*f*(*x*)) = *x* for all *x* in the domain of *f*. But that rule only applies when the function *f* is one-to-one over its *entire* domain. We had to restrict the sine and cosine functions to very small subsets of their entire domains in order for those functions to be one-to-one. That general rule, therefore, only holds for *x* in those small subsets in the case of the inverse sine and inverse cosine.

The **inverse tangent** function *y* = tan^{−1} *x* (sometimes called the **arc tangent** and denoted by *y* = arctan *x*) can be determined similarly. The function *y* = tan *x* is one-to-one over the interval \(\left(−\frac{π}{2}, \frac{π}{2} \right)\), as we see in Figure 8:

The graph of *y* = tan^{−1} *x* is shown below in Figure 9. Notice that the vertical asymptotes for *y* = tan *x* become horizontal asymptotes for *y* = tan^{−1} *x*. Note also the symmetry about the line *y* = *x* with the graph of *y* = tan *x*.

Thus, *y* = tan^{−1} *x* is a function whose domain is the set of all real numbers and whose range is the interval \(\left(−\frac{π}{2}, \frac{π}{2} \right)\). In other words:

$$\text{tan}^{-1}(\text{tan } y) = y \text{ for } −\frac{π}{2} ≤ y ≤ \frac{π}{2} \; \; \; \; (5)$$ $$\text{tan}(\text{tan}^{-1} x) = x \text{ for } \text{all real }x \; \; \; \; (6)$$

**Example 5**

Find tan^{−1} \(\left(tan \frac{π}{4}\right)\).

**Solution**: Since \(−\frac{π}{2} ≤ \frac{π}{4} ≤ \frac{π}{2}\), we know that tan^{−1} \(\left(tan \frac{π}{4}\right)\) = \(\frac{π}{4}\), by formula (5).

**Example 6**

Find tan^{−1} (tan *π*).

**Solution**: Since *π* > \(\frac{π}{2}\) , we cannot use formula (5). But we know that tan *π* = 0. Thus, tan^{−1} (tan *π*) = tan^{−1}0 is, by definition, the angle *y* such that \(−\frac{π}{2} ≤ y ≤ \frac{π}{2}\) and tan *y* = 0. That angle is y = 0. Thus, tan^{−1} (tan *π*) = 0 .

**Example 7**

Find the exact value of \(cos\left(sin^{-1} \left(−\frac{1}{4} \right)\right)\).

**Solution**: Let *θ* = \(sin^{-1} \left(−\frac{1}{4} \right)\). We know that \(−\frac{π}{2} ≤ θ ≤ \frac{π}{2}\), so since sin *θ* = \(−\frac{1}{4}\) < 0, *θ* must be in QIV. Hence cos *θ* > 0. Thus,

$$cos^2 θ = 1 - sin^2 θ = 1 − \left(−\frac{1}{4} \right)^2 = \frac{15}{16} ⇒ cos θ = \frac{\sqrt{15}}{4}$$

Note that we took the positive square root above since cos *θ* > 0. Thus, \(cos\left(sin^{-1} \left(−\frac{1}{4} \right)\right)\) = \(\boxed{\frac{\sqrt{15}}{4}}\).

**Example 8**

Show that tan (sin^{−1} *x*) = \(\frac{x}{\sqrt{1 - x^2}}\) for −1 < *x* < 1.

**Solution**: When *x* = 0, the formula holds trivially, since

$$tan (sin^{-1}0) = tan 0 = 0 = \frac{0}{\sqrt{1 - 0^2}} $$

Now suppose that 0 < *x* < 1. Let *θ* = sin^{−1} *x*. Then *θ* is in QI and sin *θ* = *x*. Draw a right triangle with an angle *θ* such that the opposite leg has length *x* and the hypotenuse has length 1, as in Figure 10 (note that this is possible since 0 < *x* < 1). Then sin *θ* = \(\frac{x}{1}\) = *x*. By the Pythagorean Theorem, the adjacent leg has length \(\sqrt{1 - x^2}\). Thus, tan *θ* = \(\frac{x}{\sqrt{1 - x^2}}\)

If −1 < *x* < 0 then *θ* = sin^{−1} *x* is in QIV. So we can draw the same triangle except that it would be “upside down” and we would again have tan *θ* = \(\frac{x}{\sqrt{1 - x^2}}\), since the tangent and sine have the same sign (negative) in QIV. Thus, tan (sin^{−1} *x*) = \(\frac{x}{\sqrt{1 - x^2}}\) for −1 < *x* < 1.

The inverse functions for cotangent, cosecant, and secant can be determined by looking at their graphs. For example, the function *y* = cot *x* is one-to-one in the interval (0,*π*), where it has a range equal to the set of all real numbers. Thus, the **inverse cotangent** *y* = cot^{−1} *x* is a function whose domain is the set of all real numbers and whose range is the interval (0,*π*). In other words:

$$\text{cot}^{-1}(\text{cot } y) = y \text{ for } 0 ≤ y ≤ π \; \; \; \; (7)$$ $$\text{cot}(\text{cot}^{-1} x) = x \text{ for } \text{all real }x \; \; \; \; (8)$$

The graph of *y* = cot^{−1} *x* is shown below in Figure 11.

Similarly, it can be shown that the **inverse cosecant** *y* = csc^{−1} *x* is a function whose domain is |*x*| ≥ 1 and whose range is \(−\frac{π}{2} ≤ y ≤ \frac{π}{2}\), y \(\ne\) 0. Likewise, the inverse secant y = sec^{−1} *x* is a function whose domain is |*x*| ≥ 1 and whose range is 0 ≤ *y* ≤ *π*, y \(\ne \frac{π}{2}\).

$$\text{csc}^{-1}(\text{csc } y) = y \text{ for } −\frac{π}{2} ≤ y ≤ \frac{π}{2}, y \ne 0 \; \; \; \; (9)$$ $$\text{csc}(\text{csc}^{-1} x) = x \text{ for } |x| ≥ 1 \; \; \; \; (10)$$

$$\text{sec}^{-1}(\text{sec } y) = y \text{ for } 0 ≤ y ≤ π, y \ne \frac{π}{2} \; \; \; \; (11)$$ $$\text{sec}(\text{sec}^{-1} x) = x \text{ for } |x| ≥ 1 \; \; \; \; (12)$$

It is also common to call cot^{−1} *x*, csc^{−1} *x*, and sec^{−1} *x* the **arc cotangent**, **arc cosecant**, and **arc secant**, respectively, of *x*. The graphs of *y* = csc^{−1} *x* and *y* = sec^{−1} *x* are shown in Figure 12:

**Example 9**

Prove the identity tan^{−1} *x* + cot^{−1} *x* = \(\frac{π}{2}\).

**Solution**: Let *θ* = cot^{−1} *x*. We have

$$tan \left(\frac{π}{2} − θ\right) = − tan \left(θ − \frac{π}{2}\right) = cot θ = cot(cot^{-1}x) = x $$

by formula (8). So since tan (tan^{−1} *x*) = *x* for all *x*, this means that tan (tan^{−1} *x*) = tan \(\left(\frac{π}{2}−θ\right)\). Thus, tan (tan^{−1} *x*) = tan (\(\frac{π}{2}\) −cot^{−1} *x*). Now, we know that 0 < cot^{−1} *x* < *π*, so −\(\frac{π}{2}\) < \(\frac{π}{2}\) − cot^{−1} *x* < \(\frac{π}{2}\), i.e. \(\frac{π}{2}\) −cot^{−1} *x* is in the restricted subset on which the tangent function is one-to-one. Hence, tan (tan^{−1} *x*) = tan (\(\frac{π}{2}\) −cot^{−1} *x*) implies that tan^{−1} *x* = \(\frac{π}{2}\) −cot^{−1} *x*, which proves the identity.

**Example 10**
Is tan^{−1} a + tan^{−1} b = tan^{−1} \(\left(\frac{a + b}{1−ab}\right)\) an identity?

**Solution**: In the tangent addition formula tan (*A* +*B*) = \(\frac{tan A + tan B}{1 − tan A tan B}\), let *A* = tan^{−1} *a* and *B* = tan^{−1} *b*. Then

$$tan(tan^{-1}a + tan^{-1}b) = \frac{tan(tan^{-1}a) + tan(tan^{-1}b)}{1 - tan(tan^{-1}a) tan(tan^{-1}b)} $$ $$\frac{a + b}{1 - ab} \text{by formula (6) so it seems that we have}$$ $$tan^{-1}a + tan^{-1}b = tan^{-1}\left(\frac{a + b}{1 - ab} \right)$$

by definition of the inverse tangent. However, recall that −\(\frac{π}{2}\) < tan^{−1} *x* < \(\frac{π}{2}\) for all real numbers *x*. So in particular, we must have −\(\frac{π}{2}\) < tan^{−1} \(\left(\frac{a + b}{1−ab}\right)\) < \(\frac{π}{2}\). But it is possible that tan^{−1} *a* + tan^{−1} *b* is *not* in the interval ¡ \(\left(−\frac{π}{2}, \frac{π}{2}\right)\). For example,

$$tan^{-1}1 + tan^{-1}2 = 1.892547 > \frac{π}{2} ≈ 1.570796 .$$

And we see that \(\left(\frac{1 + 2}{1−(1)(2)}\right)\) = tan^{−1}(−3) = −1.249045 \(\ne\) tan^{−1} 1 + tan^{−1} 2. So the formula is only true when −\(\frac{π}{2}\) < tan^{−1} *a* + tan^{−1} *b* < \(\frac{π}{2}\).