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 x1 \(\ne\) x2 then f(x1) \(\ne\) f(x2). Equivalently, f is one-to-one if f(x1) = f(x2) implies x1 = x2. 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:
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:
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−10 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 < \(\frac{π}{2}\).