Inverse Trigonometric Functions

Periodicity and Inverses

The trigonometric functions are periodic; hence, they are not one-to-one, and no inverses can be defined for the entire domain of a basic trigonometric function. By redefining each trigonometric function on a carefully chosen subset of its domain, the new function can be specified one-to-one and therefore has an inverse function.

Redefined Trigonometric Functions

The table shows domains chosen on which each trigonometric function is one-to-one:

Function $ f(x) = $ Domain Range Function $ f(x) = \(</td><td>Domain</td><td>Range</td></tr><tr><td>\) x \(</td><td>\) \(</td><td>[-1, 1]</td><td>\) x \(</td><td>\) (-, -] (0, ] \(</td><td>\) (- , -1] \(</td><td>[-1, 1]</td><td>\) x \(</td><td>\) $ R

Note that in each case, although the domain has been restricted, the entire range of the original function is retained.

Note also that in each case the restricted domain (sometimes called the principal domain) is the result of a choice. Other choices might be possible, and in the case of the secant and cosecant functions, no universal agreement exists. The choice used here is the one most commonly made in elementary calculus texts.

Definitions of Inverse Trigonometric Functions

  1. INVERSE SINE $ f(x) = ^{-1}x $ is defined by $ y = ^{-1}x $ if and only if $ x = y $ with $ -1 x $ and $ - y $ . The values the function takes on lie in quadrants I and IV.

  2. INVERSE COSINE $ f(x) = ^{-1} x $ is defined by $ y = ^{-1} x $ if and only if $ x = y $ with $ -1 x $ and $ 0 y $ . The values the function takes on lie in quadrants I and II.

  3. INVERSE TANGENT $ f(x) = ^{-1} x $ is defined by $ y = ^{-1} x $ is and only if $ x = y $ with $ x R $ and $ - < y < $ . The values the function takes on lie in quadrants I and IV.

  4. INVERSE COTANGENT $ f(x) = ^{-1} x $ is defined by $ y = ^{-1} x $ if and only if $ x = y $ with $ x R $ and $ 0 < y < $ . The values the function takes on lie in quadrants I and II.

  5. INVERSE SECANT $ f(x) = ^{-1}x $ is defined by $ y = ^{-1}x $ if and only if $ x = y $ with either $ x $ and $ 0 y < $ or $ x $ and $ y < $ . The values the function takes on lie in quadrants I and III.

  6. INVERSE COSECANT $ f(x) = ^{-1} x $ is defined by $ y = ^{-1} x $ if and only if $ x = y $ with either $ x $ and $ 0 < y $ or $ x $ and $ -< y - $ . The values the function takes on lie in quadrants I and III.

EXAMPLE 25.1 Evaluate (a) $ ^{-1} $ ; (b) $ ^{-1}(-) $ .

  1. $ y = ^{-1} $ is equivalent to $ y = $ , $ - y $ . The only solution of the equation on the interval is $ $ ; hence $ ^{-1} = $ .

  2. $ y = ^{-1}(-) $ is equivalent to $ y = - $ , $ - y $ . The only solution of the equation on the interval is $ - $ ; hence $ ^{-1}(-) = - $ . Note that this value is in quadrant IV.

EXAMPLE 25.2 Evaluate (a) $ ^{-1} $ ; (b) $ ^{-1}(-) $ .

  1. $ y = ^{-1} $ is equivalent to $ y = $ , $ 0 y $ . The only solution of the equation on the interval is $ $ ; hence $ ^{-1} = $ .

  2. $ y = ^{-1}(-) $ is equivalent to $ y = - $ , $ 0 y $ . The only solution of the equation on the interval is $ $ ; hence $ ^{-1}(-) = $ . Note that this value is in quadrant II.

Alternative Notation

The inverse trigonometric functions are also referred to as the arc functions. In this notation:

\[ \begin{aligned}\sin^{-1}x&=\arcsin x\quad&\cos^{-1}x&=\arccos x\quad&\tan^{-1}x&=\arctan x\\\csc^{-1}x&=\arccsc x\quad&\sec^{-1}x&=\arcsec x\quad&\cot^{-1}x&=\arccot x\end{aligned} \]

EXAMPLE 25.3 Evaluate arctan 1.

$ y = = ^{-1}1 $ is equivalent to $ y = 1 $ , $ - < y < $ . The only solution of the equation on the interval is $ $ ; hence $ = $ .

Phase-Shift Identity

Let A be any positive real number and B and x be any real numbers. Then

\[ A\cos bx+B\sin bx=C\cos(bx-d) \]

where $ C = $ and $ d = ^{-1} $ .

EXAMPLE 25.4 Write sinx + cosx in the form Ccos (bx - d).

Here A = B = 1; hence $ C = = = $ and $ d = ^{-1} = $ . It follows from the phase-shift identity that $ x + x = (x - ) $ .

SOLVED PROBLEMS

25.1. Sketch a graph of the sine function showing the interval of redefinition.

The redefined sine function is restricted to the domain $ [-/2, /2] $ . Draw a graph of the basic sine function, with the portion in this interval emphasized (see Fig. 25-1).

Figure 25-1

25.2. Sketch a graph of the cosine function showing the interval of redefinition.

The redefined cosine function is restricted to the domain $ [0, ] $ . Draw a graph of the basic cosine function, with the portion in this interval emphasized (see Fig. 25-2).

Figure 25-2

25.3. Sketch a graph of the inverse sine function.

The domain of the inverse sine function is the range of the (redefined) sine function: $ [-1, 1] $ . The range of the inverse sine function is the domain of the redefined sine function: $ [-/2, /2] $ . The graph of the inverse sine function is the graph of the redefined sine function, reflected in the line y = x. Form a table of values (see the table of values for the trigonometric functions in Problem 22.12) and sketch the graph (Fig. 25-3).

Figure 25-3

25.4. Sketch a graph of the inverse cosine function.

The domain of the inverse cosine function is the range of the (redefined) cosine function: $ [-1, 1] $ . The range of the inverse cosine function is the domain of the redefined cosine function: $ [0, ] $ . The graph of the inverse cosine function is the graph of the redefined cosine function, reflected in the line y = x. Form a table of values (see the table of values for the trigonometric functions in Problem 22.12) and sketch the graph (Fig. 25-4).

x -1 $ - \(</td><td>\) - \(</td><td>\) - \(</td><td>0</td></tr><tr><td>y</td><td>\) \(</td><td>\) \(</td><td>\) \(</td><td>\) \(</td><td>\) \(</td></tr><tr><td>x</td><td>1</td><td>\) \(</td><td>\) \(</td><td>\) \(</td><td></td></tr><tr><td>y</td><td>0</td><td>\) \(</td><td>\) \(</td><td>\) $

Figure 25-4

25.5. Sketch a graph of the tangent function showing the interval of redefinition.

The redefined tangent function is restricted to the domain $ (- /2, /2) $ . Draw a graph of the basic tangent function, with the portion in this interval emphasized (see Fig. 25-5).

Figure 25-5

25.6. Sketch a graph of the inverse tangent function

The domain of the inverse tangent function is the range of the (redefined) tangent function: R. The range of the inverse tangent function is the domain of the redefined tangent function: $ (- /2, /2) $ . The graph of the inverse tangent function is the graph of the redefined tangent function, reflected in the line y = x. Since the graph of the redefined tangent function has asymptotes at $ x = /2 $ , the reflected graph will have asymptotes at $ y = /2 $ . Form a table of values (see the table of values for the trigonometric functions in Problem 22.12) and sketch the graph (Fig. 25-6).

x -√3 -1 -1/√3 0
y -π/3 -π/4 -π/6 0
x √3 1 1/√3
y π/3 π/4 π/6

Figure 25-6

25.7. Sketch a graph of the secant function showing the interval of redefinition.

The redefined secant function is restricted to the domain \([0, \pi/2) \cup [\pi, 3\pi/2)\). Draw a graph of the basic secant function, with the portion in this interval emphasized.

Figure 25-7

25.8. Sketch a graph of the inverse secant function.

The domain of the inverse secant function is the range of the (redefined) secant function: $ (-∞, -1] ∪ [1, ∞) $ . The range of the inverse secant function is the domain of the redefined secant function $ [0, π/2) ∪ [π, 3π/2) $ . The graph of the inverse secant function is the graph of the redefined secant function, reflected in the line y = x. Since the graph of the redefined secant function has asymptotes at $ x = π/2 $ and $ x = 3π/2 $ , the reflected graph will have asymptotes at $ y = π/2 $ and $ y = 3π/2 $ . Form a table of values (see the table of values for the trigonometric functions in Problem 22.12) and sketch the graph (Fig. 25-8).

x -2 -√2 -2/√3 -1
y 4π/3 5π/4 7π/6 π
x 2 √2 2/√3 1
y π/3 π/4 π/6 0

Figure 25-8

25.9. Analyze the application of the function-inverse function relation to (a) the sine and inverse sine functions, (b) the cosine and inverse cosine functions, (c) the tangent and inverse tangent functions, and (d) the secant and inverse secant functions.

The function-inverse function relation (Chapter 13) states that if g is the inverse function of f, then $ g(f(x)) = x $ for all x in the domain of f, and $ f(g(y)) = y $ for all y in the domain of g. Hence

  1. $ ^{-1}(x) = x $ for all $ x, - x $ ; $ (^{-1}y) = y $ for all $ y, -1 y $ .

  2. $ ^{-1}(x) = x $ for all $ x, 0 x $ ; $ (^{-1}y) = y $ for all $ y, -1 y $ .

  3. $ ^{-1}(x)=x $ for all $ x,-<x<;(^{-1}y)=y $ for all $ yR $ .

  4. $ ^{-1}(x)=x $ for all $ x,0x< $ or $ <x $ ; $ (^{-1}y)=y $ for all $ y $ or $ y $ .

25.10. Simplify: (a) $ (^{-1}) $ (b) $ (^{-1}) $ ; (c) $ (^{-1}2) $ .

  1. Since $ -1 /2 $ , $ /2 $ is in the domain of the inverse sine function. Hence, applying the function-inverse-function relation, $ (^{-1}) = $ . Alternatively, note that $ ^{-1} = $ , hence $ (^{-1}) = = $ .

  2. Since $ -1 $ , $ $ is in the domain of the inverse sine function. Hence, applying the function-inverse-function relation, $ (^{-1}) = $ .

  3. Since 2 > 1, 2 is not in the domain of the inverse sine function. Hence, $ (^{-1}2) $ is undefined.

25.11. Simplify: (a) $ ^{-1}() $ ; (b) $ ^{-1}((-)) $ ; (c) $ ^{-1}() $

  1. Since $ -/2 < /6 < /2 $ , $ /6 $ is in the domain of the restricted tangent function. Hence, applying the function-inverse-function relationship, $ ^{-1}() = $ .

Alternatively, note that $ = $ , hence $ {-1}()={-1}= $ .

  1. Since $ -/2 < -1/4 < /2 $ , -1/4 is in the domain of the restricted tangent function. Hence, applying the function-inverse-function relationship, $ ^{-1}((-)) = - $ .

  2. Since $ 2/3 > /2 $ , $ 2/3 $ is not in the domain of the restricted tangent function, and the function-inverse-function relation cannot be used. However, $ 2/3 $ is in the domain of the general tangent function, thus:

\[ \tan^{-1}\left(\tan\frac{2\pi}{3}\right)=\tan^{-1}(-\sqrt{3})=-\frac{\pi}{3} \]

25.12. Simplify: (a) $ (^{-1}) $ ; (b) $ (^{-1}(-)) $ ; (c) $ (^{-1}(-)) $ ; (d) $ (^{-1}3) $

  1. Since $ $ is in the domain of the inverse sine function, let $ u = ^{-1} $ . Then, by the definition of the inverse sine, $ u = , - u $ . It follows from the Pythagorean identities that

\[ \cos\left(\sin^{-1}\frac{3}{5}\right)=\cos u=\sqrt{1-\sin^{2}u}=\sqrt{1-\left(\frac{3}{5}\right)^{2}}=\frac{4}{5} \]

Note that the positive sign is taken on the square root since \(u\) must be in quadrant I or IV, where the sign of the cosine is positive.

  1. Since $ - $ is the domain of the inverse cosine function, let $ u = ^{-1}(-) $ . Then, by the definition of the inverse cosine, $ u = -, 0 u $ . It follows from the Pythagorean identities that

\[ \sin\left(\cos^{-1}\left(-\frac{2}{3}\right)\right)=\sin u=\sqrt{1-\cos^{2}u}=\sqrt{1-\left(-\frac{2}{3}\right)^{2}}=\frac{\sqrt{5}}{3} \]

Note that the positive sign is taken on the square root since u must be in quadrant I or II, where the sign of the sine is positive.

  1. Since $ - $ is in the domain of the inverse secant function, let $ u = ^{-1}(-) $ . Then, by the definition of the inverse secant, $ u = -, u < $ (since $ u $ is negative). If follows from the Pythagorean identities that

\[ \tan\left(\sec^{-1}\left(-\frac{5}{2}\right)\right)=\tan u=\sqrt{\sec^{2}u-1}=\sqrt{\left(-\frac{5}{2}\right)^{2}-1}=\frac{\sqrt{21}}{2} \]

Note that the positive sign is taken on the square root since u must be in quadrant III, where the sign of the tangent is positive.

  1. Since 3 is not in the domain of the inverse cosine in function, $ (^{-1}3) $ is undefined.

25.13. Simplify: (a) $ ({-1}+{-1}) $ ; (b) $ ({-1}-{-1}) $

  1. Let $ u = ^{-1} $ and $ v = ^{-1} $ . Then $ u = $ and $ v = $ , $ - u $ , $ v $ . From the sum formula for sines, it follows that

\[ \sin\left(\sin^{-1}\frac{1}{3}+\sin^{-1}\frac{2}{3}\right)=\sin(u+v)=\sin u\cos v+\cos u\sin v \]

Now $ u $ and $ v $ are given. Proceeding as in the previous problem,

\[ \cos u=\sqrt{1-\sin^{2}u}=\sqrt{1-\left(\frac{1}{3}\right)^{2}}=\frac{2\sqrt{2}}{3}\qquad\cos v=\sqrt{1-\sin^{2}v}=\sqrt{1-\left(\frac{2}{3}\right)^{2}}=\frac{\sqrt{5}}{3} \]

Hence

\[ \sin\left(\sin^{-1}\frac{1}{3}+\sin^{-1}\frac{2}{3}\right)=\sin u\cos v+\cos u\sin v=\frac{1}{3}\cdot\frac{\sqrt{5}}{3}+\frac{2\sqrt{2}}{3}\cdot\frac{2}{3}=\frac{\sqrt{5}+4\sqrt{2}}{9} \]

  1. Let $ u = ^{-1} $ and $ v = ^{-1} $ . Then $ u = $ , $ 0 u $ , and $ v = $ , $ - v $ . From the difference formula for tangents, if follows that

\[ \tan\left(\cos^{-1}\frac{3}{5}-\sin^{-1}\frac{5}{6}\right)=\tan\left(u-v\right)=\frac{\tan u-\tan v}{1+\tan u\tan v} \]

From the Pythagorean and quotient identities

\[ \tan u=\frac{\sin u}{\cos u}=\frac{\sqrt{1-\cos^{2}u}}{\cos u}=\frac{\sqrt{1-(3/5)^{2}}}{3/5}=\frac{4}{3} \]

\[ \tan\nu=\frac{\sin\nu}{\cos\nu}=\frac{\sin\nu}{\sqrt{1-\sin^{2}\nu}}=\frac{5/6}{\sqrt{1-(5/6)^{2}}}=\frac{5}{\sqrt{11}} \]

Hence

\[ \tan\left(\cos^{-1}\frac{3}{5}-\ \sin^{-1}\frac{5}{6}\right)=\frac{\tan u-\ \tan v}{1+\ \tan u\tan v}=\frac{\frac{4}{3}-\frac{5}{\sqrt{11}}}{1+\left(\frac{4}{3}\right)\left(\frac{5}{\sqrt{11}}\right)}=\frac{4\sqrt{11}-15}{3\sqrt{11}+20}=\frac{125\sqrt{11}-432}{301} \]

25.14. Simplify (a) $ (2^{-1}) $ ; (b) $ (^{-1}(-)) $ .

  1. Let $ u = ^{-1} $ . Then $ u = , 0 u $ . From the double-angle formula for cosines, it follows that

\[ \cos\left(2\cos^{-1}\frac{5}{13}\right)=\cos2u=2\cos^{2}u-1=2\left(\frac{5}{13}\right)^{2}-1=-\frac{119}{169} \]

  1. Let $ u = ^{-1}(-) $ . Then $ u = - $ , $ - u $ (since $ u $ is negative). From the half-angle formula for sines, it follows that

\[ \sin\left(\frac{1}{2}\sin^{-1}\left(-\frac{7}{25}\right)\right)=\sin\left(\frac{1}{2}u\right)=-\sqrt{\frac{1-\cos u}{2}} \]

where the negative sign is taken on the square root since $ - $ . From the Pythagorean identity,

$ u = $ , hence

\[ \sin\left(\frac{1}{2}\sin^{-1}\left(-\frac{7}{25}\right)\right)=-\sqrt{\frac{1-\cos u}{2}}=-\sqrt{\frac{1-\frac{24}{25}}{2}}=-\frac{1}{\sqrt{50}} \]

25.15. Find an algebraic expression for sin(cos $ ^{-1}x $ )

Let \(u = \cos^{-1} x\). Then \(\cos u = x, 0 \leq u \leq \pi\). It follows from the Pythagorean identities that

\[ \sin(\cos^{-1}x)=\sin u=\sqrt{1-\cos^{2}u}=\sqrt{1-x^{2}} \]

where the positive sign is taken on the square root, since u must be in quadrant I or II, where the sign of the sine is positive.

25.16. Derive the phase-shift formula

Given a quantity of form $ A bx + B bx $ with A any positive real number and B and x any real numbers, let $ C = $ and $ d = ^{-1} $ . From the Pythagorean and quotient identities, $ d = $ and

$ d = $ . From algebra, it follows that:

\[ A\cos bx+B\sin bx=\frac{\sqrt{A^{2}+B^{2}}}{\sqrt{A^{2}+B^{2}}}\left(A\cos bx+B\sin bx\right)=\sqrt{A^{2}+B^{2}}\Bigg(\frac{A}{\sqrt{A^{2}+B^{2}}}\cos bx+\frac{B}{\sqrt{A^{2}+B^{2}}}\sin bx\Bigg) \]

Hence.

\[ A\cos bx+B\sin bx=\sqrt{A^{2}+B^{2}}(\cos d\cos bx+\sin d\sin bx) \]

By the difference formula for cosine, the quantity in parentheses must equal cos(bx - d), hence,

\[ A\cos bx+B\sin bx=\sqrt{A^{2}+B^{2}}\cos(bx-d)=C\cos(bx-d) \]

25.17. (a) Use the phase-shift formula to rewrite $ 3x + 3x $

  1. Draw a graph of $ f(x)=3x+3x $ using the result of part (a).

  2. Let $ A = B = $ , then $ = = 2 $ and

^{-1} $ $ = tan^{-1} $ $ = tan^{-1}1 = $ $ . It follows that

\[ \sqrt{2}\cos3x+\sqrt{2}\sin3x=2\cos\left(3x-\frac{\pi}{4}\right) \]

  1. To sketch $ y = f(x) = 2 ( 3x - ) $ , note amplitude = 2. The graph (Fig. 25-9) is a basic cosine curve. Period $ = $ . Phase shift $ = = $ . Divide the interval from $ $ to $ $ (= phase shift + one period) into four equal subintervals and sketch the curve with maximum height 2 and minimum height -2.

Figure 25-9

SUPPLEMENTARY PROBLEMS

25.18. Evaluate: (a) $ ^{-1}(-) $ ; (b) $ ^{-1}(-) $ ; (c) $ ^{-1}(-) $ ; (d) $ ^{-1}(-) $

Ans. (a) $ - $ ; (b) $ $ ; (c) $ - $ ; (d) $ $

25.19. Evaluate: (a) $ (^{-1}) $ ; (b) $ (^{-1}(-)) $ ; (c) $ (^{-1}0) $ ; (d) $ (^{-1}) $

Ans. (a) $ $ ; (b) $ - $ ; (c) 0; (d) not defined

25.20. Evaluate: (a) $ ^{-1}() $ ; (b) $ ^{-1}() $ ; (c) $ ^{-1}() $ ; (d) $ ^{-1}((-)) $

Ans. (a) $ $ ; (b) 1; (c) $ - $ ; (d) $ $

25.21. Evaluate: (a) $ (^{-1}) $ ; (b) $ (^{-1}2) $ ; (c) $ (^{-1}(-3)) $ ; (d) $ (^{-1}(-)) $

Ans. (a) $ $ ; (b) $ $ ; (c) $ $ ; (d) $ $

25.22. Evaluate: (a) $ ({-1}+{-1}) $ ; (b) $ ({-1}-{-1}) $ ; (c) $ ({-1}+{-1}) $

Ans. (a) $ $ ; (b) $ $ ; (c) $ $

25.23. Evaluate: (a) $ (2^{-1}) $ ; (b) $ (2^{-1}) $ ; (c) $ (^{-1}) $ ; (d) $ (^{-1}) $

Ans. (a) $ $ ; (b) $ $ ; (c) $ $ ; (d) $ $

25.24. Simplify: (a) $ (^{-1}x) $ ; (b) $ (^{-1}x) $ ; (c) $ (2^{-1}x) $ ; (d) $ (^{-1}x) $

Ans. (a) $ $ ; (b) $ $ ; (c) $ $ ; (d) $ $

25.25. (a) Show that for -1 ≤ x ≤ 1, -π/2 ≤ sin⁻¹x + cos⁻¹x ≤ π/2. (b) Show that sin(sin⁻¹x + cos⁻¹x) = 1.

  1. From (a) and (b), deduce that for $ -1 x $ , $ ^{-1}x + ^{-1}x = $ .

25.26. By making the substitution $ u = ^{-1} $ , simplify: (a) $ $ ; (b) $ $ .

Ans. (a) $ 3 u $ ; (b) $ 3 u u $

25.27. By making the substitution $ u = ^{-1} $ , simplify; (a) $ $ ; (b) $ $ .

Ans. (a) 4 secu; (b) $ $

25.28. By making the substitution $ u = ^{-1} $ , simplify: (a) $ x = x $ ; (b) $ (x^{2} - 4)^{3/2} $

Ans. (a) $ 4 u u $ ; (b) $ 8 ^{3} u $

25.29. Given $ y = 3 ^{-1}(x - 5) $ , (a) state the possible values of x and y; (b) solve for x in terms of y.

\[ Ans\quad(a)4\leq x\leq6,-3\pi/2\leq y\leq3\pi/2;(b)x=5+\sin(y/3) \]

25.30. Use the phase-shift formula to rewrite:

  1. $ 6 3x - 6 3x $ ; (b) $ 3 4x + 4x $ ; (c) $ 3 x + 4 x $

\[ Ans.\quad(a)6\sqrt{2}\cos\Big(3x+\frac{\pi}{4}\Big);(b)2\sqrt{3}\cos\Big(4x-\frac{\pi}{6}\Big);(c)5\cos\Big(\frac{1}{2}x-\tan^{-1}\frac{4}{3}\Big) \]