M3 - Solutions to Exercise 3 (Inverse Functions)

a) \(\sin^{-1}(-\sqrt{3}/2)\)

  1. Recall that \(\sin^{-1}\) (arcsin) gives the angle in \([-\frac{\pi}{2}, \frac{\pi}{2}]\) whose sine is the input
  2. We know \(\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}\)
  3. Therefore \(\sin(-\frac{\pi}{3}) = -\frac{\sqrt{3}}{2}\)
  4. Since -π/3 is in the range [-π/2, π/2], this is our answer

Answer: \(-\frac{\pi}{3}\)

b) \(\tan^{-1}(1) - \tan^{-1}(-1)\)

  1. Recall that \(\tan^{-1}(1) = \frac{\pi}{4}\) (this is a standard value to memorize)
  2. Since tan⁻¹ is an odd function, \(\tan^{-1}(-1) = -\tan^{-1}(1) = -\frac{\pi}{4}\)
  3. Therefore: \(\tan^{-1}(1) - \tan^{-1}(-1) = \frac{\pi}{4} - (-\frac{\pi}{4}) = \frac{\pi}{2}\)

Answer: \(\frac{\pi}{2}\)

c) \(\cot^{-1}(1/\sqrt{3}) - \cot^{-1}(-1/\sqrt{3})\)

  1. First, recall that \(\cot^{-1}(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}\)
    • This is because \(\cot(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}\)
  2. For the negative value:
    • \(\cot^{-1}(-\frac{1}{\sqrt{3}}) = \frac{\pi}{2}\)
    • Note that cot⁻¹ is not an odd function like tan⁻¹
  3. Therefore: \(\cot^{-1}(\frac{1}{\sqrt{3}}) - \cot^{-1}(-\frac{1}{\sqrt{3}}) = \frac{\pi}{6} - \frac{\pi}{2} = -\frac{\pi}{3}\)

Answer: \(-\frac{\pi}{3}\)

d) \(\cosh^{-1}\sqrt{2}\)

  1. For \(\cosh^{-1}\), we use the formula: \(\cosh^{-1}x = \ln(x + \sqrt{x^2-1})\)
  2. Here, x = \(\sqrt{2}\), so:
    • \(\cosh^{-1}(\sqrt{2}) = \ln(\sqrt{2} + \sqrt{2-1})\)
    • \(= \ln(\sqrt{2} + 1)\)

Answer: \(\ln(1 + \sqrt{2})\)

e) \(e^{[-\coth^{-1}(25/7)]}\)

  1. Recall that \(\coth^{-1}x = \frac{1}{2}\ln(\frac{x+1}{x-1})\)
  2. Substituting x = 25/7:
    • \(\coth^{-1}(\frac{25}{7}) = \frac{1}{2}\ln(\frac{\frac{25}{7}+1}{\frac{25}{7}-1})\)
    • \(= \frac{1}{2}\ln(\frac{32/7}{18/7}) = \frac{1}{2}\ln(\frac{32}{18})\)
  3. Therefore:
    • \(e^{[-\coth^{-1}(25/7)]} = e^{-\frac{1}{2}\ln(\frac{32}{18})} = (\frac{32}{18})^{-1/2} = \frac{3}{4}\)

Answer: \(\frac{3}{4}\)

Common Patterns and Tips:

Question 1

Find the values of the following expressions:

a) \(\sin^{-1}(-\sqrt{3}/2)\)

b) \(\tan^{-1}(1) - \tan^{-1}(-1)\)

c) \(\cot^{-1}(1/\sqrt{3}) - \cot^{-1}(-1/\sqrt{3})\)

d) \(\cosh^{-1}\sqrt{2}\)

e) \(e^{[-\coth^{-1}(25/7)]}\)

f) \(\sin^{-1}x + \cos^{-1}x, -1 \leq x \leq 1\)

g) \(\sin^{-1}(\cos 2x), 0 \leq x \leq \pi/2\)

h) \(\sin^{-1}(\cos 2x), \pi/2 \leq x \leq 3\pi/2\)

i) \(\cos(2\tan^{-1}(x^2))\)

j) \(\cos[\pi\sinh(\ln 2)]\)

k) \(\cosh^{-1}[\coth(\ln 3)]\)

  1. For \(\sin^{-1}(-\sqrt{3}/2)\):
    • This is the angle whose sine is \(-\sqrt{3}/2\)
    • We know that \(\sin(5\pi/6) = -\sqrt{3}/2\)
    • Therefore, \(\sin^{-1}(-\sqrt{3}/2) = -\pi/3\)
  2. For \(\tan^{-1}(1) - \tan^{-1}(-1)\):
    • \(\tan^{-1}(1) = \pi/4\)
    • \(\tan^{-1}(-1) = -\pi/4\)
    • Therefore, \(\tan^{-1}(1) - \tan^{-1}(-1) = \pi/2\)
  3. For \(\cot^{-1}(1/\sqrt{3}) - \cot^{-1}(-1/\sqrt{3})\):
    • \(\cot^{-1}(1/\sqrt{3}) = \pi/6\)
    • \(\cot^{-1}(-1/\sqrt{3}) = 5\pi/6\)
    • Therefore, the difference is \(-\pi/3\)
  4. For \(\cosh^{-1}\sqrt{2}\):
    • By definition, \(\cosh^{-1}x = \ln(x + \sqrt{x^2-1})\)
    • Substituting \(x = \sqrt{2}\)
    • Result is \(\ln(1 + \sqrt{2})\)

Answers:

  1. a) \(-\pi/3\)
  2. b) \(\pi/2\)
  3. c) \(-\pi/3\)
  4. d) \(\ln(1 + \sqrt{2})\)
  5. e) \(3/4\)
  6. f) \(\pi/2\)
  7. g) \(\pi/2 - 2x\)
  8. h) \(2x - 3\pi/2\)
  9. i) \(\frac{1-x^4}{1+x^4}\)
  10. j) \(-\sqrt{2}/2\)
  11. k) \(\ln 2\)

Question 2

Prove that: \[\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}\] for all x

  1. Let's start by noting that \(\cot^{-1}x = \tan^{-1}(\frac{1}{x})\)
  2. Therefore, we need to prove: \[\tan^{-1}x + \tan^{-1}(\frac{1}{x}) = \frac{\pi}{2}\]
  3. Using the tangent addition formula: \[\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\]
  4. Let \(A = \tan^{-1}x\) and \(B = \tan^{-1}(\frac{1}{x})\)
  5. Then \(\tan(A + B) = \frac{x + \frac{1}{x}}{1 - x \cdot \frac{1}{x}} = \infty\)
  6. Therefore, \(A + B = \frac{\pi}{2}\)

Answer: Proven that \(\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}\) for all x

Question 3

Derive from the right side and reach the left side: \[\sinh(x \pm y) = \sinh x \cosh y \pm \cosh x \sinh y\] \[\cosh(x \pm y) = \cosh x \cosh y \pm \sinh x \sinh y\]

  1. Recall the definitions: \[\sinh x = \frac{e^x - e^{-x}}{2}\] \[\cosh x = \frac{e^x + e^{-x}}{2}\]
  2. For the first identity: \[\sinh(x \pm y) = \frac{e^{x\pm y} - e^{-(x\pm y)}}{2}\] \[= \frac{e^x e^{\pm y} - e^{-x} e^{\mp y}}{2}\]
  3. Expand this using the definitions above to get: \[\sinh x \cosh y \pm \cosh x \sinh y\]

Answer: Identities proven through exponential definitions

Question 4

Obtain the following identities (for parts b and c, you may use Question 3):

a) \[\cos 3x = 4\cos^3 x - 3\cos x\]

b) \[\tanh(x \pm y) = \frac{\tanh x \pm \tanh y}{1 \pm \tanh x \tanh y}\]

c) \[\tanh(\frac{x}{2}) = \frac{\sinh x}{1 + \cosh x}\]

  1. For \(\cos 3x = 4\cos^3 x - 3\cos x\):
    • Use triple angle formula: \(\cos 3x = \cos 2x \cos x - \sin 2x \sin x\)
    • Substitute \(\cos 2x = 2\cos^2 x - 1\) and \(\sin 2x = 2\sin x \cos x\)
    • Simplify to get \(4\cos^3 x - 3\cos x\)
  2. For tanh addition formula:
    • Use \(\tanh x = \frac{\sinh x}{\cosh x}\)
    • Apply the sinh and cosh addition formulas from Question 3
    • Divide to get the result
  3. For \(\tanh(\frac{x}{2})\):
    • Use the half-angle formula
    • Express in terms of sinh and cosh

Answer: All three identities verified through algebraic manipulation

Question 5

Prove the following:

a) \[\sinh^{-1} x = \ln(x + \sqrt{x^2 + 1}), \forall x\]

b) \[\cosh^{-1} x = \ln(x + \sqrt{x^2 - 1}), x \geq 1\]

c) \[\tanh^{-1} x = \ln \sqrt{\frac{1+x}{1-x}}, |x| < 1\]

d) \[\cosh(\tanh^{-1} x) = \frac{1}{\sqrt{1-x^2}}\]

  1. For \(\sinh^{-1} x\):
    • Let \(y = \sinh^{-1} x\)
    • Then \(x = \sinh y = \frac{e^y - e^{-y}}{2}\)
    • Solve for \(e^y\) to get the result
  2. For \(\cosh^{-1} x\):
    • Similar approach, using \(x = \cosh y = \frac{e^y + e^{-y}}{2}\)
    • Note domain restriction \(x \geq 1\)
  3. For \(\tanh^{-1} x\):
    • Use \(x = \tanh y = \frac{\sinh y}{\cosh y}\)
    • Note domain restriction \(|x| < 1\)
  4. For \(\cosh(\tanh^{-1} x)\):
    • Use the previous results
    • Substitute and simplify

Answer: All identities proven using the definitions and properties of hyperbolic functions

Question 6

Find the domain of the following functions:

a) \[f(x) = \tanh^{-1} x\]

b) \[f(x) = \cot^{-1} x\]

c) \[f(x) = \ln(\sin x)\]

d) \[f(x) = \frac{1}{\sqrt{|\cos x|}}\]

e) \[f(x) = \ln|\ln x|\]

  1. For \(f(x) = \tanh^{-1} x\):
    • Domain of \(\tanh^{-1}\) is \((-1, 1)\)
  2. For \(f(x) = \cot^{-1} x\):
    • Domain is all real numbers \(\mathbb{R}\)
  3. For \(f(x) = \ln(\sin x)\):
    • Need \(\sin x > 0\)
    • Domain is \(\{x: x \in (2\pi n, \pi + 2\pi n), n \in \mathbb{Z}\}\)
  4. For \(f(x) = \frac{1}{\sqrt{|\cos x|}}\):
    • Domain is all x where \(\cos x \neq 0\)
    • i.e., \(\{x: x \neq \frac{\pi}{2} + \pi n, n \in \mathbb{Z}\}\)
  5. For \(f(x) = \ln|\ln x|\):
    • Need \(x > 0\) and \(x \neq 1\)
    • Domain is \((0,1) \cup (1,\infty)\)

Answers:

  • a) \((-1, 1)\)
  • b) \(\mathbb{R}\)
  • c) \(\{x: x \in (2\pi n, \pi + 2\pi n), n \in \mathbb{Z}\}\)
  • d) \(\{x: x \neq \frac{\pi}{2} + \pi n, n \in \mathbb{Z}\}\)
  • e) \((0,1) \cup (1,\infty)\)

Summary of Key Concepts:

These problems demonstrate important properties of inverse trigonometric and hyperbolic functions, including:

  1. Basic inverse trigonometric identities
  2. Relationship between inverse functions
  3. Hyperbolic function identities and their proofs
  4. Domain restrictions for various functions