M3 - Solutions to Exercise 3 (Inverse Functions)
a) \(\sin^{-1}(-\sqrt{3}/2)\)
- Recall that \(\sin^{-1}\) (arcsin) gives the angle in \([-\frac{\pi}{2}, \frac{\pi}{2}]\) whose sine is the input
- We know \(\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}\)
- Therefore \(\sin(-\frac{\pi}{3}) = -\frac{\sqrt{3}}{2}\)
- Since -π/3 is in the range [-π/2, π/2], this is our answer
Answer: \(-\frac{\pi}{3}\)
b) \(\tan^{-1}(1) - \tan^{-1}(-1)\)
- Recall that \(\tan^{-1}(1) = \frac{\pi}{4}\) (this is a standard value to memorize)
- Since tan⁻¹ is an odd function, \(\tan^{-1}(-1) = -\tan^{-1}(1) = -\frac{\pi}{4}\)
- Therefore: \(\tan^{-1}(1) - \tan^{-1}(-1) = \frac{\pi}{4} - (-\frac{\pi}{4}) = \frac{\pi}{2}\)
Answer: \(\frac{\pi}{2}\)
c) \(\cot^{-1}(1/\sqrt{3}) - \cot^{-1}(-1/\sqrt{3})\)
- First, recall that \(\cot^{-1}(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}\)
- This is because \(\cot(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}\)
- 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
- 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) \(\cosh^{-1}\sqrt{2}\)
- For \(\cosh^{-1}\), we use the formula: \(\cosh^{-1}x = \ln(x + \sqrt{x^2-1})\)
- 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) \(e^{[-\coth^{-1}(25/7)]}\)
- Recall that \(\coth^{-1}x = \frac{1}{2}\ln(\frac{x+1}{x-1})\)
- Substituting x = 25/7:
- \(\coth^{-1}(\frac{25}{7}) = \frac{1}{2}\ln(\frac{\frac{25}{7}+1}{\frac{25}{7}-1})\)
- \(= \frac{1}{2}\ln(\frac{32/7}{18/7}) = \frac{1}{2}\ln(\frac{32}{18})\)
- 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:
- For inverse trig functions, always check if the input is a standard value (like ½, √3/2, etc.)
- Remember that sin⁻¹ and tan⁻¹ are odd functions
- For inverse hyperbolic functions, use their logarithmic forms
- Pay attention to the range of each inverse function
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) \(\cosh^{-1}[\coth(\ln 3)]\)
- 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, \(\sin^{-1}(-\sqrt{3}/2) = -\pi/3\)
- For \(\tan^{-1}(1) - \tan^{-1}(-1)\):
- \(\tan^{-1}(1) = \pi/4\)
- \(\tan^{-1}(-1) = -\pi/4\)
- Therefore, \(\tan^{-1}(1) - \tan^{-1}(-1) = \pi/2\)
- 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 \(-\pi/3\)
- 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:
- a) \(-\pi/3\)
- b) \(\pi/2\)
- c) \(-\pi/3\)
- d) \(\ln(1 + \sqrt{2})\)
- e) \(3/4\)
- f) \(\pi/2\)
- g) \(\pi/2 - 2x\)
- h) \(2x - 3\pi/2\)
- i) \(\frac{1-x^4}{1+x^4}\)
- j) \(-\sqrt{2}/2\)
- k) \(\ln 2\)
Question 2
Prove that: \[\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}\] for all x
- Let's start by noting that \(\cot^{-1}x = \tan^{-1}(\frac{1}{x})\)
- Therefore, we need to prove: \[\tan^{-1}x + \tan^{-1}(\frac{1}{x}) = \frac{\pi}{2}\]
- Using the tangent addition formula: \[\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\]
- Let \(A = \tan^{-1}x\) and \(B = \tan^{-1}(\frac{1}{x})\)
- Then \(\tan(A + B) = \frac{x + \frac{1}{x}}{1 - x \cdot \frac{1}{x}} = \infty\)
- 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: \[\sinh(x \pm y) = \sinh x \cosh y \pm \cosh x \sinh y\] \[\cosh(x \pm y) = \cosh x \cosh y \pm \sinh x \sinh y\]
- Recall the definitions: \[\sinh x = \frac{e^x - e^{-x}}{2}\] \[\cosh x = \frac{e^x + e^{-x}}{2}\]
- For the first identity: \[\sinh(x \pm y) = \frac{e^{x\pm y} - e^{-(x\pm y)}}{2}\] \[= \frac{e^x e^{\pm y} - e^{-x} e^{\mp y}}{2}\]
- 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) \[\tanh(\frac{x}{2}) = \frac{\sinh x}{1 + \cosh x}\]
- 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 \(4\cos^3 x - 3\cos x\)
- 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
- 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) \[\cosh(\tanh^{-1} x) = \frac{1}{\sqrt{1-x^2}}\]
- 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
- For \(\cosh^{-1} x\):
- Similar approach, using \(x = \cosh y = \frac{e^y + e^{-y}}{2}\)
- Note domain restriction \(x \geq 1\)
- For \(\tanh^{-1} x\):
- Use \(x = \tanh y = \frac{\sinh y}{\cosh y}\)
- Note domain restriction \(|x| < 1\)
- 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) \[f(x) = \ln|\ln x|\]
- For \(f(x) = \tanh^{-1} x\):
- Domain of \(\tanh^{-1}\) is \((-1, 1)\)
- For \(f(x) = \cot^{-1} x\):
- Domain is all real numbers \(\mathbb{R}\)
- For \(f(x) = \ln(\sin x)\):
- Need \(\sin x > 0\)
- Domain is \(\{x: x \in (2\pi n, \pi + 2\pi n), n \in \mathbb{Z}\}\)
- For \(f(x) = \frac{1}{\sqrt{|\cos x|}}\):
- Domain is all x where \(\cos x \neq 0\)
- i.e., \(\{x: x \neq \frac{\pi}{2} + \pi n, n \in \mathbb{Z}\}\)
- 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:
- Basic inverse trigonometric identities
- Relationship between inverse functions
- Hyperbolic function identities and their proofs
- Domain restrictions for various functions