Inverse Trigonometric Functions Study Guide

Prerequisites and Homework Discussion

• Brain Teaser on Piecewise Function Analysis:

  • Consider the function $f(x)$ defined piecewise on $[0, 2)$ and extended periodically:     $f(x) = \begin{cases} \sqrt{x} & \text{for } 0 \le x < 1 \\ 2 - x & \text{for } 1 \le x < 2 \end{cases}$$f(x+2) = f(x) \quad \forall x \in \mathbb{R}$
  • Define $g(x) = 4f(3x) + 1$ for all $x$.
  • Objective: Compute the value of $\frac{\lfloor A \rfloor \times B \times |C|}{2}$ where:
  • $A$ denotes the sum of all solutions to $f(x) = 0.6$ in the interval $3 \le x \le 7$.
  • $B$ denotes the fundamental period of $g(x)$.
  • $C$ denotes the derivative value $g'(6.50)$.
  • Determination of $A$:
  • Fundamental solutions in the base period $[0, 2)$:
    • For $x \in [0, 1)$: $\sqrt{x} = 0.6 \implies x = 0.36$.
    • For $x \in [1, 2)$: $2 - x = 0.6 \implies x = 1.40$.
  • Solutions shifted into the target interval $[3, 7]$ by adding period multiples ($k \times 2$):
    • From $x = 1.40$: $1.40 + 2 = 3.40$ and $1.40 + 4 = 5.40$
    • From $x = 0.36$: $0.36 + 4 = 4.36$ and $0.36 + 6 = 6.36$
  • Sum of all solutions $A = 3.40 + 5.40 + 4.36 + 6.36 = 19.52$.
  • Greatest integer value $\lfloor A \rfloor = \lfloor 19.52 \rfloor = 19$
  • Determination of $B$:
  • Fundamental period of $f(x)$ is $2$.
  • For $g(x) = 4f(3x) + 1$, the fundamental period $B = \frac{2}{3}$.
  • Determination of $C$:
  • Differentiating $g(x)$ yields $g'(x) = 12 f'(3x)$.
  • At $x = 6.50$: $g'(6.50) = 12 f'(19.50)$.
  • Due to periodicity with period $2$: $19.50 = 9 \times 2 + 1.50 \implies f'(19.50) = f'(1.50)$.
  • For $x \in [1, 2)$, $f(x) = 2 - x \implies f'(x) = -1$. Thus, $f'(1.50) = -1$
  • $C = g'(6.50) = 12(-1) = -12 \implies |C| = 12$
  • Final Result Calculation:
  • Value = $\frac{19 \times (2/3) \times 12}{2} = 76$

• Proof of Periodicity for Functional Equations:

  • Problem: Show that a function satisfying $f(x + k) = 1 + \sqrt{2f(x) - f^2(x)}$ for $x \in \mathbb{R}$ ($k > 0$) is periodic and find its fundamental period.
  • Method 1:
  • Rearrange equation: $f(x+k) - 1 = \sqrt{1 - (f(x) - 1)^2}$.
  • Let $g(x) = f(x) - 1$. The equation becomes $g(x+k) = \sqrt{1 - (g(x))^2}$.
  • Squaring both sides: $(g(x+k))^2 = 1 - (g(x))^2 \implies (g(x))^2 + (g(x+k))^2 = 1$
  • Replace $x$ with $x + k$: $(g(x+k))^2 + (g(x+2k))^2 = 1$
  • Subtracting the two equations: $(g(x))^2 - (g(x+2k))^2 = 0 \implies g(x) = g(x+2k)$.
  • Since $g(x) = f(x) - 1$, it follows $f(x) = f(x+2k)$.
  • Thus, $f(x)$ is periodic with fundamental period $T = 2k$.
  • Method 2:
  • $(f(x) - 1)^2 + (f(x+k) - 1)^2 = 1$
  • Replace $x \to x + k \implies (f(x+k) - 1)^2 + (f(x+2k) - 1)^2 = 1$
  • Equating both expressions gives $(f(x) - 1)^2 = (f(x+2k) - 1)^2 \implies f(x) = f(x+2k)$, yielding period $T = 2k$

• Symmetry and Periodicity Relations:

  • Problem 1: If $f: \mathbb{R} \to \mathbb{R}$ satisfies $f(x) + f(x+2) = \sqrt{2}f(x+1)$ for all $x$, show that $f$ is periodic and find its period.
  • Result: The period of $f(x)$ is $T = 8$
  • Problem 2: If $f: \mathbb{R} \to \mathbb{R}$ is an odd function such that $f(x+1) + f(x+2) = f(x-1) + f(x)$, then find $f(2)$.
  • Setting $x = 0$: $f(1) + f(2) = f(-1) + f(0)$
  • Since $f$ is odd, $f(0) = 0$ and $f(-1) = -f(1)$
  • Substituting gives $f(1) + f(2) = -f(1) \implies f(2) = -2f(1)$
  • Evaluation gives $f(2) = 0$ (Choice A)
  • Problem 3: If the graph of a function $f$ is symmetrical about the lines $x = 2$ and $x = 8$, show that $f$ is periodic and find its period.
  • Symmetry about $x = a \implies f(a+x) = f(a-x)$
  • Symmetry about $x = 2 \implies f(2+x) = f(2-x)$
  • Symmetry about $x = 8 \implies f(8-x) = f(8+x)$
  • Period $T = 2 |8 - 2| = 12$

• Periodicity and Binomial Expansion Problem:

  • Problem: Given $f(x)$ is periodic with period $t$ such that $f(2x+3) + f(2x+7) = 2$, find the coefficient of $m^{-24}$ in the expansion of $\left(m + \frac{b^2}{m^3}\right)^{16}$.
  • Determination of Period $t$:
  • Let $g(x) = f(2x+3)$. Then $g(x+2) = f(2(x+2)+3) = f(2x+7)$.
  • Given $g(x) + g(x+2) = 2$
  • Replace $x \to x+2 \implies g(x+2) + g(x+4) = 2$
  • Subtracting yields $g(x) = g(x+4)$, so $g(x)$ has period $4$.
  • Since $f(2x+3)$ has period $4$, $f(2x)$ has period $4$, which implies $f(x)$ has period $t = 8$
  • Binomial Term Calculation:
  • General term $T_{r+1} = \,^{16}\text{C}_r m^{16-r} \left(\frac{b^2}{m^3}\right)^r = \,^{16}\text{C}_r b^{2r} m^{16-4r}$
  • For $m^{-24}$, set exponent $16 - 4r = -24 \implies 4r = 40 \implies r = 10$
  • Coefficient is $\,^{16}\text{C}_{10} b^{20} = \,^{16}\text{C}_6 b^{10}$ (Choice B)

Even and Odd Extensions of Functions

• Theoretical Definitions:

  • Assume a function $f(x)$ is defined only for $x \ge 0$.
  • Even Extension: Extends the definition of $f(x)$ to $x < 0$ assuming the function is even ($f(-x) = f(x)$).
  • Operational Rule: Replace $x$ with $-x$ in the original definition.
  • Odd Extension: Extends the definition of $f(x)$ to $x < 0$ assuming the function is odd ($f(-x) = -f(x)$).
  • Operational Rule: Multiply the definition obtained from the even extension by $-1$

• Polynomial Function Example:

  • Given $f(x) = x^3 - 6x^2 + 5x - 11$ for $x > 0$:
  • Even extension for $x < 0$: Replace $x \to -x \implies (-x)^3 - 6(-x)^2 + 5(-x) - 11 = -x^3 - 6x^2 - 5x - 11$
  • Odd extension for $x < 0$: Multiply even extension by $-1 \implies x^3 + 6x^2 + 5x + 11$

• Transcendental Function Examples:

  • Given $f(x) = x^2 + \sin(x) + \ln(x)$ for $f: (0, \infty) \to \mathbb{R}$:
  • Even extension for $x \in (-\infty, 0)$: $x^2 - \sin(x) + \ln(-x)$ (Choice C)
  • Odd extension for $x \in (-\infty, 0)$: $-x^2 + \sin(x) - \ln(-x)$ (Choice A)
  • Given $f(x) = \sin(\cos(x)) - x + \tan(\sin(x))$ for $x \in (-\infty, 0)$:
  • Odd extension in $(0, \infty)$ is $-x - \sin(\cos(x)) + \tan(\sin(x))$ (Choice D)

Definitions, Domains, and Ranges of Inverse Trigonometric Functions

• Fundamental Inverse Concepts:

  • An inverse trigonometric expression represents an angle. If $\sin(\theta) = x$, then $\theta = \sin^{-1}(x)$.
  • To define an inverse function $f^{-1}(x)$, the domain of $f(x)$ must be restricted so that it is one-one and onto (bijective).

• Comprehensive Domain and Range Table:

  • $\sin^{-1}(x)$:
  • Domain: $[-1, 1]$ ($|x| \le 1$)
  • Principal Range: $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
  • $\cos^{-1}(x)$:
  • Domain: $[-1, 1]$ ($|x| \le 1$)
  • Principal Range: $[0, \pi]$
  • $\tan^{-1}(x)$:
  • Domain: $\mathbb{R}$
  • Principal Range: $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
  • $\cot^{-1}(x)$:
  • Domain: $\mathbb{R}$
  • Principal Range: $(0, \pi)$
  • $\sec^{-1}(x)$:
  • Domain: $(-\infty, -1] \cup [1, \infty)$ ($|x| \ge 1$)
  • Principal Range: $[0, \pi] \setminus \left\{\frac{\pi}{2}\right\}$
  • $\csc^{-1}(x)$:
  • Domain: $(-\infty, -1] \cup [1, \infty)$ ($|x| \ge 1$)
  • Principal Range: $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \setminus \{0\}$

• Evaluation of Standard Angle Values:

  • $\sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$
  • $\cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$
  • $\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6}$
  • $\cot^{-1}\left(-\frac{1}{\sqrt{3}}\right) = \frac{2\pi}{3}$
  • $\sec^{-1}\left(\frac{2}{\sqrt{3}}\right) = \frac{\pi}{6}$
  • $\csc^{-1}(-2) = -\frac{\pi}{6}$
  • $\sec^{-1}(-2) = \frac{2\pi}{3}$
  • $\cot^{-1}(-\sqrt{3}) = \frac{5\pi}{6}$
  • Expressions such as $\sin^{-1}(11)$ or $\sin^{-1}(2)$ are Not Defined (N.D.) because the input values lie outside the domain $[-1, 1]$.

• Solvability and Solutions of Elementary ITF Equations:

  • $\sin^{-1}(x) = \frac{\pi}{2} \implies x = 1$
  • $\sin^{-1}(x) = \pi \implies$ No Solution (since $\pi \notin \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$)
  • $\sin^{-1}(x) = -\frac{\pi}{3} \implies x = -\frac{\sqrt{3}}{2}$
  • $\cos^{-1}(x) = \frac{2\pi}{3} \implies x = -\frac{1}{2}$
  • $\sec^{-1}(x) = \frac{2\pi}{3} \implies x = -2$
  • $\sin^{-1}(x) = 1 \implies x = \sin(1)$ (Valid, as $1 \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$)
  • $\sin^{-1}(x) = -1 \implies x = \sin(-1) = -\sin(1)$
  • $\tan^{-1}(x) = 2 \implies$ No Solution (since $2 > \frac{\pi}{2} \approx 1.57$)
  • $\cot^{-1}(x) = 3 \implies x = \cot(3)$ (Valid, as $3 \in (0, \pi)$)
  • $\cos^{-1}(x) = 2 \implies x = \cos(2)$ (Valid, as $2 \in [0, \pi]$)
  • $\sec^{-1}(x) = 3 \implies x = \sec(3)$ (Valid, as $3 \in [0, \pi]$)

Composition, Algebraic Identities, and Equation Solving

• Composition Simplification:

  • $\sin^{-1}\left(\sin\left(\frac{3\pi}{2}\right)\right) = \sin^{-1}(-1) = -\frac{\pi}{2}$
  • $\cos^{-1}\left(\sin\left(\frac{2\pi}{3}\right)\right) = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$
  • $\tan^{-1}\left(\tan\left(\frac{2\pi}{3}\right)\right) = \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3}$
  • $\cot^{-1}\left(\cos\left(\frac{17\pi}{2}\right)\right) = \cot^{-1}(0) = \frac{\pi}{2}$
  • $\tan\left(\sin^{-1}\left(\frac{1}{\sqrt{10}}\right)\right) = \frac{1}{3}$
  • $\sec(\tan^{-1}(2)) = \sqrt{5}$
  • $\cos\left(\sin^{-1}\left(\frac{1}{3}\right)\right) = \frac{2\sqrt{2}}{3}$
  • $\sin^{-1}\left(-\frac{1}{2}\right) + \cos^{-1}\left(\frac{1}{2}\right) - \tan^{-1}(-\sqrt{3}) + \cot^{-1}\left(-\frac{1}{\sqrt{3}}\right) = \frac{3\pi}{2}$
  • $\tan^{-1}\left(1 + \frac{\sqrt{3}}{6 + 3\sqrt{3}}\right) + \sec^{-1}\left(\sqrt{\frac{8 + 4\sqrt{3}}{6 + 3\sqrt{3}}}\right) = \frac{\pi}{2}$ (Choice C)
  • $\sin\left[\frac{\pi}{3} + \sin^{-1}\left(-\frac{1}{2}\right)\right] = \sin\left[\frac{\pi}{3} - \frac{\pi}{6}\right] = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$ (Choice C)
  • $\tan\left[90^\circ - \cot^{-1}\left(-\frac{1}{3}\right)\right] = \cot\left[\cot^{-1}\left(-\frac{1}{3}\right)\right] = -\frac{1}{3}$ (Choice C)
  • $\sin\left[\cos^{-1}\left(-\frac{12}{13}\right)\right] = \frac{5}{13}$ (Choice A)
  • $\sin\left[\frac{\pi}{2} - \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right] = \cos\left[\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right] = \cos\left(-\frac{\pi}{3}\right) = \frac{1}{2}$ (Choice A)

• Right-Triangle Method for Nested ITF Expressions:

  • Simplify $\cos\left[\tan^{-1}\left\{\sin\left(\cot^{-1}(x)\right)\right\}\right]$:
  • Let $\theta = \cot^{-1}(x) \implies \cot(\theta) = x$. Opposite = $1$, Adjacent = $x$, Hypotenuse = $\sqrt{1+x^2}$.
  • $\sin(\theta) = \frac{1}{\sqrt{1+x^2}}$
  • Let $\phi = \tan^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right) \implies \tan(\phi) = \frac{1}{\sqrt{1+x^2}}$. Opposite = $1$, Adjacent = $\sqrt{1+x^2}$, Hypotenuse = $\sqrt{2+x^2}$.
  • $\cos(\phi) = \frac{\sqrt{1+x^2}}{\sqrt{2+x^2}} = \sqrt{\frac{x^2+1}{x^2+2}}$ (Choice C)

• Identity-Based Evaluation and Sum of Squares:

  • Evaluate $\sec^2(\tan^{-1}(2)) + \csc^2(\cot^{-1}(3))$:
  • Using identities $\sec^2(\theta) = 1 + \tan^2(\theta)$ and $\csc^2(\phi) = 1 + \cot^2(\phi)$
  • Value = $(1 + 2^2) + (1 + 3^2) = 5 + 10 = 15$ (Choice C)
  • If $\sec^2(\tan^{-1}(\alpha)) + \csc^2(\cot^{-1}(\beta)) = 36$ with $\alpha + \beta = 8$ and $\alpha \le \beta$:
  • $(1 + \alpha^2) + (1 + \beta^2) = 36 \implies \alpha^2 + \beta^2 = 34$
  • $(\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta \implies 64 = 34 + 2\alpha\beta \implies \alpha\beta = 15$
  • Solving $t^2 - 8t + 15 = 0$ gives roots $3$ and $5$. Since $\alpha \le \beta$, $\alpha = 3$ and $\beta = 5$
  • $\alpha^2 + \beta = 3^2 + 5 = 14$

• Extremal / Boundary Value Problems:

  • If $\cos^{-1}(x) + \cos^{-1}(y) + \cos^{-1}(z) = 3\pi$:
  • Maximum value of $\cos^{-1}(\theta) = \pi$, so each term must equal $\pi$
  • $x = -1, y = -1, z = -1$
  • Value of $x^{2026} + y^{2026} + z^{2026} + \frac{1}{x^{2025} + y^{2025} + z^{2025}} = 3 - \frac{1}{3} = \frac{8}{3}$ (Choice A)
  • If $\sum_{i=1}^{2026} \cos^{-1}(x_i) = 0$:
  • Since $\cos^{-1}(x_i) \ge 0$, each $\cos^{-1}(x_i) = 0 \implies x_i = 1$
  • $\sum_{i=1}^{2026} x_i = 2026$
  • If $\sin^{-1}(\alpha) + \sin^{-1}(\beta) + \sin^{-1}(\gamma) = \frac{3\pi}{2} \implies \alpha = 1, \beta = 1, \gamma = 1$
  • $\alpha\beta + \beta\gamma + \gamma\alpha = 3$ (Choice C)

• Algebraic and Trigonometric Equations:

  • Solve $\tan(\cos^{-1}(x)) = \sin\left(\cot^{-1}\left(\frac{1}{2}\right)\right)$
  • Let $\theta = \cot^{-1}\left(\frac{1}{2}\right) \implies \sin(\theta) = \frac{2}{\sqrt{5}}$
  • Let $\phi = \cos^{-1}(x) \implies \tan(\phi) = \frac{\sqrt{1-x^2}}{x}$
  • Equating: $\frac{\sqrt{1-x^2}}{x} = \frac{2}{\sqrt{5}} \implies 9x^2 = 5 \implies x = \frac{\sqrt{5}}{3}$ (Choice B)
  • If $3\cos^{-1}\left(x^2 - 7x + \frac{25}{2}\right) = \pi \implies x^2 - 7x + \frac{25}{2} = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$
  • $x^2 - 7x + 12 = 0 \implies x = 3 \text{ or } 4$ (Choice C)
  • If $\tan(x+y) = 33$ and $x = \tan^{-1}(3)$, then $y = \tan^{-1}(0.3)$ (Choice C)
  • $\tan\left(\frac{\pi}{4} + \frac{1}{2}\cos^{-1}(x)\right) + \tan\left(\frac{\pi}{4} - \frac{1}{2}\cos^{-1}(x)\right) = \frac{2}{x}$ (Choice C)
  • Evaluate $\tan\left(\frac{1}{2}\cos^{-1}\left(\frac{\sqrt{5}}{3}\right)\right)$
  • Let $\theta = \cos^{-1}\left(\frac{\sqrt{5}}{3}\right)$
  • $\tan\left(\frac{\theta}{2}\right) = \sqrt{\frac{1-\cos(\theta)}{1+\cos(\theta)}} = \sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}}} = \frac{3-\sqrt{5}}{2}$ (Choice B)
  • A possible value of $\tan\left(\frac{1}{2}\sin^{-1}\left(\frac{\sqrt{63}}{8}\right)\right)$ is $\frac{1}{\sqrt{7}}$ (Choice A)
  • Evaluate $\tan\left(\sin^{-1}\left(\frac{3}{5}\right) - 2\cos^{-1}\left(\frac{2}{\sqrt{5}}\right)\right)$
  • Let $A = \sin^{-1}\left(\frac{3}{5}\right) \implies \tan(A) = \frac{3}{4}$
  • Let $B = \cos^{-1}\left(\frac{2}{\sqrt{5}}\right) \implies \tan(B) = \frac{1}{2} \implies \tan(2B) = \frac{4}{3}$
  • $\tan(A - 2B) = \frac{3/4 - 4/3}{1 + (3/4)(4/3)} = -\frac{7}{24}$ (Choice B)

Domain Determination for Complex Inverse Trigonometric Functions

• Core Domain Principles:

  • For $f(x) = \sin^{-1}(g(x))$ or $\cos^{-1}(g(x))$, set up inequality -1 \le g(x) \n\le 1
  • For rational inner expressions $g(x) = \frac{P(x)}{Q(x)}$ where $Q(x) > 0$ for all $x \in \mathbb{R}$, solve $-Q(x) \le P(x) \le Q(x)$

• Domain Problems:

  • Problem 1: Domain of $f(x) = \cos^{-1}\left(\frac{x^2 - 4x + 2}{x^2 + 3}\right)$
  • Inequality: $-1 \le \frac{x^2 - 4x + 2}{x^2 + 3} \le 1$
  • Since $x^2 + 3 > 0$:
    1. $x^2 - 4x + 2 \le x^2 + 3 \implies -4x \le 1 \implies x \ge -\frac{1}{4}$
    2. $x^2 - 4x + 2 \ge -(x^2 + 3) \implies 2x^2 - 4x + 5 \ge 0$ (Discriminant $D = 16 - 40 < 0$, holds for all $x \in \mathbb{R}$)
  • Domain: $\left[-\frac{1}{4}, \infty\right)$ (Choice B)
  • Problem 2: Domain of $f(x) = \frac{\cos^{-1}\left(\sqrt{x^2 - x + 1}\right)}{\sin^{-1}(2x - 1)}$
  • Numerator: $0 \le x^2 - x + 1 \le 1 \implies x^2 - x \le 0 \implies x \in [0, 1]$
  • Denominator: $-1 \le 2x - 1 \le 1 \implies x \in [0, 1]$
  • Non-zero denominator: $\sin^{-1}(2x - 1) \ne 0 \implies 2x - 1 \ne 0 \implies x \ne \frac{1}{2}$
  • Domain interval: $\left[0, \frac{1}{2}\right) \cup \left(\frac{1}{2}, 1\right]$. Sum of endpoints $\alpha + \beta = 0 + 1 = 1$ or $\frac{3}{2}$ (Choice A)
  • Problem 3: Domain of $f(x) = \sin^{-1}\left(\frac{x^2 - 3x + 2}{x^2 + 2x + 7}\right)$
  • Inequality: $-1 \le \frac{x^2 - 3x + 2}{x^2 + 2x + 7} \le 1$
  • Since $x^2 + 2x + 7 > 0$ for all $x \in \mathbb{R}$ ($D = 4 - 28 < 0$):
    1. $x^2 - 3x + 2 \le x^2 + 2x + 7 \implies -5x \le 5 \implies x \ge -1$
    2. $x^2 - 3x + 2 \ge -(x^2 + 2x + 7) \implies 2x^2 - x + 9 \ge 0$ (Holds for all $x$ since $D = 1 - 72 < 0$)
  • Domain: $[-1, \infty)$ (Choice C)
  • Problem 4: Domain of $f(x) = \cos^{-1}\left(\frac{2\sin^{-1}\left(\frac{1}{4x^2 - 1}\right)}{\pi}\right)$
  • Domain of inner inverse sine: $\left|\frac{1}{4x^2 - 1}\right| \le 1 \implies |4x^2 - 1| \ge 1$
  • Case 1: $4x^2 - 1 \ge 1 \implies 4x^2 \ge 2 \implies x^2 \ge \frac{1}{2} \implies x \in \left(-\infty, -\frac{1}{\sqrt{2}}\right] \cup \left[\frac{1}{\sqrt{2}}, \infty\right)$
  • Case 2: $4x^2 - 1 \le -1 \implies 4x^2 \le 0 \implies x = 0$
  • Total Domain: $\left(-\infty, -\frac{1}{\sqrt{2}}\right] \cup \left[\frac{1}{\sqrt{2}}, \infty\right) \cup \{0\}$ (Choice D)