Figure 3.1: Principal Branches of Inverse Trigonometric Functions
1. Reciprocal Circular Functions
In IB DP Mathematics AA HL, the three reciprocal trigonometric functions extend the standard circular functions:
$$\sec x = \frac{1}{\cos x}, \quad \csc x = \frac{1}{\sin x}, \quad \cot x = \frac{1}{\tan x} = \frac{\cos x}{\sin x}$$- Secant Function ($\sec x$):
• Domain: $x \in \mathbb{R}, x \neq \frac{\pi}{2} + k\pi, k \in \mathbb{Z}$ (vertical asymptotes where $\cos x = 0$).
• Range: $(-\infty, -1] \cup [1, \infty)$ (never takes values strictly between $-1$ and $1$).
• Parity & Period: Even function ($\sec(-x) = \sec x$); fundamental period is $2\pi$. - Cosecant Function ($\csc x$):
• Domain: $x \in \mathbb{R}, x \neq k\pi, k \in \mathbb{Z}$ (vertical asymptotes where $\sin x = 0$).
• Range: $(-\infty, -1] \cup [1, \infty)$.
• Parity & Period: Odd function ($\csc(-x) = -\csc x$); fundamental period is $2\pi$. - Cotangent Function ($\cot x$):
• Domain: $x \in \mathbb{R}, x \neq k\pi, k \in \mathbb{Z}$ (vertical asymptotes where $\sin x = 0$).
• Range: $\mathbb{R}$ (all real numbers). Zeros occur where $\cos x = 0$ ($x = \frac{\pi}{2} + k\pi$).
• Parity & Period: Odd function ($\cot(-x) = -\cot x$); fundamental period is $\pi$.
2. Extended Pythagorean Identities
Dividing the primary identity $\sin^2 x + \cos^2 x = 1$ by $\cos^2 x$ and $\sin^2 x$ respectively establishes the essential AHL Pythagorean identities:
$$1 + \tan^2 x = \sec^2 x \iff \sec^2 x - \tan^2 x = 1$$ $$1 + \cot^2 x = \csc^2 x \iff \csc^2 x - \cot^2 x = 1$$These identities are the cornerstone for reducing higher-degree non-linear trigonometric equations into quadratics in a single trigonometric ratio.
3. Inverse Trigonometric Functions & Principal Branches
Because circular functions are periodic and many-to-one, their natural domains must be restricted to standard bijective intervals so that unique inverse functions exist:
| Function | Domain | Principal Range | Key Algebraic Identities | Calculus Derivative |
|---|---|---|---|---|
| $y = \arcsin x$ | $[-1, 1]$ | $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ | $\arcsin(-x) = -\arcsin x$ (Odd) $\arcsin x + \arccos x = \frac{\pi}{2}$ |
$\frac{d}{dx}(\arcsin x) = \frac{1}{\sqrt{1 - x^2}}$ |
| $y = \arccos x$ | $[-1, 1]$ | $[0, \pi]$ | $\arccos(-x) = \pi - \arccos x$ Always non-negative output |
$\frac{d}{dx}(\arccos x) = -\frac{1}{\sqrt{1 - x^2}}$ |
| $y = \arctan x$ | $\mathbb{R}$ | $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ | $\arctan(-x) = -\arctan x$ (Odd) Horizontal Asymptotes: $y = \pm\frac{\pi}{2}$ |
$\frac{d}{dx}(\arctan x) = \frac{1}{1 + x^2}$ |
4. Right-Triangle Modeling of Composite Inverses
To evaluate or simplify composite expressions like $\tan(\arccos x)$ or $\cos(2\arcsin x)$ without a calculator, always define an angle $\theta$ and construct a reference right triangle:
Standard Construction Method:
1. Let $\theta = \arccos x \implies \cos\theta = \frac{x}{1}$.
2. In a right-angled triangle, set $\text{adjacent} = x$, $\text{hypotenuse} = 1$.
3. By Pythagoras: $\text{opposite} = \sqrt{1^2 - x^2} = \sqrt{1 - x^2}$.
4. Read off any desired ratio directly: $\sin(\arccos x) = \sqrt{1 - x^2}$ and $\tan(\arccos x) = \frac{\sqrt{1 - x^2}}{x}$.
Worked Examples
Worked Example 1: Solving Reciprocal Quadratic Equations (Paper 1 Style · 6 marks)
Solve the trigonometric equation $2\sec^2 x - 5\tan x = 5$ for $0 \le x \le 2\pi$.
Official Markscheme Solution:
Apply the Pythagorean identity $\sec^2 x = 1 + \tan^2 x$: [M1]
$2(1 + \tan^2 x) - 5\tan x = 5 \implies 2 + 2\tan^2 x - 5\tan x - 5 = 0$ [A1]
$2\tan^2 x - 5\tan x - 3 = 0$.
Factor the quadratic in $\tan x$: [M1]
$(2\tan x + 1)(\tan x - 3) = 0$.
• Branch 1: $\tan x = 3$
Reference angle $\alpha = \arctan(3) \approx 1.25$ rad [A1].
Since tangent is positive in Quadrants 1 and 3:
$x_1 = 1.25$ rad
$x_2 = 1.25 + \pi \approx 4.39$ rad [A1].
• Branch 2: $\tan x = -\frac{1}{2}$
Reference angle $\beta = \arctan(0.5) \approx 0.464$ rad.
Since tangent is negative in Quadrants 2 and 4:
$x_3 = \pi - 0.464 \approx 2.68$ rad [A1]
$x_4 = 2\pi - 0.464 \approx 5.82$ rad [A1].
All solutions in $[0, 2\pi]$: $x \in \{1.25, 2.68, 4.39, 5.82\}$ rad (to 3 s.f.).
Worked Example 2: Exact Value of Compound Inverses (Paper 1 Style · 6 marks)
Find the exact value of $\sin\left(\arccos\left(-\frac{4}{5}\right) + \arcsin\left(\frac{5}{13}\right)\right)$ without using a calculator.
Official Markscheme Solution:
Let $A = \arccos\left(-\frac{4}{5}\right)$ and $B = \arcsin\left(\frac{5}{13}\right)$ [M1].
By definition of principal branches:
• $A \in [0, \pi]$: Since $\cos A = -\frac{4}{5} < 0$, $A$ lies in Quadrant 2.
Therefore $\sin A > 0$: $\sin A = \sqrt{1 - \left(-\frac{4}{5}\right)^2} = \sqrt{\frac{9}{25}} = \frac{3}{5}$ [A1].
• $B \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$: Since $\sin B = \frac{5}{13} > 0$, $B$ lies in Quadrant 1.
Therefore $\cos B > 0$: $\cos B = \sqrt{1 - \left(\frac{5}{13}\right)^2} = \sqrt{\frac{144}{169}} = \frac{12}{13}$ [A1].
Expand using the sine addition formula: [M1]
$\sin(A + B) = \sin A \cos B + \cos A \sin B$ [M1]
$= \left(\frac{3}{5}\right)\left(\frac{12}{13}\right) + \left(-\frac{4}{5}\right)\left(\frac{5}{13}\right)$ [A1]
$= \frac{36}{65} - \frac{20}{65} = \frac{16}{65}$ [A1].
Worked Example 3: Algebraic Modeling of Double-Angle Inverse (Paper 1 Style · 5 marks)
Express $\cos(2\arctan x)$ as a rational algebraic function of $x$, valid for all $x \in \mathbb{R}$.
Official Markscheme Solution:
Let $\theta = \arctan x \implies \tan\theta = x$ with $\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ [M1].
Recall the double angle formula for cosine in terms of tangent: [M1]
$\cos(2\theta) = \frac{\cos^2\theta - \sin^2\theta}{\cos^2\theta + \sin^2\theta} = \frac{1 - \tan^2\theta}{1 + \tan^2\theta}$ [M1].
Substitute $\tan\theta = x$ into the identity: [A1]
$\cos(2\arctan x) = \frac{1 - x^2}{1 + x^2}$ [A1][AG].
Alternative Method (Right Triangle):
Draw a right triangle with angle $\theta$, opposite side $x$, and adjacent side $1$.
Hypotenuse $= \sqrt{1 + x^2}$ [M1].
Then $\cos\theta = \frac{1}{\sqrt{1 + x^2}}$ and $\sin\theta = \frac{x}{\sqrt{1 + x^2}}$ [A1].
$\cos(2\theta) = \cos^2\theta - \sin^2\theta = \frac{1}{1 + x^2} - \frac{x^2}{1 + x^2} = \frac{1 - x^2}{1 + x^2}$ [A1].
Worked Example 4: Domain, Range, and Extrema of Inverse Composite (Paper 2 Style · 7 marks)
Let $f(x) = \arccos(x^2 - 1)$.
(a) Find the largest possible real domain of $f$. [3 marks]
(b) Determine the range of $f$. [2 marks]
(c) State the coordinates of the absolute maximum and minimum points of $y = f(x)$. [2 marks]
Official Markscheme Solution:
(a) Domain: [3 marks]
The domain of $\arccos u$ is $-1 \le u \le 1$.
Therefore, $-1 \le x^2 - 1 \le 1$ [M1].
Add 1 to all parts: $0 \le x^2 \le 2$ [M1].
Since $x^2 \ge 0$ is trivially satisfied for all real $x$, this reduces to $x^2 \le 2 \implies -\sqrt{2} \le x \le \sqrt{2}$ [A1].
Domain: $[-\sqrt{2}, \sqrt{2}]$.
(b) Range: [2 marks]
As $x$ varies over $[-\sqrt{2}, \sqrt{2}]$, $u = x^2 - 1$ ranges continuously from $-1$ (at $x = 0$) to $1$ (at $x = \pm\sqrt{2}$) [M1].
Because $\arccos u$ is continuous and strictly decreasing on $[-1, 1]$, it maps $[-1, 1]$ onto $[0, \pi]$.
Range: $[0, \pi]$ [A1].
(c) Extrema: [2 marks]
• Absolute Maximum: Occurs when $x^2 - 1 = -1 \implies x = 0$, giving $(0, \arccos(-1)) = (0, \pi)$ [A1].
• Absolute Minima: Occur when $x^2 - 1 = 1 \implies x = \pm\sqrt{2}$, giving $(\sqrt{2}, 0)$ and $(-\sqrt{2}, 0)$ [A1].
Worked Example 5: Reciprocal Identity Proof & Equation Solving (Paper 2 Style · 7 marks)
(a) Show that $\cot x + \tan x = 2\csc(2x)$ for all valid $x$. [3 marks]
(b) Hence solve the equation $\cot x + \tan x = 4$ for $0 < x < \pi$. [4 marks]
Official Markscheme Solution:
(a) Identity Proof: [3 marks]
Start with the Left-Hand Side (LHS):
$\text{LHS} = \cot x + \tan x = \frac{\cos x}{\sin x} + \frac{\sin x}{\cos x}$ [M1]
$= \frac{\cos^2 x + \sin^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x}$ [A1]
Multiply numerator and denominator by 2: [M1]
$= \frac{2}{2\sin x \cos x} = \frac{2}{\sin(2x)} = 2\csc(2x) = \text{RHS}$ [AG].
(b) Equation Solving: [4 marks]
Substitute the identity into the equation: [M1]
$2\csc(2x) = 4 \implies \csc(2x) = 2 \implies \sin(2x) = \frac{1}{2}$ [A1].
Since $0 < x < \pi$, the argument satisfies $0 < 2x < 2\pi$ [M1].
The solutions for $2x$ in $(0, 2\pi)$ where sine equals $1/2$ are:
$2x = \frac{\pi}{6} \implies x_1 = \frac{\pi}{12}$ [A1]
$2x = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \implies x_2 = \frac{5\pi}{12}$ [A1].
Solutions: $x \in \left\{\frac{\pi}{12}, \frac{5\pi}{12}\right\}$.
Practice Exercises (10 Tiered Questions)
Work through these 10 examination-calibrated exercises ranging from core foundational skills to advanced Paper 1 and Paper 2 problems. Expand each exercise to view the full IB markscheme and examiner advice.
Find the exact value in radians of each expression without a calculator:
(a) $\arcsin\left(\frac{\sqrt{3}}{2}\right)$
(b) $\arccos\left(-\frac{1}{2}\right)$
(c) $\arctan(-1)$
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(a) In principal range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \implies \frac{\pi}{3}$ [A1].
(b) In principal range $[0, \pi]$, $\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \implies \frac{2\pi}{3}$ [A1].
(c) In principal range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, $\tan\left(-\frac{\pi}{4}\right) = -1 \implies -\frac{\pi}{4}$ [A1].
Examiner Advice: Never write $\frac{3\pi}{4}$ for $\arctan(-1)$; the principal range requires negative angles in $\left(-\frac{\pi}{2}, 0\right)$.
Given that $\tan\theta = -\frac{4}{3}$ and $\frac{\pi}{2} < \theta < \pi$, find the exact values of $\sec\theta$, $\csc\theta$, and $\cot\theta$.
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• In Quadrant 2, $\cos\theta < 0$ and $\sin\theta > 0$ [R1].
• $\cot\theta = \frac{1}{\tan\theta} = -\frac{3}{4}$ [A1].
• Use $1 + \tan^2\theta = \sec^2\theta$: $1 + \left(-\frac{4}{3}\right)^2 = 1 + \frac{16}{9} = \frac{25}{9}$ [M1].
Since $\theta$ is in Quadrant 2, $\sec\theta = -\sqrt{\frac{25}{9}} = -\frac{5}{3}$ [A1].
• $\cos\theta = -\frac{3}{5} \implies \sin\theta = \frac{4}{5} \implies \csc\theta = \frac{5}{4}$ [A1].
Find the domain of the function $f(x) = \arcsin(3x - 1)$.
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The argument must satisfy $-1 \le 3x - 1 \le 1$ [M1].
Add 1: $0 \le 3x \le 2$ [M1].
Divide by 3: $0 \le x \le \frac{2}{3}$ [A1].
Domain: $\left[0, \frac{2}{3}\right]$.
Find the exact value of $\tan\left(2\arctan\left(\frac{1}{3}\right)\right)$.
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Let $\theta = \arctan(1/3) \implies \tan\theta = \frac{1}{3}$ [M1].
Use the tangent double angle formula: [M1]
$\tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta} = \frac{2(1/3)}{1 - (1/3)^2} = \frac{2/3}{1 - 1/9} = \frac{2/3}{8/9} = \frac{2}{3} \times \frac{9}{8} = \frac{3}{4}$ [A1].
Solve $2\csc^2 x + \cot x - 3 = 0$ for $0 < x < \pi$.
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Substitute $\csc^2 x = 1 + \cot^2 x$: [M1]
$2(1 + \cot^2 x) + \cot x - 3 = 0 \implies 2\cot^2 x + \cot x - 1 = 0$ [A1].
Factor the quadratic: $(2\cot x - 1)(\cot x + 1) = 0$ [M1].
• $\cot x = \frac{1}{2} \implies \tan x = 2 \implies x = \arctan(2) \approx 1.11$ rad [A1].
• $\cot x = -1 \implies \tan x = -1 \implies x = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \approx 2.36$ rad [A1].
Solutions: $x \in \{1.11, 2.36\}$ rad.
Simplify $\sin(\arccos(2x))$ for $-\frac{1}{2} \le x \le \frac{1}{2}$ into an algebraic expression in $x$.
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Let $\theta = \arccos(2x) \implies \cos\theta = 2x$ with $\theta \in [0, \pi]$ [M1].
Since $\theta \in [0, \pi]$, $\sin\theta \ge 0$.
$\sin\theta = \sqrt{1 - \cos^2\theta} = \sqrt{1 - (2x)^2} = \sqrt{1 - 4x^2}$ [A1].
Result: $\sqrt{1 - 4x^2}$ [A1].
Let $f(x) = \sec\left(x - \frac{\pi}{4}\right)$ for $0 \le x \le 2\pi$.
(a) State the equations of all vertical asymptotes of $y = f(x)$. [2 marks]
(b) Find the $y$-intercept of the graph. [1 mark]
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(a) Vertical asymptotes: [2 marks]
Vertical asymptotes occur where the cosine argument is $\frac{\pi}{2} + k\pi$ [M1]:
$x - \frac{\pi}{4} = \frac{\pi}{2} \implies x = \frac{3\pi}{4}$ [A1].
$x - \frac{\pi}{4} = \frac{3\pi}{2} \implies x = \frac{7\pi}{4}$ [A1].
Asymptotes: $x = \frac{3\pi}{4}$ and $x = \frac{7\pi}{4}$.
(b) $y$-intercept: [1 mark]
$f(0) = \sec\left(-\frac{\pi}{4}\right) = \frac{1}{\cos(-\pi/4)} = \frac{1}{\sqrt{2}/2} = \sqrt{2}$ [A1].
Prove that $\arcsin x + \arccos x = \frac{\pi}{2}$ for all $x \in [-1, 1]$.
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Let $\theta = \arcsin x \implies \sin\theta = x$ with $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ [M1].
Using the co-function identity: $\cos\left(\frac{\pi}{2} - \theta\right) = \sin\theta = x$ [M1].
Check the range of $\frac{\pi}{2} - \theta$:
Since $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$, multiplying by $-1$ and adding $\frac{\pi}{2}$ gives:
$0 \le \frac{\pi}{2} - \theta \le \pi$, which is precisely the principal range of $\arccos$ [R1]!
Therefore, by definition of arccos: $\arccos x = \frac{\pi}{2} - \theta$ [A1].
$\theta + \arccos x = \frac{\pi}{2} \implies \arcsin x + \arccos x = \frac{\pi}{2}$ [AG].
Prove that $\arctan(2) + \arctan(3) = \frac{3\pi}{4}$.
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Let $\alpha = \arctan(2)$ and $\beta = \arctan(3)$ [M1].
Since $2 > 1$ and $3 > 1$, both $\alpha, \beta \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right)$, so $\alpha + \beta \in \left(\frac{\pi}{2}, \pi\right)$ [R1].
Apply the tangent compound angle formula: [M1]
$\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = \frac{2 + 3}{1 - (2)(3)} = \frac{5}{1 - 6} = \frac{5}{-5} = -1$ [A1].
Since $\alpha + \beta \in \left(\frac{\pi}{2}, \pi\right)$ (Quadrant 2) and $\tan(\alpha + \beta) = -1$:
$\alpha + \beta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$ [A1][AG].
Examiner Advice: A common blunder is stating $\alpha + \beta = -\frac{\pi}{4}$. Always establish quadrant bounds for the sum of angles!
Let $f(x) = \arctan\left(\frac{x}{2}\right) + \arctan\left(\frac{2}{x}\right)$ for $x > 0$.
(a) Show that $f'(x) = 0$ for all $x > 0$. [3 marks]
(b) Hence deduce the constant value of $f(x)$ for all $x > 0$. [2 marks]
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(a) Derivative Proof: [3 marks]
Use the chain rule: $\frac{d}{dx}(\arctan u) = \frac{u'}{1 + u^2}$ [M1].
• First term: $\frac{1/2}{1 + (x/2)^2} = \frac{1/2}{1 + x^2/4} = \frac{2}{4 + x^2}$ [A1].
• Second term: $\frac{-2/x^2}{1 + (2/x)^2} = \frac{-2/x^2}{1 + 4/x^2} = \frac{-2}{x^2 + 4}$ [A1].
Sum: $f'(x) = \frac{2}{4 + x^2} - \frac{2}{4 + x^2} = 0$ [AG].
(b) Constant Value: [2 marks]
Since $f'(x) = 0$ on the connected interval $(0, \infty)$, $f(x)$ is constant [R1].
Evaluate at any convenient value, e.g., $x = 2$:
$f(2) = \arctan(1) + \arctan(1) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$ [A1].
Therefore, $f(x) = \frac{\pi}{2}$ for all $x > 0$.
IB Formula Booklet References (Topic 3)
Syllabus 3.10: Reciprocal & Inverse Trigonometry
The official IB DP Mathematics AA Formula Booklet provides:
- $\sec\theta = \frac{1}{\cos\theta}, \quad \csc\theta = \frac{1}{\sin\theta}, \quad \cot\theta = \frac{1}{\tan\theta}$
- $1 + \tan^2\theta = \sec^2\theta, \quad 1 + \cot^2\theta = \csc^2\theta$
- Inverse derivatives: $\frac{d}{dx}(\arcsin x) = \frac{1}{\sqrt{1 - x^2}}, \quad \frac{d}{dx}(\arccos x) = -\frac{1}{\sqrt{1 - x^2}}, \quad \frac{d}{dx}(\arctan x) = \frac{1}{1 + x^2}$
Key Formulas & Identities
1. Principal Domain and Range Restrictions
- $\arcsin x$: Domain $[-1, 1]$, Range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ (Odd function: $\arcsin(-x) = -\arcsin x$).
- $\arccos x$: Domain $[-1, 1]$, Range $[0, \pi]$ ($\arccos(-x) = \pi - \arccos x$).
- $\arctan x$: Domain $\mathbb{R}$, Range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ (Horizontal asymptotes $y = \pm\frac{\pi}{2}$).
- Co-function identity: $\arcsin x + \arccos x = \frac{\pi}{2}$ for all $x \in [-1, 1]$.
2. Double-Angle & Right Triangle Rational Identities
- $\cos(2\arctan x) = \frac{1 - x^2}{1 + x^2}$
- $\sin(2\arctan x) = \frac{2x}{1 + x^2}$
- $\sin(\arccos x) = \sqrt{1 - x^2}, \quad \tan(\arccos x) = \frac{\sqrt{1 - x^2}}{x}$
- $\cos(\arcsin x) = \sqrt{1 - x^2}, \quad \tan(\arcsin x) = \frac{x}{\sqrt{1 - x^2}}$
Common Exam Pitfalls
Pitfall 1: $\arccos(-x)$ Symmetry Misconception
$\arccos(-x) \neq -\arccos x$! Since the range of $\arccos$ is strictly $[0, \pi]$, $\arccos(-x) = \pi - \arccos x$. For example, $\arccos\left(-\frac{1}{2}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$, NOT $-\frac{\pi}{3}$.
Pitfall 2: Reciprocal vs Inverse Notation Confusion
Never confuse $\sin^{-1}(x)$ (inverse sine) with $(\sin x)^{-1}$ (cosecant $\csc x = \frac{1}{\sin x}$). They are completely different mathematical operations with different domains and graphs!
Pitfall 3: Inverse Tangent Horizontal Asymptotes
When sketching $y = \arctan x$, students often forget to draw and label the dashed horizontal asymptotes at $y = \frac{\pi}{2}$ and $y = -\frac{\pi}{2}$. Failure to show asymptotes costs communication and accuracy marks.
Topic Diagnostic Test
Complete this timed diagnostic test to verify your conceptual command of reciprocal and inverse circular functions.
Section A: Diagnostic Multiple Choice Questions
What is the exact principal value of $\arccos\left(-\frac{\sqrt{3}}{2}\right)$?
Which value is undefined for the function $y = \sec x$?
What is the exact value of $\cos(\arctan(\sqrt{3}))$?
What are the equations of the horizontal asymptotes of $y = \arctan x$?
Section B: Examination-Style Free Response Questions
Consider the equation $3\sec^2 x = 4\tan x + 5$ for $0 \le x \le 2\pi$.
(a) Show that the equation can be written in the form $3\tan^2 x - 4\tan x - 2 = 0$. [2 marks]
(b) Hence find the number of real solutions in the interval $[0, 2\pi]$. [4 marks]
Show Full Markscheme
(a) Quadratic Form: [2 marks]
Substitute $\sec^2 x = 1 + \tan^2 x$: [M1]
$3(1 + \tan^2 x) = 4\tan x + 5 \implies 3 + 3\tan^2 x - 4\tan x - 5 = 0$
$3\tan^2 x - 4\tan x - 2 = 0$ [A1][AG].
(b) Number of Solutions: [4 marks]
Discriminant of quadratic in $\tan x$: [M1]
$\Delta = (-4)^2 - 4(3)(-2) = 16 + 24 = 40 > 0$ [A1].
Therefore, there are two distinct real roots for $\tan x$:
$\tan x = \frac{4 \pm \sqrt{40}}{6} = \frac{2 \pm \sqrt{10}}{3}$ [M1].
• $\tan x = \frac{2 + \sqrt{10}}{3} > 0$: positive tangent gives 2 solutions in $[0, 2\pi]$ (Quadrants 1 and 3) [A1].
• $\tan x = \frac{2 - \sqrt{10}}{3} < 0$ (since $\sqrt{10} \approx 3.16$): negative tangent gives 2 solutions in $[0, 2\pi]$ (Quadrants 2 and 4) [A1].
Total number of real solutions in $[0, 2\pi]$ is $2 + 2 = 4$ [A1].
Let $f(x) = \arcsin(2x) + \arccos(x)$.
(a) Find the largest possible real domain of $f(x)$. [3 marks]
(b) Find the coordinates of the $y$-intercept of $y = f(x)$. [1 mark]
(c) Determine the exact value of $x$ for which $f(x) = \frac{\pi}{2}$. [3 marks]
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(a) Domain: [3 marks]
• For $\arcsin(2x)$: $-1 \le 2x \le 1 \implies -\frac{1}{2} \le x \le \frac{1}{2}$ [A1].
• For $\arccos(x)$: $-1 \le x \le 1$ [A1].
The domain of $f$ is the intersection: $\left[-\frac{1}{2}, \frac{1}{2}\right]$ [A1].
(b) $y$-intercept: [1 mark]
$f(0) = \arcsin(0) + \arccos(0) = 0 + \frac{\pi}{2} = \frac{\pi}{2}$. Point: $\left(0, \frac{\pi}{2}\right)$ [A1].
(c) Solving $f(x) = \frac{\pi}{2}$: [3 marks]
$\arcsin(2x) + \arccos(x) = \frac{\pi}{2} \implies \arcsin(2x) = \frac{\pi}{2} - \arccos(x)$ [M1].
Use identity $\frac{\pi}{2} - \arccos(x) = \arcsin(x)$ [M1]:
$\arcsin(2x) = \arcsin(x)$
Since $\arcsin$ is strictly one-to-one on its domain:
$2x = x \implies x = 0$ [A1].
Check: $x = 0$ lies in $\left[-\frac{1}{2}, \frac{1}{2}\right]$, so $x = 0$ is the unique solution.