IB Syllabus Alignment

Standard Level Topic 3.7: The circular functions $\sin x, \cos x, \tan x$; their domains and ranges; amplitude, their periodic nature; and their graphs. Composite transformations: $f(x) = a\sin(b(x - c)) + d$ and $f(x) = a\cos(b(x - c)) + d$; finding parameters from given features or data points. Modeling periodic phenomena such as tides, rotating wheels, temperature cycles, and waves.

1. Anatomy of the Sinusoidal Wave Model

A general sinusoidal wave can be written in either sine or cosine form: $$y = a\sin(b(x - c)) + d \qquad \text{or} \qquad y = a\cos(b(x - c)) + d$$ Each parameter controls a specific geometric transformation of the parent curve.

  • Amplitude ($|a|$): The vertical distance from the midline to a peak (or trough): $$\text{Amplitude } = |a| = \frac{\text{Maximum} - \text{Minimum}}{2}$$ If $a < 0$, the wave is reflected vertically across its midline.
  • Principal Axis / Midline ($y = d$): The horizontal center line about which the curve oscillates: $$d = \frac{\text{Maximum} + \text{Minimum}}{2}$$ The range of the function is $[d - |a|, \; d + |a|]$.
  • Period ($T$) and Frequency ($b$): The horizontal distance for one complete repeating cycle: $$T = \frac{2\pi}{|b|} \quad (\text{radians}) \qquad \text{or} \qquad T = \frac{360^\circ}{|b|} \quad (\text{degrees})$$ Conversely, the coefficient $b$ is given by $b = \frac{2\pi}{T}$.
  • Phase Shift ($c$): The horizontal translation. If written as $b(x - c)$, the curve is shifted $c$ units to the right. If written as $bx - k$, you must factor out $b$: $b\left(x - \frac{k}{b}\right)$, giving a horizontal shift of $\frac{k}{b}$.
    • For $y = a\cos(b(x - c)) + d$, $c$ represents the $x$-coordinate of a maximum (peak).
    • For $y = a\sin(b(x - c)) + d$, $c$ represents the $x$-coordinate of an inflection point on the midline where the curve is increasing.
  • The Quarter-Period Rule: Every sinusoidal wave exhibits four key geometric intervals per cycle, spaced exactly $\frac{T}{4}$ apart: $$\text{Midline Crossing} \xrightarrow{+T/4} \text{Peak} \xrightarrow{+T/4} \text{Midline Crossing} \xrightarrow{+T/4} \text{Trough} \xrightarrow{+T/4} \text{Midline Crossing}$$
Worked Example 1.1 — Reading Wave Parameters & Range

A function is defined by $f(x) = 5\cos\left(3x - \frac{\pi}{2}\right) + 4$.
(a) State the amplitude and midline equation of $f(x)$.
(b) Find the period of $f(x)$.
(c) Find the horizontal phase shift and direction.
(d) State the range of $f(x)$.

  1. Step 1: Rewrite in factored form: Factor out $3$ from the horizontal argument: $$3x - \frac{\pi}{2} = 3\left(x - \frac{\pi}{6}\right)$$ Thus, $f(x) = 5\cos\left(3\left(x - \frac{\pi}{6}\right)\right) + 4$.
  2. Step 2: Identify parameters: - Amplitude: $a = 5$.
    - Midline: $y = 4$.
    - Period: $T = \frac{2\pi}{b} = \frac{2\pi}{3}$.
    - Phase shift: $c = \frac{\pi}{6}$ (shifted $\frac{\pi}{6}$ units right).
  3. Step 3: Determine range: $$\text{Max} = d + a = 4 + 5 = 9$$ $$\text{Min} = d - a = 4 - 5 = -1$$ $$\text{Range: } [-1, 9]$$
Answers: (a) Amplitude $= 5$, Midline: $y = 4$; (b) Period $= \frac{2\pi}{3}$; (c) Phase shift $= \frac{\pi}{6}$ units right; (d) Range: $[-1, 9]$.

2. The Tangent Function & Vertical Asymptotes

The tangent function $y = \tan x = \frac{\sin x}{\cos x}$ differs substantially from sine and cosine: it is unbounded, has no amplitude, and possesses an infinite sequence of vertical asymptotes.

  • Fundamental Period: The parent tangent function repeats every $\pi$ radians ($180^\circ$): $$\text{Period of } y = a\tan(bx) = \frac{\pi}{|b|}$$
  • Vertical Asymptotes: Since $\tan x = \frac{\sin x}{\cos x}$, vertical asymptotes occur wherever the denominator $\cos x = 0$: $$x = \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z} \qquad \left(\dots, -\frac{\pi}{2}, \; \frac{\pi}{2}, \; \frac{3\pi}{2}, \; \frac{5\pi}{2}, \dots\right)$$ For the transformed function $y = a\tan(b(x - c)) + d$: $$b(x - c) = \frac{\pi}{2} + k\pi \implies x = c + \frac{\pi}{2b} + \frac{k\pi}{b}$$
  • Key Properties of Tangent:
    • Domain: $\{x \in \mathbb{R} \mid x \neq \frac{\pi}{2} + k\pi, k \in \mathbb{Z}\}$.
    • Range: $(-\infty, \infty)$ (all real numbers $\mathbb{R}$).
    • Amplitude: None (or undefined), because the curve extends indefinitely toward $\pm\infty$.
    • Inflection points / Midline intercepts occur halfway between consecutive asymptotes.
Worked Example 1.2 — Asymptotes and Period of a Transformed Tangent

Let $g(x) = 2\tan\left(2x - \frac{\pi}{3}\right)$ for $x \in [0, \pi]$.
(a) Find the period of $g(x)$.
(b) Find the equations of all vertical asymptotes of $g(x)$ in the interval $[0, \pi]$.
(c) Find the coordinates of all $x$-intercepts of $g(x)$ in $[0, \pi]$.

  1. Step 1: Calculate period: $$T = \frac{\pi}{b} = \frac{\pi}{2}$$
  2. Step 2: Solve for vertical asymptotes: Asymptotes occur where the argument equals $\frac{\pi}{2} + k\pi$: $$2x - \frac{\pi}{3} = \frac{\pi}{2} + k\pi \implies 2x = \frac{\pi}{2} + \frac{\pi}{3} + k\pi = \frac{5\pi}{6} + k\pi$$ $$x = \frac{5\pi}{12} + \frac{k\pi}{2}$$ For $k = 0$: $x = \frac{5\pi}{12} \in [0, \pi]$.
    For $k = 1$: $x = \frac{5\pi}{12} + \frac{6\pi}{12} = \frac{11\pi}{12} \in [0, \pi]$.
    For $k = 2$: $x > \pi$.
    Thus, asymptotes in $[0, \pi]$ are $x = \frac{5\pi}{12}$ and $x = \frac{11\pi}{12}$.
  3. Step 3: Solve for $x$-intercepts: Set $g(x) = 0 \implies \tan\left(2x - \frac{\pi}{3}\right) = 0$: $$2x - \frac{\pi}{3} = k\pi \implies 2x = \frac{\pi}{3} + k\pi \implies x = \frac{\pi}{6} + \frac{k\pi}{2}$$ For $k = 0$: $x = \frac{\pi}{6}$.
    For $k = 1$: $x = \frac{\pi}{6} + \frac{3\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3}$.
    The intercepts are $\left(\frac{\pi}{6}, 0\right)$ and $\left(\frac{2\pi}{3}, 0\right)$.
Answers: (a) Period $= \frac{\pi}{2}$; (b) Asymptotes: $x = \frac{5\pi}{12}, x = \frac{11\pi}{12}$; (c) Intercepts: $\left(\frac{\pi}{6}, 0\right), \left(\frac{2\pi}{3}, 0\right)$.

Anatomy of Sinusoidal Wave Transformations

The diagram below highlights the geometric relationships governing a full wave cycle: Amplitude $a$, Midline $y = d$, Period $T = \frac{2\pi}{b}$, and Phase Shift $c$.

x y O y = d (Midline) d + a d - a Phase Shift c One Full Period T = 2π / b Amplitude a Peak (c+T/4, d+a) (c+T/2, d) Trough (c+3T/4, d-a) Sinusoidal Formula y = a sin(b(x - c)) + d
Figure 3.4 — Anatomy of a Transformed Sinusoidal Function: Visualizing Amplitude $a = \frac{\text{Max}-\text{Min}}{2}$, Midline $y = d$, Period $T = \frac{2\pi}{b}$, and horizontal phase displacement $c$.

3. Modeling Periodic Real-World Phenomena

In IB DP examinations, cyclic scenarios such as Ferris wheels, ocean tides, predator-prey populations, and seasonal temperatures are modeled using sine and cosine functions.

  1. Extracting Extrema: Identify the maximum value ($M$) and minimum value ($m$) from the physical context. $$a = \frac{M - m}{2}, \qquad d = \frac{M + m}{2}$$
  2. Determining the Period & $b$: Identify the time taken for one full revolution or cycle ($T$). $$b = \frac{2\pi}{T}$$
  3. Selecting Sine vs. Cosine:
    • Cosine with $a > 0$: Use when the scenario starts ($t = 0$) at its highest point: $y = a\cos(bt) + d$.
    • Cosine with $a < 0$ (or negative cosine): Use when the scenario starts ($t = 0$) at its lowest point (e.g., boarding a Ferris wheel at the bottom): $$y = -a\cos(bt) + d$$
    • Sine with $a > 0$: Use when the scenario starts at the midline and is increasing: $y = a\sin(bt) + d$.
Worked Example 1.3 — The Ferris Wheel Model

A Ferris wheel of diameter $60\text{ m}$ rotates at a constant speed, completing one full revolution every $80\text{ seconds}$. Passengers board their capsules at the bottom of the wheel, $4\text{ m}$ above the ground.
(a) Find the maximum height reached by a passenger.
(b) Formulate a trigonometric model for height $h(t)$ in the form $h(t) = a\cos(bt) + d$, where $a < 0$.
(c) Calculate the passenger's height after $20\text{ seconds}$.
(d) Find the first two times during the first revolution when the passenger is at a height of $49\text{ m}$.

  1. Step 1: Determine extrema and parameters: - Minimum height: $m = 4\text{ m}$.
    - Maximum height: $M = 4 + 60 = 64\text{ m}$.
    - Amplitude: $|a| = \frac{64 - 4}{2} = 30\text{ m}$. Since boarding is at the bottom, $a = -30$.
    - Midline: $d = \frac{64 + 4}{2} = 34\text{ m}$.
    - Period: $T = 80\text{ s} \implies b = \frac{2\pi}{80} = \frac{\pi}{40}$.
  2. Step 2: State the model: $$h(t) = -30\cos\left(\frac{\pi}{40}t\right) + 34$$
  3. Step 3: Height at $t = 20\text{ s}$: $$h(20) = -30\cos\left(\frac{20\pi}{40}\right) + 34 = -30\cos\left(\frac{\pi}{2}\right) + 34 = -30(0) + 34 = 34\text{ m}$$ (The passenger is exactly at the midline height after one quarter of a turn).
  4. Step 4: Solve for $h(t) = 49\text{ m}$: $$-30\cos\left(\frac{\pi}{40}t\right) + 34 = 49 \implies -30\cos\left(\frac{\pi}{40}t\right) = 15 \implies \cos\left(\frac{\pi}{40}t\right) = -\frac{1}{2}$$ Recall that $\cos\theta = -\frac{1}{2}$ at $\theta = \frac{2\pi}{3}$ and $\theta = \frac{4\pi}{3}$: $$\frac{\pi}{40}t_1 = \frac{2\pi}{3} \implies t_1 = 40 \times \frac{2}{3} = \frac{80}{3} \approx 26.7\text{ seconds}$$ $$\frac{\pi}{40}t_2 = \frac{4\pi}{3} \implies t_2 = 40 \times \frac{4}{3} = \frac{160}{3} \approx 53.3\text{ seconds}$$
Answers: (a) Max height $= 64\text{ m}$; (b) $h(t) = -30\cos\left(\frac{\pi}{40}t\right) + 34$; (c) $h(20) = 34\text{ m}$; (d) $t = 26.7\text{ s}$ and $t = 53.3\text{ s}$.

4. Tidal Modeling & Parameter Extraction

Worked Example 1.4 — Tidal Water Depth Modeling

The depth of water $D(t)$ metres in a harbor is modeled by the function $D(t) = A\cos(Bt) + C$, where $t$ is the time in hours after midnight.
High tide occurs at 04:00 with depth $15\text{ m}$. The subsequent low tide occurs at 10:00 with depth $7\text{ m}$.
(a) Find the values of $A$, $B$, and $C$. [4 marks]
(b) Calculate the depth of water in the harbor at 12:00 noon. [2 marks]
(c) A cruise ship requires a minimum water depth of $13\text{ m}$ to enter the harbor. Find the total number of hours between 00:00 and 12:00 during which the ship can safely enter. [4 marks]

  1. Part (a): Find model parameters: $$\text{Amplitude } A = \frac{\text{Max} - \text{Min}}{2} = \frac{15 - 7}{2} = 4\text{ m} \quad [A1]$$ $$\text{Vertical shift } C = \frac{\text{Max} + \text{Min}}{2} = \frac{15 + 7}{2} = 11\text{ m} \quad [A1]$$ $$\text{Half-period} = 10 - 4 = 6\text{ h} \implies \text{Full Period } P = 12\text{ h} \quad [M1]$$ $$B = \frac{2\pi}{P} = \frac{2\pi}{12} = \frac{\pi}{6} \quad [A1]$$

    Since maximum occurs at $t = 4$, shift cosine: $D(t) = 4\cos\left(\frac{\pi}{6}(t - 4) \right) + 11$.

  2. Part (b): Depth at $t = 12$: $$D(12) = 4\cos\left(\frac{\pi}{6}(12 - 4) \right) + 11 = 4\cos\left(\frac{8\pi}{6} \right) + 11 = 4\cos\left(\frac{4\pi}{3} \right) + 11 \quad [M1]$$ $$\cos\left(\frac{4\pi}{3} \right) = -\frac{1}{2} \implies D(12) = 4\left(-\frac{1}{2} \right) + 11 = -2 + 11 = 9\text{ m} \quad [A1]$$
  3. Part (c): Safe entry window ($D(t) \ge 13$): $$4\cos\left(\frac{\pi}{6}(t - 4) \right) + 11 \ge 13 \implies 4\cos\left(\frac{\pi}{6}(t - 4) \right) \ge 2 \implies \cos\left(\frac{\pi}{6}(t - 4) \right) \ge \frac{1}{2} \quad [M1]$$ $$\frac{\pi}{6}(t - 4) \in \left[-\frac{\pi}{3}, \frac{\pi}{3} \right] \implies t - 4 \in [-2, 2] \implies 2 \le t \le 6 \quad [M1][A1]$$ $$\text{Total duration} = 6 - 2 = 4\text{ hours} \quad [A1]$$
Answers: (a) $A = 4, B = \frac{\pi}{6}, C = 11$; (b) $9\text{ m}$; (c) $4\text{ hours}$ (from 02:00 to 06:00).
Worked Example 1.5 — Tangent Graph Asymptotes and Intercepts

The function $f(x) = 3\tan\left(2x - \frac{\pi}{4} \right)$ is defined on the interval $0 \le x \le \pi$.
(a) State the period of $f(x)$. [1 mark]
(b) Find the equations of all vertical asymptotes of $f(x)$ on this interval. [3 marks]
(c) Find the $x$-coordinates of all $x$-intercepts of $f(x)$ on this interval. [3 marks]

  1. Part (a): Period of tangent function: $$\text{Period} = \frac{\pi}{b} = \frac{\pi}{2} \quad [A1]$$
  2. Part (b): Vertical asymptotes occur where argument $= \frac{\pi}{2} + k\pi$: $$2x - \frac{\pi}{4} = \frac{\pi}{2} \implies 2x = \frac{3\pi}{4} \implies x = \frac{3\pi}{8} \quad [M1][A1]$$ $$2x - \frac{\pi}{4} = \frac{3\pi}{2} \implies 2x = \frac{7\pi}{4} \implies x = \frac{7\pi}{8} \quad [A1]$$ Both $\frac{3\pi}{8}$ and $\frac{7\pi}{8}$ lie in $[0, \pi]$.
  3. Part (c): $x$-intercepts where $\tan(2x - \frac{\pi}{4}) = 0 \implies 2x - \frac{\pi}{4} = k\pi$: $$k = 0 \implies 2x - \frac{\pi}{4} = 0 \implies 2x = \frac{\pi}{4} \implies x = \frac{\pi}{8} \quad [M1][A1]$$ $$k = 1 \implies 2x - \frac{\pi}{4} = \pi \implies 2x = \frac{5\pi}{4} \implies x = \frac{5\pi}{8} \quad [A1]$$
Answers: (a) $\frac{\pi}{2}$; (b) $x = \frac{3\pi}{8}, x = \frac{7\pi}{8}$; (c) $x = \frac{\pi}{8}, x = \frac{5\pi}{8}$.

10 Tiered Practice Exercises

Test your understanding across procedural graph parameter readings, tangent asymptote derivations, authentic harmonic modeling, and calculus extrema.

Tier 1: Core Fluency (Problems 1–4)

Exercise 1 Sinusoidal Parameters

For the curve $y = 6\sin(4x) - 1$, state:
(a) The amplitude.
(b) The period in radians.
(c) The equation of the midline (principal axis).
(d) The minimum and maximum values of the curve.

View Full Solution

(a) Amplitude: $|a| = 6$.

(b) Period: $T = \frac{2\pi}{b} = \frac{2\pi}{4} = \frac{\pi}{2}\text{ radians}$.

(c) Midline: $y = -1$.

(d) Extrema:
$$\text{Max} = d + a = -1 + 6 = 5$$ $$\text{Min} = d - a = -1 - 6 = -7$$

Final Answers: (a) $6$; (b) $\frac{\pi}{2}$; (c) $y = -1$; (d) $\text{Min} = -7, \text{Max} = 5$.

Exercise 2 Tangent Period & Asymptotes

Consider the function $f(x) = \tan\left(3x\right)$ for $0 \le x \le \pi$.
(a) State the period of $f(x)$.
(b) Find the equations of all vertical asymptotes of $f(x)$ in the interval $[0, \pi]$.

View Full Solution

(a) Period: For tangent, $T = \frac{\pi}{b} = \frac{\pi}{3}$.

(b) Vertical asymptotes:
Set $3x = \frac{\pi}{2} + k\pi \implies x = \frac{\pi}{6} + \frac{k\pi}{3}$.
For $k = 0$: $x = \frac{\pi}{6}$.
For $k = 1$: $x = \frac{\pi}{6} + \frac{2\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2}$.
For $k = 2$: $x = \frac{\pi}{6} + \frac{4\pi}{6} = \frac{5\pi}{6}$.
All three values lie within $[0, \pi]$.

Final Answers: (a) $\frac{\pi}{3}$; (b) $x = \frac{\pi}{6}, x = \frac{\pi}{2}, x = \frac{5\pi}{6}$.

Exercise 3 Formulating Model from Extrema

A cosine curve $y = a\cos(b(x - c)) + d$ has its first peak at $(2, 11)$ and the very next trough at $(8, -3)$, with $a > 0$ and $b > 0$. Find the values of $a, b, c$, and $d$.

View Full Solution

Step 1: Calculate $a$ and $d$:
$$a = \frac{\text{Max} - \text{Min}}{2} = \frac{11 - (-3)}{2} = \frac{14}{2} = 7$$ $$d = \frac{\text{Max} + \text{Min}}{2} = \frac{11 + (-3)}{2} = \frac{8}{2} = 4$$

Step 2: Calculate Period and $b$:
The horizontal distance from peak to trough is half a period: $$\frac{T}{2} = 8 - 2 = 6 \implies T = 12$$ $$b = \frac{2\pi}{T} = \frac{2\pi}{12} = \frac{\pi}{6}$$

Step 3: Determine phase shift $c$:
A standard unshifted cosine peak occurs at $x = 0$. Here the peak is at $x = 2$, so $c = 2$.

Final Answers: $a = 7, b = \frac{\pi}{6}, c = 2, d = 4$.

Exercise 4 Evaluating Intercepts

Find all $x$-intercepts of $y = 2\sin\left(x + \frac{\pi}{4}\right) - \sqrt{2}$ in the domain $0 \le x \le 2\pi$.

View Full Solution

Step 1: Set $y = 0$:
$$2\sin\left(x + \frac{\pi}{4}\right) = \sqrt{2} \implies \sin\left(x + \frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$$

Step 2: Solve inside interval:
Since $0 \le x \le 2\pi$, the argument satisfies $\frac{\pi}{4} \le x + \frac{\pi}{4} \le \frac{9\pi}{4}$.
The angles where sine equals $\frac{\sqrt{2}}{2}$ are $\frac{\pi}{4}, \frac{3\pi}{4}$, and $\frac{9\pi}{4}$: $$x + \frac{\pi}{4} = \frac{\pi}{4} \implies x = 0$$ $$x + \frac{\pi}{4} = \frac{3\pi}{4} \implies x = \frac{\pi}{2}$$ $$x + \frac{\pi}{4} = \frac{9\pi}{4} \implies x = 2\pi$$

Final Answer: $x = 0, \frac{\pi}{2}, 2\pi$.

Tier 2: Exam Modeling (Problems 5–8)

Exercise 5 Tidal Water Depth Modeling

The depth of water, $D$ meters, in a harbour $t$ hours after midnight is modeled by: $$D(t) = 4\cos\left(\frac{\pi}{6}(t - 2)\right) + 7, \qquad 0 \le t \le 24$$ (a) State the maximum and minimum water depths.
(b) Find the time of the first high tide after midnight.
(c) A cargo ship requires at least $9\text{ m}$ of water to safely enter the harbour. Find the number of hours during the first 12 hours of the day when the ship can enter safely.

View Full Solution

(a) Maximum and minimum depths:
$$\text{Max} = 7 + 4 = 11\text{ m}, \qquad \text{Min} = 7 - 4 = 3\text{ m}$$

(b) First high tide:
High tide occurs at the cosine peak, which happens when $\frac{\pi}{6}(t - 2) = 0 \implies t = 2$ hours ($02:00$).

(c) Safe entry window ($D(t) \ge 9$):
$$4\cos\left(\frac{\pi}{6}(t - 2)\right) + 7 \ge 9 \implies \cos\left(\frac{\pi}{6}(t - 2)\right) \ge \frac{2}{4} = \frac{1}{2}$$ For $\cos\theta \ge \frac{1}{2}$, $-\frac{\pi}{3} \le \theta \le \frac{\pi}{3}$: $$-\frac{\pi}{3} \le \frac{\pi}{6}(t - 2) \le \frac{\pi}{3}$$ Multiply through by $\frac{6}{\pi}$: $$-2 \le t - 2 \le 2 \implies 0 \le t \le 4$$ The period is $T = \frac{2\pi}{\pi/6} = 12\text{ hours}$.
During the first 12 hours ($0 \le t \le 12$), the safe window is $t \in [0, 4]$.
Total duration $= 4 - 0 = 4\text{ hours}$.

Final Answers: (a) $\text{Max} = 11\text{ m}, \text{Min} = 3\text{ m}$; (b) $02:00$ ($t = 2$); (c) $4\text{ hours}$.

Exercise 6 Phase Shift Form Equivalence

Show that the function $f(x) = -4\sin(2x) + 3$ can be expressed in the form $f(x) = 4\cos(2(x - c)) + 3$, finding the smallest positive value of $c$.

View Full Solution

Step 1: Use cosine phase identity:
Recall that $\cos\left(\theta + \frac{\pi}{2}\right) = -\sin\theta$.
Therefore: $$-4\sin(2x) = 4\cos\left(2x + \frac{\pi}{2}\right)$$

Step 2: Factor out 2:
$$4\cos\left(2\left(x + \frac{\pi}{4}\right)\right) + 3$$ To write in the form $x - c$ with $c > 0$, recall that cosine has period $2\pi$, so the argument can be shifted by $\pm 2\pi$: $$2\left(x + \frac{\pi}{4}\right) - 2\pi = 2\left(x + \frac{\pi}{4} - \pi\right) = 2\left(x - \frac{3\pi}{4}\right)$$ Thus, $c = \frac{3\pi}{4}$.

Final Answer: $c = \frac{3\pi}{4}$

Exercise 7 Seasonal Temperature Model

The average daily temperature $T$ in a city over a 365-day year is modeled by: $$T(d) = -12\cos\left(\frac{2\pi}{365}(d - 15)\right) + 16$$ where $d$ is the day of the year ($d = 1$ is January 1st).
(a) Find the maximum and minimum average temperatures and the days on which they occur.
(b) Find the temperature on day $d = 106$ (mid-April).

View Full Solution

(a) Extrema and days:
Because the coefficient of cosine is $-12$ (negative):
- Minimum occurs when $\cos(\dots) = 1$: $$d - 15 = 0 \implies d = 15\text{ (January 15th)}, \quad T_{\min} = -12(1) + 16 = 4^\circ\text{C}$$ - Maximum occurs half a year later ($\frac{365}{2} = 182.5$ days later): $$d = 15 + 182.5 = 197.5 \implies \text{Day } 197 \text{ or } 198\text{ (mid-July)}, \quad T_{\max} = -12(-1) + 16 = 28^\circ\text{C}$$

(b) Temperature on day $d = 106$:
$$T(106) = -12\cos\left(\frac{2\pi}{365}(106 - 15)\right) + 16 = -12\cos\left(\frac{2\pi \times 91}{365}\right) + 16$$ Note that $\frac{91}{365} \approx \frac{1}{4}$, so the argument is $\approx \frac{2\pi}{4} = \frac{\pi}{2}$. $$\cos\left(\frac{\pi}{2}\right) \approx 0 \implies T(106) \approx -12(0) + 16 = 16^\circ\text{C}$$

Final Answers: (a) Min $4^\circ\text{C}$ on day 15; Max $28^\circ\text{C}$ on day 197/198; (b) $16.0^\circ\text{C}$.

Exercise 8 Reconstructing Tangent Parameters

The graph of $y = a\tan(bx)$ has adjacent vertical asymptotes at $x = -\frac{\pi}{4}$ and $x = \frac{\pi}{4}$, and passes through the point $\left(\frac{\pi}{8}, 3\right)$. Determine the exact values of $a$ and $b$ (with $a, b > 0$).

View Full Solution

Step 1: Determine period and $b$:
The distance between adjacent asymptotes is one period: $$T = \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) = \frac{\pi}{2}$$ Since $T = \frac{\pi}{b}$, we have $\frac{\pi}{b} = \frac{\pi}{2} \implies b = 2$.

Step 2: Determine $a$:
Substitute point $\left(\frac{\pi}{8}, 3\right)$ into $y = a\tan(2x)$: $$3 = a\tan\left(2 \times \frac{\pi}{8}\right) = a\tan\left(\frac{\pi}{4}\right) = a(1) \implies a = 3$$

Final Answers: $a = 3, b = 2$.

Tier 3: Synthesis & Problem-Solving (Problems 9–10)

Exercise 9 Calculus Rate of Change on a Wave

A buoy moves vertically according to $h(t) = 3\sin(2t) + 5$, where $h$ is height in meters and $t$ is time in seconds.
(a) Find the derivative $\frac{dh}{dt}$ representing the vertical velocity of the buoy.
(b) Find the maximum speed of the buoy and state the heights at which this maximum speed occurs.
(c) Find the acceleration of the buoy when it is at its highest point.

View Full Solution

(a) Derivative:
$$\frac{dh}{dt} = 3 \times 2\cos(2t) = 6\cos(2t)\text{ m/s}$$

(b) Maximum speed:
$$\text{Speed} = \left|\frac{dh}{dt}\right| = 6|\cos(2t)| \le 6\text{ m/s}$$ Maximum speed occurs when $|\cos(2t)| = 1 \implies \sin(2t) = 0$.
When $\sin(2t) = 0$, $h = 3(0) + 5 = 5\text{ m}$ (at the midline!).

(c) Acceleration at highest point:
$$\frac{d^2h}{dt^2} = -12\sin(2t)$$ At the highest point, $\sin(2t) = 1$: $$\frac{d^2h}{dt^2} = -12(1) = -12\text{ m/s}^2$$

Final Answers: (a) $6\cos(2t)$; (b) Maximum speed is $6\text{ m/s}$, occurring at height $h = 5\text{ m}$; (c) Acceleration is $-12\text{ m/s}^2$.

Exercise 10 Multi-Cycle Intersection Analysis

Let $f(x) = 2\sin(2x) + 3$ and $g(x) = 4$ for $x \in [0, 2\pi]$.
(a) Find all exact points of intersection between the two curves.
(b) Determine the set of values of $x$ in $[0, 2\pi]$ for which $f(x) \ge 4$.
(c) Calculate the total length of the intervals where $f(x) \ge 4$.

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(a) Intersections:
$$2\sin(2x) + 3 = 4 \implies \sin(2x) = \frac{1}{2}$$ For $x \in [0, 2\pi]$, $2x \in [0, 4\pi]$.
Cycle 1: $2x = \frac{\pi}{6}, \frac{5\pi}{6} \implies x = \frac{\pi}{12}, \frac{5\pi}{12}$.
Cycle 2: $2x = \frac{\pi}{6} + 2\pi = \frac{13\pi}{6}, \frac{5\pi}{6} + 2\pi = \frac{17\pi}{6} \implies x = \frac{13\pi}{12}, \frac{17\pi}{12}$.

(b) Solution set for $f(x) \ge 4$:
Since sine is at or above $\frac{1}{2}$ between the lower and upper angles: $$x \in \left[\frac{\pi}{12}, \frac{5\pi}{12}\right] \cup \left[\frac{13\pi}{12}, \frac{17\pi}{12}\right]$$

(c) Total length of intervals:
$$\Delta x_1 = \frac{5\pi}{12} - \frac{\pi}{12} = \frac{4\pi}{12} = \frac{\pi}{3}$$ $$\Delta x_2 = \frac{17\pi}{12} - \frac{13\pi}{12} = \frac{4\pi}{12} = \frac{\pi}{3}$$ $$\text{Total Length} = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3}\text{ radians}$$

Final Answers: (a) $x = \frac{\pi}{12}, \frac{5\pi}{12}, \frac{13\pi}{12}, \frac{17\pi}{12}$; (b) $\left[\frac{\pi}{12}, \frac{5\pi}{12}\right] \cup \left[\frac{13\pi}{12}, \frac{17\pi}{12}\right]$; (c) $\frac{2\pi}{3}$.

Trigonometric Functions Reference Sheet

In the IB Formula Booklet (Topic 3.7)

The official IB Mathematics AA SL formula booklet references the periodic relations:

Period of Sine & Cosine $$T = \frac{2\pi}{|b|}$$
Period of Tangent $$T = \frac{\pi}{|b|}$$
Pythagorean Identity $$\cos^2\theta + \sin^2\theta = 1$$
Tangent Definition $$\tan\theta = \frac{\sin\theta}{\cos\theta}$$
Must Memorize & Harmonic Modeling Keys
  • Parameter Extraction Formulas: $$a = \frac{\text{Max} - \text{Min}}{2}, \qquad d = \frac{\text{Max} + \text{Min}}{2}, \qquad b = \frac{2\pi}{T}$$
  • Phase Shift Factoring Rule: In $y = \sin(bx - c)$, you must factor out $b$: $y = \sin(b(x - c/b))$, giving a shift of $\frac{c}{b}$.
  • Tangent Asymptotes Formula: For $y = \tan(bx)$, asymptotes occur at: $$x = \frac{\pi}{2b} + \frac{k\pi}{b}, \quad k \in \mathbb{Z}$$
  • Quarter-Cycle Principle: Adjacent key features (midline crossing to peak) are separated by $\frac{T}{4}$.
Critical IB Examination Pitfalls
  • Pitfall 1: Confusing Frequency $b$ with Period $T$: Writing $b$ as the period instead of $T = \frac{2\pi}{b}$. For example, in $y = \sin(4x)$, the period is $\frac{2\pi}{4} = \frac{\pi}{2}$, NOT $4$.
  • Pitfall 2: Claiming Tangent Has an Amplitude: Answering that the amplitude of $y = 3\tan x$ is $3$. Tangent is unbounded ($-\infty, \infty$), so amplitude is undefined!
  • Pitfall 3: Tangent Period Confusion: Using $\frac{2\pi}{b}$ for tangent instead of $\frac{\pi}{b}$.
  • Pitfall 4: Neglecting Inverted Sine/Cosine Sign: Writing $a = 30$ instead of $a = -30$ when modeling a ride starting at the minimum bottom height.

Section A: Diagnostic Multiple Choice

Answer the questions to check your understanding and score.

Score: 0 / 4

Question 1: What are the amplitude and midline equation of $y = -5\cos(2x) + 3$?

Question 2: What is the period of the function $y = \tan(3x)$?

Question 3: What is the maximum value attained by $f(t) = 6\sin(4t - 1) - 2$?

Question 4: What is the horizontal phase shift of $y = \sin(4x - \pi)$?

Section B: Exam-Style Free-Response Questions

Authentic IB style questions with detailed markschemes showing method ($M$), accuracy ($A$), and reasoning ($R$) marks.

Question 5 (Paper 1 Style) 6 Marks

The diagram shows the graph of $f(x) = a\sin(bx) + d$ for $0 \le x \le 2\pi$. The graph has a local maximum at $\left(\frac{\pi}{4}, 8\right)$ and a local minimum at $\left(\frac{3\pi}{4}, 2\right)$.
(a) Find the values of $a$ and $d$. [2]
(b) Find the period of $f$ and hence determine the value of $b$. [2]
(c) Find the exact solutions to $f(x) = 5$ in the interval $[0, \pi]$. [2]

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(a) Find $a$ and $d$: [2 marks]
$$a = \frac{\text{Max} - \text{Min}}{2} = \frac{8 - 2}{2} = 3 \quad (A1)$$ $$d = \frac{\text{Max} + \text{Min}}{2} = \frac{8 + 2}{2} = 5 \quad (A1)$$

(b) Period and $b$: [2 marks]
Half a period is $\frac{3\pi}{4} - \frac{\pi}{4} = \frac{2\pi}{4} = \frac{\pi}{2} \implies T = \pi$ (M1).
$$b = \frac{2\pi}{T} = \frac{2\pi}{\pi} = 2 \quad (A1)$$

(c) Exact solutions to $f(x) = 5$: [2 marks]
$$3\sin(2x) + 5 = 5 \implies 3\sin(2x) = 0 \implies \sin(2x) = 0 \quad (M1)$$ For $x \in [0, \pi]$, $2x \in [0, 2\pi]$: $$2x = 0, \pi, 2\pi \implies x = 0, \frac{\pi}{2}, \pi \quad (A1)$$

Question 6 (Paper 2 Style) 8 Marks

The depth of water, $h$ meters, at the entrance to a marina is given by: $$h(t) = 2.5\cos\left(\frac{\pi}{6}(t - 1)\right) + 4.5, \qquad 0 \le t \le 24$$ where $t$ is the number of hours after midnight.
(a) Find the maximum water depth and the time of day at which it first occurs. [2]
(b) Find the depth of the water at $08:00$ ($t = 8$). [2]
(c) A boat requires a minimum depth of $5.75\text{ m}$ to enter the marina. Find the time intervals during the 24-hour day when the boat can safely enter. [4]

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(a) Maximum depth and time: [2 marks]
$$\text{Max Depth} = 4.5 + 2.5 = 7.0\text{ m} \quad (A1)$$ First occurs when $\frac{\pi}{6}(t - 1) = 0 \implies t = 1$ ($01:00\text{ AM}$) (A1).

(b) Depth at $t = 8$: [2 marks]
$$h(8) = 2.5\cos\left(\frac{\pi}{6}(8 - 1)\right) + 4.5 = 2.5\cos\left(\frac{7\pi}{6}\right) + 4.5 \quad (M1)$$ Since $\cos(7\pi/6) = -\frac{\sqrt{3}}{2} \approx -0.8660$: $$h(8) = 2.5(-0.8660) + 4.5 = -2.165 + 4.5 \approx 2.33\text{ m} \quad (A1)$$

(c) Safe intervals ($h(t) \ge 5.75$): [4 marks]
$$2.5\cos\left(\frac{\pi}{6}(t - 1)\right) + 4.5 \ge 5.75 \implies 2.5\cos\left(\frac{\pi}{6}(t - 1)\right) \ge 1.25 \quad (M1)$$ $$\cos\left(\frac{\pi}{6}(t - 1)\right) \ge \frac{1.25}{2.5} = 0.5$$ Cosine equals $0.5$ at $\pm\frac{\pi}{3}$: $$-\frac{\pi}{3} \le \frac{\pi}{6}(t - 1) \le \frac{\pi}{3} \quad (M1)$$ $$-2 \le t - 1 \le 2 \implies -1 \le t \le 3$$ Since $0 \le t \le 24$ and period is $T = 12\text{ hours}$:
First cycle: $t \in [0, 3]$ and $t \in [11, 15]$ (since $-1 \equiv 11$).
Second cycle ($+12\text{ hours}$): $t \in [23, 24]$.
Thus, safe intervals are: $$[0, 3], \quad [11, 15], \quad [23, 24] \quad (A2)$$