SPM Paper 1 & Paper 2 (8 - 10 Mark Sketch Questions)
Trigonometric Graphs, Identities & Number of Solutions
Trigonometry forms the architectural bedrock of physical oscillations, periodic cycles, and calculus integration. Master identity proofs, double angle substitutions, sketching sinusoidal waves with absolute value reflections, and superimposing straight lines to determine the exact number of roots without algebraic solving.
1. Fundamental Formulas & Trigonometric Identities
Pythagorean Identities
$\sin^2\theta + \cos^2\theta = 1$
$1 + \tan^2\theta = \sec^2\theta$
$1 + \cot^2\theta = \csc^2\theta$
Reciprocals: $\sec\theta = \frac{1}{\cos\theta}$, $\csc\theta = \frac{1}{\sin\theta}$, $\cot\theta = \frac{1}{\tan\theta}$.
Double Angle Formulas
$\sin 2A = 2\sin A \cos A$
$\cos 2A = \cos^2 A - \sin^2 A$
$\cos 2A = 2\cos^2 A - 1 = 1 - 2\sin^2 A$
$\tan 2A = \frac{2\tan A}{1 - \tan^2 A}$
Choose the $\cos 2A$ form that cancels out constants in proof questions!
Graph Characteristics
$y = a\sin(bx) + c$
Amplitude $= |a|$
Period $= \frac{2\pi}{b}$ or $\frac{360^\circ}{b}$
Cycles in $2\pi$ $= b$
For modulus $y = |f(x)|$, reflect any portions below the $x$-axis upwards.
The 4-Step SPM Protocol for "Number of Solutions":
- Sketch the primary curve $y = f(x)$ accurately with key endpoints, peaks, and troughs clearly labeled.
- Rearrange the given exam equation until the LHS strictly matches the sketched curve: $f(x) = mx + c$.
- Compute at least two coordinates (usually at domain boundaries $x=0$ and $x=\pi$ or $2\pi$) for the straight line $y = mx + c$, and plot them directly on the same axes.
- Count the number of intersection points between the curve and the straight line. State: "Number of solutions = [N]".
2. Progressive SPM Worked Examples with Marking Schemes
Example 1 • Proving Identity & Equation Solving
[6 Marks]
(a) Prove that $\frac{\sin 2x}{1 + \cos 2x} = \tan x$. [3 marks]
(b) Hence, solve the equation $\frac{2\sin 2x}{1 + \cos 2x} = 3\cot x$ for $0^\circ \le x \le 360^\circ$. [3 marks]
Step-by-Step Marking Solution
(a) Proof:
$\text{LHS} = \frac{\sin 2x}{1 + \cos 2x}$
Substitute $\sin 2x = 2\sin x \cos x$ and $\cos 2x = 2\cos^2 x - 1$:
$\text{LHS} = \frac{2\sin x \cos x}{1 + (2\cos^2 x - 1)} = \frac{2\sin x \cos x}{2\cos^2 x}$
$= \frac{\sin x}{\cos x} = \tan x = \text{RHS}$ (Proven)
$\text{LHS} = \frac{\sin 2x}{1 + \cos 2x}$
Substitute $\sin 2x = 2\sin x \cos x$ and $\cos 2x = 2\cos^2 x - 1$:
$\text{LHS} = \frac{2\sin x \cos x}{1 + (2\cos^2 x - 1)} = \frac{2\sin x \cos x}{2\cos^2 x}$
$= \frac{\sin x}{\cos x} = \tan x = \text{RHS}$ (Proven)
[K1] $\sin 2x = 2\sin x\cos x$
[K1] $\cos 2x = 2\cos^2 x - 1$
[N1] $\tan x = \text{RHS}$
(b) Solving equation:
$2\left(\frac{\sin 2x}{1 + \cos 2x}\right) = 3\cot x \implies 2\tan x = \frac{3}{\tan x}$
$2\tan^2 x = 3 \implies \tan^2 x = \frac{3}{2} = 1.5$
$\tan x = \pm \sqrt{1.5} \approx \pm 1.2247$
Reference angle $\alpha = \tan^{-1}(1.2247) = 50.77^\circ$
Quadrant 1: $x = 50.77^\circ$
Quadrant 2: $x = 180^\circ - 50.77^\circ = 129.23^\circ$
Quadrant 3: $x = 180^\circ + 50.77^\circ = 230.77^\circ$
Quadrant 4: $x = 360^\circ - 50.77^\circ = 309.23^\circ$
$2\left(\frac{\sin 2x}{1 + \cos 2x}\right) = 3\cot x \implies 2\tan x = \frac{3}{\tan x}$
$2\tan^2 x = 3 \implies \tan^2 x = \frac{3}{2} = 1.5$
$\tan x = \pm \sqrt{1.5} \approx \pm 1.2247$
Reference angle $\alpha = \tan^{-1}(1.2247) = 50.77^\circ$
Quadrant 1: $x = 50.77^\circ$
Quadrant 2: $x = 180^\circ - 50.77^\circ = 129.23^\circ$
Quadrant 3: $x = 180^\circ + 50.77^\circ = 230.77^\circ$
Quadrant 4: $x = 360^\circ - 50.77^\circ = 309.23^\circ$
[P1] replace $\tan x$
[K1] $\alpha = 50.77^\circ$
[N1] all 4 angles
Example 2 • SPM Paper 2 Classic Graph & Number of Solutions
[8 Marks]
(a) Sketch the graph of $y = |2\cos 2x| - 1$ for $0 \le x \le 2\pi$. [4 marks]
(b) Hence, using the same axes, sketch a suitable straight line to find the number of solutions to the equation $\pi|2\cos 2x| - \pi = 2x - 2\pi$. State the number of solutions. [4 marks]
Step-by-Step Marking Solution
(a) Graph Analysis:
• Base curve: $2\cos 2x$ has amplitude 2 and period $\frac{2\pi}{2} = \pi$ (so 2 complete cycles in $0 \le x \le 2\pi$).
• Modulus $|2\cos 2x|$ reflects negative troughs from $-2$ to $+2$, giving 4 humps oscillating between $0$ and $2$.
• Vertical shift $-1$: graph oscillates between minimum value $-1$ and maximum value $2 - 1 = 1$.
• Key coordinate checkpoints:
• Base curve: $2\cos 2x$ has amplitude 2 and period $\frac{2\pi}{2} = \pi$ (so 2 complete cycles in $0 \le x \le 2\pi$).
• Modulus $|2\cos 2x|$ reflects negative troughs from $-2$ to $+2$, giving 4 humps oscillating between $0$ and $2$.
• Vertical shift $-1$: graph oscillates between minimum value $-1$ and maximum value $2 - 1 = 1$.
• Key coordinate checkpoints:
- $x = 0 \implies y = |2(1)| - 1 = 1$
- $x = \frac{\pi}{4} \implies y = |0| - 1 = -1$
- $x = \frac{\pi}{2} \implies y = |-2| - 1 = 1$
- $x = \pi \implies y = 1$, $x = 2\pi \implies y = 1$
(b) Straight Line Derivation:
Given equation: $\pi|2\cos 2x| - \pi = 2x - 2\pi$
Divide entire equation by $\pi$:
$|2\cos 2x| - 1 = \frac{2}{\pi}x - 2$
Notice LHS is exactly $y$: $\therefore y = \frac{2}{\pi}x - 2$ [K1]
Find two coordinates to draw the line:
By counting the intersections between the line and the periodic curve:
Given equation: $\pi|2\cos 2x| - \pi = 2x - 2\pi$
Divide entire equation by $\pi$:
$|2\cos 2x| - 1 = \frac{2}{\pi}x - 2$
Notice LHS is exactly $y$: $\therefore y = \frac{2}{\pi}x - 2$ [K1]
Find two coordinates to draw the line:
- When $x = 0 \implies y = -2$
- When $x = \pi \implies y = \frac{2}{\pi}(\pi) - 2 = 2 - 2 = 0 \implies (\pi, 0)$
- When $x = 2\pi \implies y = \frac{2}{\pi}(2\pi) - 2 = 4 - 2 = 2 \implies (2\pi, 2)$
By counting the intersections between the line and the periodic curve:
The straight line intersects the curve at exactly 3 points. $\therefore$ Number of solutions $= 3$.
[N1 line drawn] [N1 3 solutions]
Example 3 • KBAT SPM Oceanography Modeling
[5 Marks]
The water depth $h(t)$, in meters, at a commercial harbor quay in Port Klang over a 24-hour cycle is modeled by the periodic function: $$h(t) = 3.5\sin\left(\frac{\pi}{6}t\right) + 7.0$$ where $t$ is the elapsed time in hours after midnight ($0 \le t \le 24$).
(a) State the maximum and minimum water depth at the quay. [2 marks]
(b) A container cargo vessel requires a minimum water depth of $8.75\text{ m}$ to safely berth. Determine the time intervals during the 24-hour cycle when the ship can safely berth. [3 marks]
(a) Since $-1 \le \sin\left(\frac{\pi}{6}t\right) \le 1$:
$h_{\text{max}} = 3.5(1) + 7.0 = 10.5\text{ m}$
$h_{\text{min}} = 3.5(-1) + 7.0 = 3.5\text{ m}$
$h_{\text{max}} = 3.5(1) + 7.0 = 10.5\text{ m}$
$h_{\text{min}} = 3.5(-1) + 7.0 = 3.5\text{ m}$
[N1] $10.5\text{ m}$
[N1] $3.5\text{ m}$
(b) Safe depth condition: $h(t) \ge 8.75$
$3.5\sin\left(\frac{\pi}{6}t\right) + 7.0 \ge 8.75 \implies 3.5\sin\left(\frac{\pi}{6}t\right) \ge 1.75 \implies \sin\left(\frac{\pi}{6}t\right) \ge 0.5$
Reference angle $\theta = \sin^{-1}(0.5) = \frac{\pi}{6}$.
First cycle ($0 \le t \le 12$, period $= \frac{2\pi}{\pi/6} = 12\text{ hours}$):
$\frac{\pi}{6} \le \frac{\pi}{6}t \le \pi - \frac{\pi}{6} = \frac{5\pi}{6} \implies 1 \le t \le 5\text{ hours}$ (01:00 to 05:00)
Second cycle ($12 \le t \le 24$): Add period $12$:
$1 + 12 \le t \le 5 + 12 \implies 13 \le t \le 17\text{ hours}$ (13:00 to 17:00)
Safe berthing windows: 01:00 to 05:00 and 13:00 to 17:00. [K1, N1]