SPM KSSM Diagnostic Drill
Form 5 Chapter 6: Trigonometric Functions (Fungsi Trigonometri)
Total Marks: 40 Marks • Time Allowed: 50 Minutes
Examiner Score
____ / 40
Student Name
Class / School
Date
A Bahagian A: SPM Paper 1 Format [16 Marks]
Answer all questions
Soalan 1
[4 Marks]
Given that $\sin \theta = -\frac{3}{5}$ and $180^\circ \le \theta \le 270^\circ$, without using a calculator, find the exact value of:
(a) $\cos \theta$. [1 mark]
(b) $\tan 2\theta$. [3 marks]
$\theta$ lies in Quadrant III: $\sin \theta < 0, \cos \theta < 0, \tan \theta > 0$.
Reference triangle: opposite $= 3$, hypotenuse $= 5 \implies \text{adjacent} = \sqrt{5^2 - 3^2} = 4$.
(a) $\cos \theta = -\frac{4}{5}$
[N1] $-\frac{4}{5}$
(b) $\tan \theta = \frac{-3}{-4} = \frac{3}{4}$ [P1]
$\tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta} = \frac{2\left(\frac{3}{4}\right)}{1 - \left(\frac{3}{4}\right)^2} = \frac{\frac{3}{2}}{1 - \frac{9}{16}} = \frac{\frac{3}{2}}{\frac{7}{16}}$ [K1]
$\tan 2\theta = \frac{3}{2} \times \frac{16}{7} = \frac{24}{7}$ [N1] $\frac{24}{7}$
Soalan 2
[4 Marks]
Prove that: $$\frac{1 + \sin 2x - \cos 2x}{1 + \sin 2x + \cos 2x} = \tan x$$
$\text{LHS} = \frac{1 + \sin 2x - \cos 2x}{1 + \sin 2x + \cos 2x}$
Substitute double angle identities:
Numerator: $\sin 2x = 2\sin x \cos x$, and $\cos 2x = 1 - 2\sin^2 x$ so $1 - \cos 2x = 2\sin^2 x$. [K1]
Denominator: $\cos 2x = 2\cos^2 x - 1$ so $1 + \cos 2x = 2\cos^2 x$. [K1]
$\text{LHS} = \frac{2\sin^2 x + 2\sin x \cos x}{2\cos^2 x + 2\sin x \cos x} = \frac{2\sin x(\sin x + \cos x)}{2\cos x(\cos x + \sin x)}$ [K1]
$= \frac{\sin x}{\cos x} = \tan x = \text{RHS}$ (Proven) [N1]
Soalan 3
[4 Marks]
Solve the equation $2\cos^2 x - 3\sin x = 0$ for $0^\circ \le x \le 360^\circ$.
Replace $\cos^2 x = 1 - \sin^2 x$:
$2(1 - \sin^2 x) - 3\sin x = 0 \implies 2 - 2\sin^2 x - 3\sin x = 0$ [P1]
$2\sin^2 x + 3\sin x - 2 = 0$
$(2\sin x - 1)(\sin x + 2) = 0$ [K1]
$\sin x = \frac{1}{2}$ or $\sin x = -2$ (no solution since $-1 \le \sin x \le 1$).
For $\sin x = \frac{1}{2}$, reference angle $\alpha = 30^\circ$:
Quadrant 1: $x = 30^\circ$
Quadrant 2: $x = 180^\circ - 30^\circ = 150^\circ$
$x = 30^\circ, 150^\circ$ [N1, N1]
Soalan 4
[4 Marks]
A trigonometric function is given by $f(x) = 3\sin(2x) - 1$ for $0 \le x \le 2\pi$.
(a) State the amplitude and the period of $f(x)$. [2 marks]
(b) Find the range of $f(x)$. [2 marks]
(a) Amplitude $= |3| = 3$. Period $= \frac{2\pi}{2} = \pi$ radians (or $180^\circ$).
[N1] Amp 3[N1] Period $\pi$
(b) Since $-1 \le \sin(2x) \le 1$:
Minimum value $= 3(-1) - 1 = -4$
Maximum value $= 3(1) - 1 = 2$
$\therefore$ Range: $-4 \le f(x) \le 2$
[K1] boundsMinimum value $= 3(-1) - 1 = -4$
Maximum value $= 3(1) - 1 = 2$
$\therefore$ Range: $-4 \le f(x) \le 2$
[N1] $-4 \le f(x) \le 2$
B Bahagian B: SPM Paper 2 Format [24 Marks]
Detailed solutions required
Soalan 5
[10 Marks]
(a) Prove that $\cot x + \tan x = 2\csc 2x$. [3 marks]
(b) Hence, solve the equation $\cot x + \tan x = 4$ for $0 \le x \le 2\pi$, expressing answers in terms of $\pi$ or correct to 3 decimal places. [3 marks]
(c) Solve the equation $3\sec^2 x = 4 + 2\tan x$ for $0^\circ \le x \le 360^\circ$. [4 marks]
(a) Proof:
$\text{LHS} = \frac{\cos x}{\sin x} + \frac{\sin x}{\cos x} = \frac{\cos^2 x + \sin^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x}$ [K1]
Multiply numerator and denominator by 2: $\frac{2}{2\sin x \cos x} = \frac{2}{\sin 2x} = 2\csc 2x = \text{RHS}$ [K1, N1]
$\text{LHS} = \frac{\cos x}{\sin x} + \frac{\sin x}{\cos x} = \frac{\cos^2 x + \sin^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x}$ [K1]
Multiply numerator and denominator by 2: $\frac{2}{2\sin x \cos x} = \frac{2}{\sin 2x} = 2\csc 2x = \text{RHS}$ [K1, N1]
(b) Solving $\cot x + \tan x = 4$:
$2\csc 2x = 4 \implies \csc 2x = 2 \implies \sin 2x = \frac{1}{2}$ [P1]
Given $0 \le x \le 2\pi \implies 0 \le 2x \le 4\pi$.
Reference angle $\alpha = \frac{\pi}{6}$.
First cycle: $2x = \frac{\pi}{6}, \frac{5\pi}{6}$
Second cycle (add $2\pi$): $2x = \frac{\pi}{6} + 2\pi = \frac{13\pi}{6}, \frac{5\pi}{6} + 2\pi = \frac{17\pi}{6}$ [K1]
Divide by 2: $x = \frac{\pi}{12}, \frac{5\pi}{12}, \frac{13\pi}{12}, \frac{17\pi}{12}$ [N1]
$2\csc 2x = 4 \implies \csc 2x = 2 \implies \sin 2x = \frac{1}{2}$ [P1]
Given $0 \le x \le 2\pi \implies 0 \le 2x \le 4\pi$.
Reference angle $\alpha = \frac{\pi}{6}$.
First cycle: $2x = \frac{\pi}{6}, \frac{5\pi}{6}$
Second cycle (add $2\pi$): $2x = \frac{\pi}{6} + 2\pi = \frac{13\pi}{6}, \frac{5\pi}{6} + 2\pi = \frac{17\pi}{6}$ [K1]
Divide by 2: $x = \frac{\pi}{12}, \frac{5\pi}{12}, \frac{13\pi}{12}, \frac{17\pi}{12}$ [N1]
(c) $3\sec^2 x = 4 + 2\tan x$:
Identity $\sec^2 x = 1 + \tan^2 x$:
$3(1 + \tan^2 x) = 4 + 2\tan x \implies 3\tan^2 x - 2\tan x - 1 = 0$ [P1]
$(3\tan x + 1)(\tan x - 1) = 0$ [K1]
Case 1: $\tan x = 1 \implies \alpha = 45^\circ \implies x = 45^\circ, 225^\circ$ [N1]
Case 2: $\tan x = -\frac{1}{3} \implies \alpha = \tan^{-1}\left(\frac{1}{3}\right) = 18.43^\circ$
Quadrant 2: $x = 180^\circ - 18.43^\circ = 161.57^\circ$
Quadrant 4: $x = 360^\circ - 18.43^\circ = 341.57^\circ$ [N1]
Identity $\sec^2 x = 1 + \tan^2 x$:
$3(1 + \tan^2 x) = 4 + 2\tan x \implies 3\tan^2 x - 2\tan x - 1 = 0$ [P1]
$(3\tan x + 1)(\tan x - 1) = 0$ [K1]
Case 1: $\tan x = 1 \implies \alpha = 45^\circ \implies x = 45^\circ, 225^\circ$ [N1]
Case 2: $\tan x = -\frac{1}{3} \implies \alpha = \tan^{-1}\left(\frac{1}{3}\right) = 18.43^\circ$
Quadrant 2: $x = 180^\circ - 18.43^\circ = 161.57^\circ$
Quadrant 4: $x = 360^\circ - 18.43^\circ = 341.57^\circ$ [N1]
Soalan 6 • SPM Paper 2 & KBAT
[14 Marks]
Full SPM Graph & Root Superposition Protocol
Trigonometric Graphing & Real-World Power Surge Modeling
(a) Sketch the graph of $y = 3\sin\left(\frac{3}{2}x\right) - 1$ 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:
$$6\pi\sin\left(\frac{3}{2}x\right) - 2\pi = 3x$$
State the equation of the straight line and determine the number of solutions. [4 marks]
(c) KBAT Real-World Electrical Engineering Application:
The voltage fluctuation $V(t)$, in volts, in a green energy inverter circuit is given by:
$$V(t) = 120\sqrt{2}\cos(100\pi t) + 15$$
where $t$ is measured in seconds ($t \ge 0$).
(i) Find the peak (maximum) voltage and the root-mean-square amplitude. [2 marks]
(ii) Find the frequency of the oscillation in Hertz ($\text{Hz} = \frac{1}{\text{Period}}$). [2 marks]
(iii) Determine the earliest instant $t > 0$ when the voltage drops to zero. [2 marks]
(i) Find the peak (maximum) voltage and the root-mean-square amplitude. [2 marks]
(ii) Find the frequency of the oscillation in Hertz ($\text{Hz} = \frac{1}{\text{Period}}$). [2 marks]
(iii) Determine the earliest instant $t > 0$ when the voltage drops to zero. [2 marks]
(a) Graph of $y = 3\sin\left(\frac{3}{2}x\right) - 1$ for $0 \le x \le 2\pi$:
• Period $= \frac{2\pi}{3/2} = \frac{4\pi}{3}$. Cycles in $2\pi = \frac{2\pi}{4\pi/3} = 1.5$ cycles.
• Maximum value $= 3(1) - 1 = 2$. Minimum value $= 3(-1) - 1 = -4$.
• Checkpoint coordinates:
• Period $= \frac{2\pi}{3/2} = \frac{4\pi}{3}$. Cycles in $2\pi = \frac{2\pi}{4\pi/3} = 1.5$ cycles.
• Maximum value $= 3(1) - 1 = 2$. Minimum value $= 3(-1) - 1 = -4$.
• Checkpoint coordinates:
- $x = 0 \implies y = -1$
- $x = \frac{\pi}{3} \implies y = 3\sin\left(\frac{\pi}{2}\right) - 1 = 2$ (Peak 1)
- $x = \frac{2\pi}{3} \implies y = 3\sin(\pi) - 1 = -1$
- $x = \pi \implies y = 3\sin\left(\frac{3\pi}{2}\right) - 1 = -4$ (Trough 1)
- $x = \frac{4\pi}{3} \implies y = 3\sin(2\pi) - 1 = -1$ (Complete cycle 1)
- $x = \frac{5\pi}{3} \implies y = 3\sin\left(\frac{5\pi}{2}\right) - 1 = 2$ (Peak 2)
- $x = 2\pi \implies y = 3\sin(3\pi) - 1 = -1$
(b) Straight line equation:
$6\pi\sin\left(\frac{3}{2}x\right) - 2\pi = 3x$
Factor out $2\pi$: $2\pi\left[3\sin\left(\frac{3}{2}x\right) - 1\right] = 3x$
$\implies 3\sin\left(\frac{3}{2}x\right) - 1 = \frac{3}{2\pi}x$
Since LHS is $y$, the line equation is: $y = \frac{3}{2\pi}x$ [K1]
Coordinates for line:
Intersections: The line passes from $(0,0)$ (above $-1$) across the first wave peak/fall, intersecting the curve at 3 locations in $[0, 2\pi]$.
$\therefore$ Number of solutions $= 3$. [N1 line drawn, N1 3 solutions]
$6\pi\sin\left(\frac{3}{2}x\right) - 2\pi = 3x$
Factor out $2\pi$: $2\pi\left[3\sin\left(\frac{3}{2}x\right) - 1\right] = 3x$
$\implies 3\sin\left(\frac{3}{2}x\right) - 1 = \frac{3}{2\pi}x$
Since LHS is $y$, the line equation is: $y = \frac{3}{2\pi}x$ [K1]
Coordinates for line:
- When $x = 0 \implies y = 0 \implies (0, 0)$
- When $x = 2\pi \implies y = \frac{3}{2\pi}(2\pi) = 3 \implies (2\pi, 3)$
Intersections: The line passes from $(0,0)$ (above $-1$) across the first wave peak/fall, intersecting the curve at 3 locations in $[0, 2\pi]$.
$\therefore$ Number of solutions $= 3$. [N1 line drawn, N1 3 solutions]
(c) Electrical Engineering KBAT:
(i) Peak Voltage $V_{\text{max}} = 120\sqrt{2}(1) + 15 \approx 169.71 + 15 = 184.71\text{ V}$. Amplitude $= 120\sqrt{2} \approx 169.71\text{ V}$. [N1, N1]
(ii) Period $T = \frac{2\pi}{100\pi} = \frac{1}{50} = 0.02\text{ s}$. Frequency $f = \frac{1}{T} = 50\text{ Hz}$. [K1, N1]
(iii) $V(t) = 0 \implies 120\sqrt{2}\cos(100\pi t) + 15 = 0 \implies \cos(100\pi t) = -\frac{15}{120\sqrt{2}} \approx -0.08839$.
Reference angle $\alpha = \cos^{-1}(0.08839) \approx 1.6591\text{ rad}$.
Earliest instant $t > 0$ occurs in Quadrant 2: $100\pi t = \pi - 1.6591 \dots$ wait, $\cos^{-1}(-0.08839) = 1.6591\text{ rad}$ directly (since $\cos(\pi/2)=0$).
$100\pi t = 1.6591 \implies t = \frac{1.6591}{100\pi} \approx 0.00528\text{ s}$ (or $5.28\text{ ms}$). [K1, N1]
(i) Peak Voltage $V_{\text{max}} = 120\sqrt{2}(1) + 15 \approx 169.71 + 15 = 184.71\text{ V}$. Amplitude $= 120\sqrt{2} \approx 169.71\text{ V}$. [N1, N1]
(ii) Period $T = \frac{2\pi}{100\pi} = \frac{1}{50} = 0.02\text{ s}$. Frequency $f = \frac{1}{T} = 50\text{ Hz}$. [K1, N1]
(iii) $V(t) = 0 \implies 120\sqrt{2}\cos(100\pi t) + 15 = 0 \implies \cos(100\pi t) = -\frac{15}{120\sqrt{2}} \approx -0.08839$.
Reference angle $\alpha = \cos^{-1}(0.08839) \approx 1.6591\text{ rad}$.
Earliest instant $t > 0$ occurs in Quadrant 2: $100\pi t = \pi - 1.6591 \dots$ wait, $\cos^{-1}(-0.08839) = 1.6591\text{ rad}$ directly (since $\cos(\pi/2)=0$).
$100\pi t = 1.6591 \implies t = \frac{1.6591}{100\pi} \approx 0.00528\text{ s}$ (or $5.28\text{ ms}$). [K1, N1]