Sec 3 A Math: Trigonometric Equations & Graphs, practice questions & worked solutions

Q: Do these A Math questions include worked solutions?

Yes. Every question includes the worked solution for all the parts shown.

Q: Which year are the questions from?

All the selected questions are from 2025 Singapore school Preliminary Examination papers.

Q: How should I use this topic page?

Attempt a question fully on paper, then open the worked solution and compare your method and answer.

Trigonometric Equations & Graphs practice questions selected from 2025 Singapore school A Math prelim papers, each with a full worked solution.

About this topic & key methods

These questions practise Sec 3 topics; the source papers are Sec 4 prelims. Questions requiring later Sec 4 methods are excluded.

Attempt each question on paper before opening its worked solution. Keep the source question number when checking against the original paper.

Key methods

Read or calculate amplitude, period and vertical displacement.
Sketch a trigonometric graph over the stated interval.
Find every solution of a trigonometric equation in the given range.
Use a sinusoidal model to interpret a practical situation.

For guided practice, see our Sec 3 A Maths tuition programme.

Questions & worked solutions

Unless the question specifies otherwise, give numerical answers to 3 significant figures and angles in degrees to 1 decimal place. Angles in radians are stated explicitly.

Q1Ahmad Ibrahim Secondary School2025Prelim P1 Q9●●○ Standard6 marks

Trigonometric Graphs

It is given that $\mathrm{f}(x) = 4 + \cos\left(\dfrac{x}{2}\right)$ and $\mathrm{g}(x) = -2\sin x$ .

State the period and amplitude of $\mathrm{f}(x)$ .
State the period and amplitude of $\mathrm{g}(x)$ .
Sketch, on the same axes, the graphs of $y = \mathrm{f}(x)$ and $y = \mathrm{g}(x)$ for $0^\circ \leq x \leq 360^\circ$ .

Show worked solution▾

(a)

$\begin{aligned} \text{Period} &= 720^\circ \text{ or } 4\pi\\ \text{Amplitude} &= 1 \end{aligned}$

(b)

$\begin{aligned} \text{Period} &= 360^\circ \text{ or } 2\pi\\ \text{Amplitude} &= 2 \end{aligned}$

(c)

Ahmad Ibrahim Secondary School 2025 P1 Q9: worked solution diagram

	$\mathrm{f}(x)$	$\mathrm{g}(x)$
Shape	cosine	negative sine
Amplitude	1	2
No. of cycle	Half	1
Shift	4	-

Q2Catholic High School2025Prelim P1 Q11●●○ Standard6 marks

Trigonometric Equations

Solve $2\cot^2 2x = 1 - \dfrac{5}{\sin 2x}$ for $-180^{\circ} \le x \le 180^{\circ}$ .

Show worked solution▾

$\begin{aligned} 2\cot^2 2x &= 1-\frac{5}{\sin 2x}\\ 2\cot^2 2x &= 1-5\,\mathrm{cosec}\,2x\\ 2(\mathrm{cosec}^2\,2x-1) &= 1-5\,\mathrm{cosec}\,2x\\ 2\,\mathrm{cosec}^2\,2x+5\,\mathrm{cosec}\,2x-3 &= 0 \end{aligned}$ Let $y=\mathrm{cosec}\,2x$ . $\begin{aligned} 2y^2+5y-3 &= 0\\ (2y-1)(y+3) &= 0\\ y=-3 \;&\text{or}\; y=\tfrac{1}{2}\\ \mathrm{cosec}\,2x=-3 \;&\text{or}\; \mathrm{cosec}\,2x=\tfrac{1}{2}\\ \sin 2x=-\tfrac{1}{3} \;&\text{or}\; \sin 2x=2\ \text{(rej.)} \end{aligned}$ $\begin{aligned} \alpha &= \sin^{-1}\tfrac{1}{3}\\ &= 19.471^\circ \end{aligned}$ $\begin{aligned} 2x &= 199.471^\circ,\ 340.529^\circ\\ x &= 99.7^\circ,\ 170.3^\circ (1\text{ d.p})\\[4pt] 2x &= -19.471^\circ,\ -160.529^\circ\\ x &= -9.7^\circ,\ -80.3^\circ (1\text{ d.p}) \end{aligned}$

Q3Catholic High School2025Prelim P1 Q2●●○ Standard6 marks

Trigonometric Functions

A function is given by $f(x) = 4\cos ax + b$ , where $a$ and $b$ are constants. The maximum value of $f(x)$ is 6 and the period of $f(x)$ is $4\pi$ .

State the amplitude of $f(x)$ .
Write down the value of $a$ and of $b$ .
Sketch the graph of $y = f(x)$ for $0 \le x \le 3\pi$ , indicating clearly the coordinates of the maximum and minimum points.

Show worked solution▾

(a)

Amplitude $= 4$ .

(b)

$\begin{aligned} \text{Period} &= 4\pi\\ \frac{2\pi}{a} &= 4\pi\\ a &= \frac{1}{2} \end{aligned}$ $\begin{aligned} 4+b &= 6\\ b &= 2 \end{aligned}$

(c)

$f(x)=4\cos\tfrac{1}{2}x+2$ for $0\le x\le 3\pi$ .

Catholic High School 2025 P1 Q2: worked solution diagram

Q4Crescent Girls’ School2025Prelim P1 Q7●●○ Standard6 marks

Trigonometric Graphs

Crescent Girls’ School 2025 P1 Q7: question diagram

The diagram shows the curve $y = a\cos bx + c$ for $0 \leq x \leq \dfrac{3\pi}{2}$ radians. A minimum point $\left(\dfrac{\pi}{2}, -2\right)$ and a maximum point $\left(\dfrac{5\pi}{4}, 4\right)$ are indicated on the diagram.

Explain why $b = 4$ .
Explain why $c = 1$ .
Hence find the equation of the curve.

Show worked solution▾

(a)

$\begin{aligned} \text{Period} &= \frac{\pi}{2}\\ \text{Period, } \frac{2\pi}{b} &= \frac{\pi}{2}\\ b &= 2\pi\div\frac{\pi}{2}=4 \end{aligned}$

(b)

$y=c$ is the axis of curve $\begin{aligned} c &= \frac{\text{maximum}+\text{minimum}}{2}\\ &= \frac{4+(-2)}{2}\\ c &= 1 \end{aligned}$

(c)

By observation, $a<0$ . $\begin{gathered} a=-(4-1) \text{OR} a=-[1-(-2)]\\ \text{OR} a=-\!\left(\frac{\text{maximum}-\text{minimum}}{2}\right)=-\!\left(\frac{4-(-2)}{2}\right)=-3 \end{gathered}$ Equation of curve is $y=-3\cos 4x+1$ .

Q5Dunman Secondary School2025Prelim P1 Q11●●○ Standard6 marks

Trigonometric Graphs

Sketch the graphs of $y = -2\cos 3x + 1$ and $y = -\dfrac{2}{\pi}x + 1$ on the same axes, for $0 \le x \le \pi$ .
Explain how the number of solutions for $2\cos 3x = \dfrac{2}{\pi}x$ can be found using the graphs.

Show worked solution▾

(a)

Dunman Secondary School 2025 P1 Q11: worked solution diagram

(b)

The $2\cos 3x=\dfrac{2}{\pi}x$ is equivalent to $-2\cos 3x+1=-\dfrac{2}{\pi}x+1$ and so its solutions can be found from the interception points of the two graphs.

Q6Jurongville Secondary School2025Prelim P2 Q8●●○ Standard6 marks

Trigonometric Graphs

Jurongville Secondary School 2025 P2 Q8: question diagram

The diagram shows the curve $y = a\sin bx + c$ for $0 \leq x \leq \dfrac{\pi}{2}$ radians. The curve has a minimum point at $\left(\dfrac{\pi}{8}, -8\right)$ and a maximum point at $\left(\dfrac{3\pi}{8}, 2\right)$ .

Explain why $c = -3$ .
Explain why $b = 4$ .
Hence find the equation of the curve.

Show worked solution▾

(a)

$y = c$ is axis of curve $\begin{aligned} c &= \frac{\text{maximum} + \text{minimum}}{2}\\ &= \frac{2 + (-8)}{2}\\ &= -3 \end{aligned}$

(b)

$\begin{aligned} \frac{2\pi}{b} &= \text{period}\\ \frac{2\pi}{b} &= \frac{\pi}{2}\\ b &= 4 \end{aligned}$

(c)

$y = a\sin bx + c$ Since this is a negative sine graph and the amplitude is 5, $a = -5$ .

The equation of the curve is $y = -5\sin 4x - 3$ .

Q7Anglican High School2025Prelim P2 Q7●●○ Standard7 marks

Trigonometric Graphs

A waterwheel rotates at 1 revolution per 12 seconds. Initially, Point $A$ , marked on the highest point of the rim of the wheel is 6 m above the water surface. It reaches 2 m below the water surface 6 seconds later.

The motion of point $A$ can be modelled using the trigonometric equation $h = b\cos(kt) + 2$ , where the height of point $A$ above the water surface, $h$ , is a function of time $t$ , in seconds.

Show that $b = 4$ and $k = \dfrac{\pi}{6}$ .
On the axes below, sketch the graph of $h$ , for $0 \leq t \leq 18$ .

A similar waterwheel at a different location has a point, $B$ , marked on its rim. The motion of point $B$ is modelled by the equation $h_B = 4\cos\!\left(\dfrac{\pi}{6}t\right) + 4$ .

By drawing a suitable straight line on the same axes above, determine the number of time(s) when point $B$ touches the water surface from $0 \leq t \leq 18$ .

Show worked solution▾

(a)

$b = \frac{6 + 2}{2} = 4$ $k = \frac{2\pi}{12} = \frac{\pi}{6}$

(b)

Anglican High School 2025 P2 Q7: worked solution diagram

(c)

2 times

Q8Maris Stella High School2025Prelim P1 Q10●●● Challenging8 marks

Trigonometric Graphs

State the amplitude and period of $2\sin 3x + 1$ .
Sketch the graph of $y = 2\sin 3x + 1$ for $0 \leq x \leq \pi$ .
By drawing a suitable straight line on your sketch, determine the number of solutions of the equation $\pi\sin 3x - x = 0$ .

Show worked solution▾

(a)

$\begin{aligned} \text{Amplitude} &= 2\\ \text{Period} &= \frac{2\pi}{3}\ (\text{or } 120^\circ) \end{aligned}$

(b)

$y=2\sin 3x+1$ for $0\le x\le\pi$ :

Maris Stella High School 2025 P1 Q10: worked solution diagram

(c)

$\begin{aligned} \pi\sin 3x-x &= 0\\ 2\sin 3x-\frac{2x}{\pi} &= 0\\ 2\sin 3x+1 &= \frac{2x}{\pi}+1 \end{aligned}$ Draw $y=\dfrac{2x}{\pi}+1$ .

Number of solutions $=4$ .

Q9Singapore Chinese Girls’ School2025Prelim P1 Q7●●● Challenging8 marks

Trigonometric Graphs

State the amplitude and period of the graph of $y = 3\sin\dfrac{x}{2} + 1$ .
Sketch the graph of $y = 3\sin\dfrac{x}{2} + 1$ for $0 \leq x \leq 4\pi$ .
By adding another suitable curve on your sketch, determine the number of solutions of the equation $\sin\dfrac{x}{2} = -\dfrac{1}{3} - \dfrac{2}{3}\cos\dfrac{x}{2}$ .

Show worked solution▾

(a)

Amplitude $= 3$

Period $= 4\pi$ or $720^\circ$

(b)

Singapore Chinese Girls’ School 2025 P1 Q7: worked solution diagram

(c)

Singapore Chinese Girls’ School 2025 P1 Q7: worked solution diagram

$\begin{aligned} 3\sin\frac{x}{2} + 1 &= -2\cos\frac{x}{2} \end{aligned}$ Sketch $y = -2\cos\dfrac{x}{2}$

Number of solutions $= 2$

Q10Methodist Girls’ School2025Prelim P1 Q10●●● Challenging9 marks

Trigonometric Graphs

The maximum height that a roller coaster can reach is 90 meters above sea level, and the lowest possible height is 10 meters above sea level. The roller coaster takes 10 seconds to move from its highest position to its lowest position. The height, $h$ m, of the roller coaster above sea level can be modelled by the function $h = a\cos\left(\dfrac{\pi}{b}t\right) + c$ , where $t$ is the time in seconds.

Show that $h = 40\cos\left(\dfrac{\pi}{10}t\right) + 50$ .
Sketch the graph of $h = 40\cos\left(\dfrac{\pi}{10}t\right) + 50$ for $0 \leq t \leq 30$ .
Find the time when the roller coaster first reaches a height of 62 m.
1. On your graph in part (b), draw the graph of $h = t + 15$ .
2. State for the interval $0 \leq t \leq 30$ , the number of solutions of $t = 40\cos\left(\dfrac{\pi}{10}t\right) + 35$ .

Show worked solution▾

(a)

$\begin{aligned} \text{Period: } \frac{2\pi}{\frac{\pi}{b}} &= 20\\ b &= 10 \end{aligned}$ Max $h = 90$

Min $h = 10$ $\begin{aligned} a + c &= 90\\ -a + c &= 10\\ 2c &= 100\\ c &= 50\\ a = 90 - 50 = 40 &\text{or} a = \frac{90 - 10}{2} = 40 \end{aligned}$ $\therefore h = 40\cos\left(\frac{\pi}{10}t\right) + 50$

(b)

(d)

Methodist Girls’ School 2025 P1 Q10: worked solution diagram

(c)

$\begin{aligned} 62 &= 40\cos\left(\frac{\pi}{10}t\right) + 50\\ 40\cos\left(\frac{\pi}{10}t\right) &= 12\\ \cos\left(\frac{\pi}{10}t\right) &= \frac{12}{40}\\ \alpha &= 1.26610\\ \frac{\pi}{10}t &= 1.26610\\ t &= 4.03 \text{ s (3sf)} \end{aligned}$

(d)

(ii) $\begin{aligned} t &= 40\cos\left(\frac{\pi}{10}t\right) + 35\\ t + 15 &= 40\cos\left(\frac{\pi}{10}t\right) + 50 \end{aligned}$ $3$ solutions

Need help with trigonometric equations & graphs? See our Sec 3 A Maths tuition programme.

Frequently asked questions

Do these A Math questions include worked solutions?▾

Yes. Every question includes the worked solution for all the parts shown.

Which year are the questions from?▾

All the selected questions are from 2025 Singapore school Preliminary Examination papers.

How should I use this topic page?▾

Attempt a question fully on paper, then open the worked solution and compare your method and answer.

Sec 3 A Math: Trigonometric Equations & Graphs, practice questions & worked solutions

Key methods

Questions & worked solutions

Frequently asked questions

Related A Math topics