Trigonometric Functions Worked Examples (Easy)

Q: How do I remember which reciprocal ratio goes with which?

Look at the third letter: cosec pairs with sine (cosec⁡θ=1sin⁡θ\operatorname{cosec}\theta=\tfrac{1}{\sin\theta}cosecθ=sinθ1), sec pairs with cosine (sec⁡θ=1cos⁡θ\sec\theta=\tfrac{1}{\cos\theta}secθ=cosθ1), and cot pairs with tan (cot⁡θ=1tan⁡θ\cot\theta=\tfrac{1}{\tan\theta}cotθ=tanθ1). The odd one out is that sec goes with cos, not sin.

Q: How do I read the period from an equation like y=3\sin 2x?

The period of y=asin⁡bx+cy=a\sin bx+cy=asinbx+c is 360∘b\dfrac{360^{\circ}}{b}b360∘ in degrees. Here b=2b=2b=2, so the period is 180∘180^{\circ}180∘ and the curve fits two complete waves into a full turn. The number aaa sets the amplitude, and ccc shifts the whole curve up or down.

What these examples cover

These easy Trigonometric Functions examples build the everyday moves that open almost every question in the chapter: turning a known ratio into the reciprocal ratios $\operatorname{cosec}\theta=\dfrac{1}{\sin\theta}$ , $\sec\theta=\dfrac{1}{\cos\theta}$ and $\cot\theta=\dfrac{1}{\tan\theta}$ ; reading an exact value such as $\sin 150^{\circ}$ from its quadrant and basic angle; reading the amplitude and period straight off a graph of the shape $y=a\sin bx+c$ ; and solving a first equation like $\sin x=\dfrac{1}{2}$ over a full turn. Each one uses small, clean numbers so you can follow every line while your calculator does only the arithmetic.

The single most important habit is the sign rule for quadrants, all ratios positive in the first quadrant, then sine only, tangent only and cosine only positive in the second, third and fourth. Use the set the honest way: cover the solution, attempt the question in full on paper, and only then check line by line against our working.

Where your answer differs, find the exact step where the two solutions part company; that single line is usually where the real learning is.

Worked examples

Work through all four. Attempt each fully before you read the matching solution, and notice how the same discipline, write the rule, substitute one value at a time, then simplify, runs through reciprocal ratios, exact values, graphs and equations alike.

Q1[4 marks]

Given that $\sin\theta=\dfrac{3}{5}$ and $\theta$ is an acute angle, find the exact values of $\cos\theta$ , $\tan\theta$ , $\operatorname{cosec}\theta$ , $\sec\theta$ and $\cot\theta$ .

Show worked solution

Because $\theta$ is acute it lies in the first quadrant, where every trigonometric ratio is positive. Start from the Pythagorean identity to find $\cos\theta$ :

Given in the exam

\sin^{2}\theta+\cos^{2}\theta=1

Make $\cos^{2}\theta$ the subject and substitute $\sin\theta=\dfrac{3}{5}$ :

\cos^{2}\theta=1-\left(\dfrac{3}{5}\right)^{2}=1-\dfrac{9}{25}=\dfrac{16}{25}

Take the positive square root, since cosine is positive in the first quadrant:

\cos\theta=\sqrt{\dfrac{16}{25}}=\dfrac{4}{5}

Tangent is sine divided by cosine:

\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{3/5}{4/5}=\dfrac{3}{4}

The last three ratios are simply the reciprocals of the first three:

\operatorname{cosec}\theta=\dfrac{1}{\sin\theta}=\dfrac{5}{3},\quad \sec\theta=\dfrac{1}{\cos\theta}=\dfrac{5}{4},\quad \cot\theta=\dfrac{1}{\tan\theta}=\dfrac{4}{3}

Answer

$\cos\theta=\dfrac{4}{5}$ , $\tan\theta=\dfrac{3}{4}$ , $\operatorname{cosec}\theta=\dfrac{5}{3}$ , $\sec\theta=\dfrac{5}{4}$ , $\cot\theta=\dfrac{4}{3}$ . Quick check with the identity: $\sin^{2}\theta+\cos^{2}\theta=\dfrac{9}{25}+\dfrac{16}{25}=\dfrac{25}{25}=1$ , so the pair $\left(\tfrac{3}{5},\tfrac{4}{5}\right)$ is consistent.

Q2[3 marks]

Find the exact value of (a) $\sin 150^{\circ}$ , (b) $\cos 300^{\circ}$ , (c) $\tan 225^{\circ}$ .

Show worked solution

Each angle sits outside the first quadrant, so use two steps: first find the basic angle (the acute angle to the horizontal axis), then attach the sign for that quadrant. The sign rule is that all ratios are positive in the first quadrant, then only sine stays positive in the second, only tangent in the third, and only cosine in the fourth.

(a) $150^{\circ}$ is in the second quadrant; its basic angle is $180^{\circ}-150^{\circ}=30^{\circ}$ . Sine is positive in the second quadrant:

\sin 150^{\circ}=+\sin 30^{\circ}=\dfrac{1}{2}

(b) $300^{\circ}$ is in the fourth quadrant; its basic angle is $360^{\circ}-300^{\circ}=60^{\circ}$ . Cosine is positive in the fourth quadrant:

\cos 300^{\circ}=+\cos 60^{\circ}=\dfrac{1}{2}

(c) $225^{\circ}$ is in the third quadrant; its basic angle is $225^{\circ}-180^{\circ}=45^{\circ}$ . Tangent is positive in the third quadrant:

\tan 225^{\circ}=+\tan 45^{\circ}=1

Answer

(a) $\dfrac{1}{2}$ , (b) $\dfrac{1}{2}$ , (c) $1$ . Each answer matches the calculator, but writing the quadrant and basic angle is what earns the method marks, a bare decimal does not show your reasoning.

Q3[3 marks]

The graph of $y=3\sin 2x$ is drawn for $0^{\circ}\le x\le 360^{\circ}$ . State (a) its amplitude, (b) its period, and (c) the number of complete cycles in this range.

Show worked solution

Compare the equation with the standard shape $y=a\sin bx+c$ . Reading off the constants, $a=3$ , $b=2$ and $c=0$ .

(a) The amplitude is the size of $a$ , the greatest distance the curve reaches above or below its mid-line:

\text{amplitude}=|a|=3

(b) For a sine or cosine curve the period (the width of one full wave) is $\dfrac{360^{\circ}}{b}$ :

\text{period}=\dfrac{360^{\circ}}{b}=\dfrac{360^{\circ}}{2}=180^{\circ}

\dfrac{360^{\circ}}{180^{\circ}}=2

Answer

Amplitude $3$ , period $180^{\circ}$ , and $2$ complete cycles. As a sense check, the curve runs from a maximum of $+3$ down to a minimum of $-3$ and back, twice, between $0^{\circ}$ and $360^{\circ}$ .

Q4[3 marks]

Solve $\sin x=\dfrac{1}{2}$ for $0^{\circ}\le x\le 360^{\circ}$ .

Show worked solution

First find the basic angle by taking the inverse sine of the positive value:

\text{basic angle}=\sin^{-1}\dfrac{1}{2}=30^{\circ}

The value of $\sin x$ is positive, so $x$ lies in the two quadrants where sine is positive, the first and the second. Read the angle in each quadrant using the basic angle $30^{\circ}$ :

x=30^{\circ}\quad\text{or}\quad x=180^{\circ}-30^{\circ}=150^{\circ}

Answer

$x=30^{\circ}$ or $x=150^{\circ}$ . Check both: $\sin 30^{\circ}=\dfrac{1}{2}$ and $\sin 150^{\circ}=\sin 30^{\circ}=\dfrac{1}{2}$ .

Over one full turn a sine equation with a value strictly between $0$ and $1$ has exactly two answers, so we have found them all.

Q5[3 marks]

Convert (a) $150^{\circ}$ to radians, giving your answer as a fraction of $\pi$ , and (b) $\dfrac{7\pi}{6}$ radians to degrees.

Show worked solution

Degrees and radians are linked by one conversion fact: a straight-line turn of $180^{\circ}$ equals $\pi$ radians. Multiply by $\dfrac{\pi}{180}$ to go from degrees to radians, and by $\dfrac{180^{\circ}}{\pi}$ to go back.

180^{\circ}=\pi\text{ rad}

(a) Multiply $150^{\circ}$ by $\dfrac{\pi}{180}$ :

150^{\circ}=150\times\dfrac{\pi}{180}=\dfrac{5\pi}{6}\text{ rad}

(b) Multiply $\dfrac{7\pi}{6}$ by $\dfrac{180^{\circ}}{\pi}$ :

\dfrac{7\pi}{6}\times\dfrac{180^{\circ}}{\pi}=7\times30^{\circ}=210^{\circ}

Answer

(a) $\dfrac{5\pi}{6}$ rad, (b) $210^{\circ}$ . As a check, $\dfrac{5\pi}{6}$ rad converts straight back to $150^{\circ}$ , so the two conversions undo each other correctly.

Q6[3 marks]

The point $P(-3,4)$ lies on the terminal arm of angle $\theta$ , measured from the positive $x$ -axis in the usual way. Find the exact values of $\sin\theta$ , $\cos\theta$ and $\tan\theta$ .

Show worked solution

For any point $P(x,y)$ on the terminal arm of $\theta$ , let $r$ be the distance from the origin to $P$ . Find $r$ first, using Pythagoras' theorem:

r=\sqrt{x^{2}+y^{2}}

Substitute $x=-3$ and $y=4$ :

r=\sqrt{(-3)^{2}+4^{2}}=\sqrt{9+16}=\sqrt{25}=5

The three ratios are then defined directly from $x$ , $y$ and $r$ , keeping the sign of $x$ and $y$ :

\sin\theta=\dfrac{y}{r}=\dfrac{4}{5},\quad\cos\theta=\dfrac{x}{r}=-\dfrac{3}{5},\quad\tan\theta=\dfrac{y}{x}=-\dfrac{4}{3}

Answer

$\sin\theta=\dfrac{4}{5}$ , $\cos\theta=-\dfrac{3}{5}$ , $\tan\theta=-\dfrac{4}{3}$ . $P$ has a negative $x$ -coordinate and a positive $y$ -coordinate, so it sits in the second quadrant, where only sine is positive, exactly what the three signs above show.

Q7[3 marks]

Given that $\tan\theta=\dfrac{5}{12}$ and $\theta$ is acute, find the exact value of $\sec\theta$ .

Show worked solution

Tangent and secant are linked directly by a Pythagorean identity, so there is no need to find $\sin\theta$ or $\cos\theta$ first:

Given in the exam

\sec^{2}\theta=1+\tan^{2}\theta

Substitute $\tan\theta=\dfrac{5}{12}$ :

\sec^{2}\theta=1+\left(\dfrac{5}{12}\right)^{2}=1+\dfrac{25}{144}=\dfrac{169}{144}

Take the positive square root, since $\theta$ is acute and secant is positive in the first quadrant:

\sec\theta=\sqrt{\dfrac{169}{144}}=\dfrac{13}{12}

Answer

$\sec\theta=\dfrac{13}{12}$ . The numbers $5$ , $12$ , $13$ form a Pythagorean triple, so a right triangle with opposite $5$ and adjacent $12$ has hypotenuse $13$ , giving $\cos\theta=\dfrac{12}{13}$ and confirming $\sec\theta=\dfrac{13}{12}$ .

Q8[3 marks]

The graph of $y=2\cos x+3$ is drawn for $0^{\circ}\le x\le360^{\circ}$ . State (a) the maximum value of $y$ and the corresponding value of $x$ , (b) the minimum value of $y$ and the corresponding value of $x$ .

Show worked solution

Compare with $y=a\cos x+c$ : here $a=2$ and $c=3$ . Since $\cos x$ only ever takes values from $-1$ to $1$ , $y$ is largest when $\cos x=1$ and smallest when $\cos x=-1$ .

(a) $\cos x=1$ at $x=0^{\circ}$ (and again at $x=360^{\circ}$ ):

y_{\max}=2(1)+3=5\quad\text{at }x=0^{\circ}

(b) $\cos x=-1$ at $x=180^{\circ}$ :

y_{\min}=2(-1)+3=1\quad\text{at }x=180^{\circ}

Answer

Maximum $y=5$ at $x=0^{\circ}$ (also at $x=360^{\circ}$ ); minimum $y=1$ at $x=180^{\circ}$ . Both values sit inside the expected range $c-a$ to $c+a$ , that is $1$ to $5$ , so the two answers check out.

Notice how short each solution stays once the plan is set. Reciprocal ratios come from the identity plus a single square root; exact values need only a quadrant and a basic angle; a graph hands you its amplitude and period directly from $a$ and $b$ ; and a basic equation reduces to one basic angle placed in the correct quadrants.

Name what the question is really asking, pick the matching move, and the arithmetic stays clean.

Key method points

These four examples rehearse the tools that open almost every Trigonometric Functions question in Add Math. Keep the following points in mind as you practise more.

The three reciprocal ratios are defined as $\operatorname{cosec}\theta=\dfrac{1}{\sin\theta}$ , $\sec\theta=\dfrac{1}{\cos\theta}$ and $\cot\theta=\dfrac{1}{\tan\theta}$ , pair sec with cos, not sin.
The identity $\sin^{2}\theta+\cos^{2}\theta=1$ turns any one of sine or cosine into the other; take the sign of the root from the quadrant.
For an exact value, find the basic angle to the horizontal axis, then attach the sign using ASTC: All, Sine, Tangent, Cosine positive in quadrants 1 to 4.
For $y=a\sin bx+c$ or $y=a\cos bx+c$ , the amplitude is $|a|$ and the period is $\dfrac{360^{\circ}}{b}$ ; the constant $c$ just shifts the mid-line.
To solve a basic equation, work out the basic angle first, then place it in every quadrant where the ratio has the required sign, within the stated range.
Keep every substitution line, with analytic marking a clear method line still earns method marks even if the final digit slips.

How a teacher helps

When a student drops a mark on questions like these, it is nearly always a small, fixable habit, pairing $\sec\theta$ with sine instead of cosine, forgetting the sign in a second-quadrant angle, or giving only one answer to an equation that has two. In a one-to-one lesson our teacher watches the exact line where the slip happens and corrects it on the spot, before it settles into a routine.

Because our teachers are experienced, you work with someone who explains why the sign changes by quadrant, not just which button to press. Lessons are taught in English, while SPM papers are set in both Malay and English, so the notation reads the same to you either way.

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Frequently asked questions

How do I remember which reciprocal ratio goes with which?

Look at the third letter: cosec pairs with sine ( $\operatorname{cosec}\theta=\tfrac{1}{\sin\theta}$ ), sec pairs with cosine ( $\sec\theta=\tfrac{1}{\cos\theta}$ ), and cot pairs with tan ( $\cot\theta=\tfrac{1}{\tan\theta}$ ). The odd one out is that sec goes with cos, not sin.

What is a basic angle and why do I need it?

The basic angle is the acute angle between the arm of the angle and the horizontal axis. Every angle in the four quadrants shares a ratio's size with its basic angle; you find the size from the basic angle, then set the sign from the quadrant.

This two-step split keeps exact values reliable.

How do I read the period from an equation like

y=3\sin 2x

The period of $y=a\sin bx+c$ is $\dfrac{360^{\circ}}{b}$ in degrees. Here $b=2$ , so the period is $180^{\circ}$ and the curve fits two complete waves into a full turn.

The number $a$ sets the amplitude, and $c$ shifts the whole curve up or down.

Why does

\sin x=\tfrac{1}{2}

have two answers between

0^{\circ}

and

360^{\circ}

Sine is positive in both the first and second quadrants, so a positive value strictly between $0$ and $1$ is reached twice in one turn. The basic angle gives the first answer; $180^{\circ}$ minus the basic angle gives the second.

Always scan the whole stated range.

Source:SRC-DSKP-EN

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

Trigonometric Functions, Worked Examples (easy)

What these examples cover

Worked examples

Key method points

How a teacher helps

Frequently asked questions

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