O Level (Sec 3 & 4)Additional Mathematics
Trigonometric Functions
Master O-Level A Math trigonometric functions with clear notes on sine, cosine, tangent, identities, compound and double-angle formulae, R-formula and trigonometric equations. Includes a free worksheet and worked solutions for Singapore students.
Timothy Gan
14 May 2026
Understanding Trigonometric Functions
Trigonometric functions connect angles to ratios. In O-Level Additional Mathematics, $\sin x$, $\cos x$ and $\tan x$ appear in identities, equations, graphs and later in calculus.
The main skill is not memorising formulas in isolation. You need to recognise which identity changes the form of an expression, which angle range affects the answer, and when a trigonometric equation has more than one solution.
This guide gives you a compact revision path: start with the three core functions, then build towards identities, compound-angle formulae, double-angle formulae, R-formula and equation solving.
Trigonometric Functions Study Guide
The Three Basic Trigonometric Functions
For a right-angled triangle, the three main ratios are:
$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$
$\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$
$\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$
A useful memory aid is SOH-CAH-TOA. In A Math, these ratios are extended beyond acute angles, so you must also understand signs in different quadrants and how angles repeat across cycles.
Core Identities
The most important Pythagorean identity is:
$\sin^2 x + \cos^2 x = 1$
Dividing by $\cos^2 x$ gives:
$\tan^2 x + 1 = \sec^2 x$
Dividing by $\sin^2 x$ gives:
$1 + \cot^2 x = \csc^2 x$
At O-Level, the first two forms are especially useful. They help you simplify expressions, prove identities and convert a question into a single trigonometric function.
Compound and Double-Angle Formulae
Compound-angle formulae allow you to expand angles such as $A+B$ and $A-B$:
$\sin(A+B)=\sin A\cos B+\cos A\sin B$
$\cos(A+B)=\cos A\cos B-\sin A\sin B$
$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$
Double-angle formulae come from setting $A=B$:
$\sin 2A=2\sin A\cos A$
$\cos 2A=\cos^2 A-\sin^2 A=2\cos^2 A-1=1-2\sin^2 A$
$\tan 2A=\frac{2\tan A}{1-\tan^2 A}$
R-Formula
The R-formula rewrites expressions such as $a\sin x+b\cos x$ into one trigonometric function:
$a\sin x+b\cos x=R\sin(x+\alpha)$
where $R=\sqrt{a^2+b^2}$.
After expanding $R\sin(x+\alpha)$, compare coefficients:
$R\cos\alpha=a$ and $R\sin\alpha=b$
This method is often used to find maximum and minimum values, solve equations, or simplify a graphing problem.
Solving Trigonometric Equations
A trigonometric equation can have several answers in a given interval. The safest process is:
1. Rearrange into a standard form such as $\sin x=k$, $\cos x=k$ or $\tan x=k$.
2. Find the basic angle.
3. Use quadrant signs and the required interval to list all valid answers.
4. Check whether the question uses degrees or radians.
For example, if $\sin x=\frac{1}{2}$ for $0^\circ \le x \le 360^\circ$, the answers are $30^\circ$ and $150^\circ$ because sine is positive in Quadrants I and II.
Practice Questions with Worked Solutions
Work through each question carefully, then compare your method with the step-by-step solution.
Question 1
Using the Pythagorean Identity
Question
Given that $\sin x=\frac{3}{5}$ and $x$ is acute, find $\cos x$ and $\tan x$.
Step-by-Step Solution
- 1Use $\sin^2 x+\cos^2 x=1$.
- 2Since $\sin x=\frac{3}{5}$, $\cos^2 x=1-\left(\frac{3}{5}\right)^2=\frac{16}{25}$.
- 3Because $x$ is acute, $\cos x$ is positive, so $\cos x=\frac{4}{5}$.
- 4Then $\tan x=\frac{\sin x}{\cos x}=\frac{3/5}{4/5}=\frac{3}{4}$.
Answer:$\cos x=\frac{4}{5}$ and $\tan x=\frac{3}{4}$
Question 2
Simplifying a Trigonometric Expression
Question
Simplify $1-2\sin^2 x$.
Step-by-Step Solution
- 1Recall the double-angle identity $\cos 2x=1-2\sin^2 x$.
- 2The expression is already in the exact form of this identity.
- 3Therefore, $1-2\sin^2 x=\cos 2x$.
Answer:$\cos 2x$
Question 3
Applying R-Formula
Question
Write $3\sin x+4\cos x$ in the form $R\sin(x+\alpha)$, where $R>0$ and $\alpha$ is acute.
Step-by-Step Solution
- 1Let $3\sin x+4\cos x=R\sin(x+\alpha)$.
- 2Expand: $R\sin(x+\alpha)=R\sin x\cos\alpha+R\cos x\sin\alpha$.
- 3Compare coefficients: $R\cos\alpha=3$ and $R\sin\alpha=4$.
- 4Find $R=\sqrt{3^2+4^2}=5$.
- 5Then $\cos\alpha=\frac{3}{5}$ and $\sin\alpha=\frac{4}{5}$, so $\alpha\approx53.1^\circ$.
Answer:$3\sin x+4\cos x=5\sin(x+53.1^\circ)$
Question 4
Solving a Basic Equation
Question
Solve $2\cos x=1$ for $0^\circ \le x \le 360^\circ$.
Step-by-Step Solution
- 1Divide both sides by 2 to get $\cos x=\frac{1}{2}$.
- 2The basic angle is $60^\circ$.
- 3Cosine is positive in Quadrants I and IV.
- 4Hence $x=60^\circ$ or $x=360^\circ-60^\circ=300^\circ$.
Answer:$x=60^\circ$ or $x=300^\circ$
Key Formulas to Remember
Pythagorean Identity
sin^2 x + cos^2 x = 1Note: Use this to convert between sine and cosine.
Tangent Ratio
tan x = sin x / cos xNote: Useful when a question gives sine and cosine.
Double-Angle Formula
sin 2x = 2 sin x cos xDouble-Angle Formula
cos 2x = cos^2 x - sin^2 x = 2cos^2 x - 1 = 1 - 2sin^2 xR-Formula
a sin x + b cos x = R sin(x + alpha), where R = sqrt(a^2 + b^2)Special Identities to Memorize
sin^2 x + cos^2 x = 1tan x = sin x / cos x1 + tan^2 x = sec^2 xsin 2x = 2sin x cos xcos 2x = 1 - 2sin^2 xcos 2x = 2cos^2 x - 1
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