Secondary 3 Additional Mathematics | Singapore G3
Trigonometric Identities & Equations

Why the Same Angle Can Have More Than One Answer

If sin θ = 1/2, θ is not automatically 30°. The unit circle reaches the same y-coordinate again.

This is where trigonometry stops behaving like a one-answer triangle calculation and becomes a function problem.

For example, in the interval 0° ≤ θ ≤ 360°:

sin θ = 1/2 → θ = 30° or 150°.

The calculator’s sin⁻¹ button gives the principal value 30°. It does not automatically give every angle in the required interval.

inverse trig gives a starting angle; periodicity and symmetry give the complete interval answer.

This page owns the identity-and-equation machinery that follows the Trigonometric Functions foundation.

SEAB 2027 G3 Additional Mathematics syllabus (K341) →

The Topic Job

Transform trigonometric expressions into useful forms, prove that two forms are identical, and solve equations for every valid angle in a stated interval.

Equation or Identity?

An equation is true only for particular values.

sin θ = 1/2

is true only for selected θ.

An identity is true for every value for which both sides are defined.

sin²θ + cos²θ ≡ 1.

This difference controls how you work. You solve equations. You transform or prove identities.

The Fundamental Identity

On the unit circle, a point has coordinates (cos θ, sin θ). Because its distance from the origin is 1:

cos²θ + sin²θ = 1.

This is simply the circle equation x²+y²=1 with x=cos θ and y=sin θ.

So the most-used trig identity is geometry in disguise.

Two More Identities Come From Division

Divide sin²θ+cos²θ=1 by cos²θ:

tan²θ + 1 = sec²θ.

Divide by sin²θ:

1 + cot²θ = cosec²θ.

They are not three disconnected formulas. Two are consequences of the first.

Reciprocal and Quotient Relationships

• tan θ = sin θ / cos θ
• cot θ = cos θ / sin θ
• sec θ = 1 / cos θ
• cosec θ = 1 / sin θ

These are often the first conversion tools when an identity contains too many different trig functions.

A Strong Identity Strategy

When asked to prove an identity, do not manipulate both sides randomly.

1. Choose the more complicated side.
2. Convert sec, cosec, cot or tan if that reduces the number of function types.
3. Look for sin²+cos²=1.
4. Factorise or combine fractions using ordinary algebra.
5. Stop when the target side appears.

Identity proof = controlled transformation, not equation solving.

Worked Proof 1

Prove:

(1−cos²θ)/sin θ ≡ sin θ.

Start with the left side:

(1−cos²θ)/sin θ
= sin²θ/sin θ
= sin θ.

The decisive recognition was 1−cos²θ = sin²θ.

Why “Cross-Multiplying” an Identity Can Be Dangerous

If you multiply or divide by an expression that can be zero, you may silently change the domain on which the statement is valid.

At this level, keep transformations algebraically legitimate and avoid cancelling a trig factor unless you know what happens when that factor is zero.

Compound-Angle Formulae

The syllabus includes:

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)=(tan A ± tan B)/(1 ∓ tan A tan B).

The cosine sign pattern is easy to misremember. A useful check is to set B=0. The formula must reduce to cos A.

Exact Values From Compound Angles

Find sin 15° exactly.

Write 15°=45°−30°:

sin15° = sin(45°−30°)
= sin45°cos30°−cos45°sin30°
= (√2/2)(√3/2)−(√2/2)(1/2)
= (√6−√2)/4.

This is one reason surds reappear naturally inside trigonometry.

Double-Angle Formulae

Set B=A in the compound-angle formulae:

sin2A = 2sin A cos A

cos2A = cos²A−sin²A = 2cos²A−1 = 1−2sin²A

tan2A = 2tan A/(1−tan²A).

The three cosine forms are useful because different questions contain different raw material.

Choose the Cos 2A Form That Removes the Unwanted Function

If the expression contains only sin²A, use:

cos2A = 1−2sin²A.

If it contains only cos²A, use:

cos2A = 2cos²A−1.

The formula choice should reduce complexity, not increase it.

The R-Formula

An expression such as:

a cos θ + b sin θ

can be written as one shifted sine or cosine:

R cos(θ−α)

or an equivalent allowed form.

Expand:

R cos(θ−α)=R cosθ cosα + R sinθ sinα.

Compare coefficients:

• R cosα = a;
• R sinα = b.

Square and add:

R=√(a²+b²).

Then tanα=b/a, with quadrant chosen from the signs of a and b.

Why R-Form Is Useful

Once:

3cosθ+4sinθ = 5cos(θ−α),

we immediately know the expression lies between −5 and 5.

A two-function expression has become one sinusoid with amplitude 5 and phase shift α.

Solving a Basic Trigonometric Equation

Solve sin θ = √3/2 for 0°≤θ≤360°.

Reference angle = 60°.

Sine is positive in Quadrants I and II.

Therefore:

θ=60°,120°.

When the Angle Is Multiplied

Solve:

cos 2x = 1/2,   0°≤x≤360°.

The angle inside cosine is 2x.

Since x ranges from 0° to 360°, 2x ranges from 0° to 720°.

In 0°≤2x≤720°:

2x=60°,300°,420°,660°.

Therefore:

x=30°,150°,210°,330°.

A common error is solving 2x over the original x-interval and losing solutions.

Solve the Internal Angle Over Its Own Interval

before solving, transform the stated x-interval into the corresponding interval for the trig argument.

This one habit prevents a large class of missing-solution errors.

Equations That Must Be Simplified First

Solve:

2sin²x−sin x−1=0.

Treat sin x as one algebraic object:

(2sin x+1)(sin x−1)=0.

So:

• sin x=−1/2;
• sin x=1.

Now solve each trig equation over the specified interval.

The structure is:

trig equation → ordinary algebra in one trig function → angle solutions.

Do Not Divide by a Trig Function Too Early

If:

sin x(2cos x−1)=0,

dividing both sides by sin x loses every solution with sin x=0.

Instead use the zero-product rule:

• sin x=0;
• 2cos x−1=0.

Never divide by an expression that might be zero unless you handle that zero case separately.

The Earliest Weak Link

What you see	Likely weak link
Calculator gives one angle and student stops	principal value mistaken for interval solution
Missing answers in cos 2x equation	internal-angle interval not transformed
Proof manipulates both sides until they meet	identity proof structure weak
Wrong sign in cos(A±B)	compound formula not checked structurally
Loses sin x=0 solution	divides by a possibly zero factor
Uses wrong R or α quadrant	coefficient comparison not linked to geometry

Common Mistakes to Repair

• Identity = equation. They require different tasks.
• Changing both sides in a proof. It becomes difficult to show a valid chain of equivalence.
• Ignoring function domains when cancelling.
• Using the principal inverse-trig value as the only solution.
• Forgetting the changed interval for 2x or 3x.
• Dividing by sin x or cos x and losing zero solutions.
• Rounding exact special-angle values too early.

Retrieval Check

1. State sin²A+cos²A.
2. Derive sec²A=1+tan²A from the fundamental identity.
3. State sin(A+B).
4. State two equivalent forms of cos2A.
5. What is R for 5cosθ+12sinθ?
6. Why can sin θ=1/2 have two answers in 0°≤θ≤360°?

Transfer Set

1. Prove (sec²x−1)/tan x ≡ tan x.
2. Find cos15° exactly.
3. Express 8cosθ+6sinθ in the form Rcos(θ−α), giving R exactly and α appropriately.
4. Solve 2sin x cos x=1/2 for 0°≤x≤360°.
5. Solve 2cos²x−3cos x+1=0 for 0°≤x≤360°.
6. Solve tan(2x)=1 for 0°≤x≤180°.

Answer outline — open only after attempting

1. sec²x−1=tan²x, so tan²x/tan x=tan x where defined.
2. cos(45°−30°)=(√6+√2)/4.
3. R=10; 10cos(θ−α) with cosα=4/5, sinα=3/5.
4. sin2x=1/2. Since 0°≤2x≤720°, obtain 2x=30°,150°,390°,510°, hence x=15°,75°,195°,255°.
5. (2cos x−1)(cos x−1)=0. So cos x=1/2 or 1; x=0°,60°,300°,360° subject to endpoint convention in the stated interval.
6. 0°≤2x≤360°. tan=1 at 45°,225°. Hence x=22.5°,112.5°.

Independent Checks

• Substitute a convenient angle into both sides of an identity as a quick error check—without treating that as a proof.
• Check every final angle lies inside the stated interval.
• Substitute solutions into the original trig equation.
• For an R-form, expand it back and compare coefficients.
• Check exact answers against a calculator only after the exact structure is complete.

For Parents and Tutors — What This Topic Is Really Testing

The main transition is from remembering identities to choosing transformations.

Ask:

• What function types are present?
• Which identity reduces the number of function types?
• Is this an equation to solve or an identity to prove?
• What is the true interval for the internal angle?
• Could any factor you are dividing by be zero?
• Did the inverse-trig button give one principal angle or all required answers?

The teaching target is transformation with solution control.

Where This Connects Next

• Trigonometric Functions
• Surds
• Differentiation
• Integration
• Secondary 3 A-Math Complete Map
• Additional Mathematics Directory

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Bukit Timah Tutor Mathematics
A trigonometric equation is not finished when the calculator produces one angle. It is finished when every valid angle in the stated interval has been accounted for and checked.