Additional Mathematics Classroom | Chapter 8: Trigonometric Functions, Graphs, Identities, Formulae and Equations | SEC G2 K232 / G3 K341

The 2027 SEC position

Both G2 K232 and G3 K341 list the same core trigonometric content. Students are expected to work with:

• six trigonometric functions for angles of any magnitude, in degrees or radians;
• principal values of sin⁻¹x, cos⁻¹x and tan⁻¹x;
• exact values at 30°, 45°, 60° or π/6, π/4, π/3;
• amplitude, periodicity and symmetries of sine and cosine;
• graphs of transformed sine, cosine and tangent functions of the forms specified by the syllabus;
• fundamental trigonometric identities involving tan, cot, sec and cosec;
• addition and subtraction formulae for sine, cosine and tangent;
• double-angle formulae;
• expressions such as a cos θ + b sin θ in R-form;
• simplification of trigonometric expressions;
• simple trigonometric equations over a stated interval;
• proofs of simple trigonometric identities;
• trigonometric functions used as models.

The current syllabus explicitly excludes general solution as the examination target for simple trigonometric equations. That matters when adapting the old textbook, which was written for a different examination structure. We will still use periodicity to understand why multiple solutions occur, but the assessed route will focus on finding all solutions in the interval actually given.

Modern SEC trigonometry is not a list of ratios. It is a connected system linking angle, function, graph, identity, equation and model.

Official references: SEAB 2027 G2 Additional Mathematics K232 · SEAB 2027 G3 Additional Mathematics K341.

---

What this chapter is really teaching

Trigonometry becomes difficult when students keep several disconnected memories: SOHCAHTOA for triangles, a graph sketch from another lesson, identities from a formula sheet, a calculator procedure for equations and an R-formula learned near the end. The subject becomes much more manageable when those pieces are seen as different views of the same periodic functions.

The governing cycle is:

1. Angle: where are we around the circle, in degrees or radians?
2. Function: what are sin θ, cos θ, tan θ and their reciprocal functions?
3. Graph: how does the function behave as the angle varies continuously?
4. Identity: which different-looking expressions are equal for every admissible angle?
5. Equation: for which angles in the stated interval does a relationship become true?
6. Model: what repeating or oscillating phenomenon is the function describing?

A strong learner can move around this cycle in both directions. The graph helps solve an equation; an identity simplifies a graph expression; an exact triangle gives a special-angle value; an R-formula reveals amplitude and maximum value; a model gives physical meaning to period and phase.

Chapter map

1. Angles of any magnitude
2. Degrees and radians
3. Six trigonometric functions
4. Quadrants and signs
5. Exact special-angle values
6. Principal inverse-trigonometric values
7. Graphs of sin, cos and tan
8. Amplitude, period, shifts and symmetry
9. Transformed trigonometric graphs
10. Fundamental identities
11. Proving identities
12. Addition and subtraction formulae
13. Double-angle formulae
14. R-formulae
15. Solving trigonometric equations in intervals
16. Equations requiring identities or algebraic substitution
17. Trigonometric modelling
18. Error diagnosis and SEC transfer
19. Original guided practice and worked answers

---

1 · Angles do not stop at 90°

Right-triangle trigonometry is a starting point, not the complete definition. Additional Mathematics needs trigonometric functions for angles larger than 90°, negative angles and angles that have completed several revolutions.

Use a coordinate-circle view. Let a point P(x,y) lie on a circle of radius r centred at the origin. For an angle θ measured from the positive x-axis:

• sin θ = y/r;
• cos θ = x/r;
• tan θ = y/x where x ≠ 0.

These definitions continue to work around the entire circle. The signs of x and y determine the signs of sine, cosine and tangent in each quadrant.

Quadrant signs

• Quadrant I: sin, cos, tan all positive.
• Quadrant II: sin positive; cos and tan negative.
• Quadrant III: tan positive; sin and cos negative.
• Quadrant IV: cos positive; sin and tan negative.

Memorable acronyms can help retrieval, but the coordinate definitions are more powerful because they explain the signs. In Quadrant II, y is positive and x is negative, so sin = y/r is positive, cos = x/r is negative and tan = y/x is negative.

Reference angles

A reference angle is the acute angle between the terminal arm and the x-axis. It allows the exact or calculator magnitude to be found from an acute angle while the quadrant supplies the sign.

For example, sin150° = sin30° = 1/2 because 150° lies in Quadrant II where sine is positive. cos150° = −cos30° = −√3/2 because cosine is negative there.

---

2 · Degrees and radians are two coordinate systems for angle

Degrees divide a full revolution into 360 parts. Radians measure angle by comparing arc length to radius. One full revolution is 2π radians, so

180° = π radians.

Therefore:

• degrees → radians: multiply by π/180;
• radians → degrees: multiply by 180/π.

Special-angle equivalences

• 30° = π/6;
• 45° = π/4;
• 60° = π/3;
• 90° = π/2;
• 180° = π;
• 270° = 3π/2;
• 360° = 2π.

Students should become bilingual rather than repeatedly converting every angle before thinking. When an interval is written 0 ≤ θ ≤ 2π, recognise immediately that it is one full revolution.

Calculator-mode discipline

A correct calculation in the wrong angle mode gives a wrong result. Before calculator work, check whether the question is in degrees or radians. This is not a “careless mistake” in the trivial sense; it is a representation error that changes the mathematical meaning of the input.

---

3 · The six trigonometric functions

The primary three functions are sine, cosine and tangent. The reciprocal functions are cosecant, secant and cotangent:

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

Also:

tan θ = sin θ / cos θ, where cos θ ≠ 0.

These definitions carry domain restrictions. sec θ is undefined where cos θ = 0. cosec θ is undefined where sin θ = 0. cot θ is undefined where tan θ = 0, equivalently where sin θ = 0 while cos θ is nonzero.

Worked route · all six from one ratio

Given tan θ = 3/4 and θ is in Quadrant III, find sin θ, cos θ, sec θ, cosec θ and cot θ.

Use a 3–4–5 triangle for the reference angle. In Quadrant III, x and y are both negative, so sine and cosine are negative while tangent is positive.

• sin θ = −3/5;
• cos θ = −4/5;
• tan θ = 3/4;
• cosec θ = −5/3;
• sec θ = −5/4;
• cot θ = 4/3.

The quadrant is not an afterthought. It determines the signs of the exact values.

---

4 · Exact values of 30°, 45° and 60°

The syllabus explicitly requires exact values at the standard special angles. These should be understood from geometry rather than learned as an isolated table.

45° from an isosceles right triangle

Take a right triangle with legs 1 and 1. The hypotenuse is √2. Therefore:

• sin45° = 1/√2 = √2/2;
• cos45° = √2/2;
• tan45° = 1.

30° and 60° from an equilateral triangle

Bisect an equilateral triangle of side 2. The half-triangle has sides 1, √3 and 2. Thus:

• sin30° = 1/2;
• cos30° = √3/2;
• tan30° = 1/√3 = √3/3;
• sin60° = √3/2;
• cos60° = 1/2;
• tan60° = √3.

The reciprocal-function exact values then follow immediately.

Exact means exact

If the question asks for an exact value, √3/2 is better than 0.866. The decimal is only an approximation. The Surds chapter is therefore an active dependency of trigonometry, not a finished topic left behind.

---

5 · Principal values of inverse trigonometric functions

The notation sin⁻¹x, cos⁻¹x and tan⁻¹x denotes inverse functions, not reciprocals. In particular:

• sin⁻¹x is arcsine, not cosec x;
• cos⁻¹x is arccosine, not sec x;
• tan⁻¹x is arctangent, not cot x.

Trigonometric functions repeat, so an equation such as sin θ = 1/2 has infinitely many angles in the unrestricted real line. To make an inverse function single-valued, a principal-value range is chosen.

• sin⁻¹x returns a principal angle in [−90°,90°], or [−π/2,π/2].
• cos⁻¹x returns a principal angle in [0°,180°], or [0,π].
• tan⁻¹x returns a principal angle in (−90°,90°), or (−π/2,π/2).

Worked route

sin⁻¹(1/2) = 30°. That does not mean sin θ = 1/2 has only one solution. It means the inverse function returns the principal one. On 0° ≤ θ ≤ 360°, the equation has θ = 30° and 150°.

Principal inverse value and complete interval solution are different jobs.

---

6 · The basic graphs: periodic functions leave fingerprints

The graph of a trigonometric function is a map of how its value changes continuously with angle. Once the graph is understood, many equations and transformations become easier to reason about.

y = sin x

• amplitude = 1;
• period = 360° or 2π;
• range = [−1,1];
• zeros at integer multiples of 180° or π;
• maximum 1 at 90° + 360°k;
• minimum −1 at 270° + 360°k.

y = cos x

• amplitude = 1;
• period = 360° or 2π;
• range = [−1,1];
• maximum 1 at 0° + 360°k;
• minimum −1 at 180° + 360°k.

y = tan x

• period = 180° or π;
• range = all real values;
• zeros at integer multiples of 180° or π;
• vertical asymptotes where cos x = 0: 90° + 180°k or π/2 + kπ.

Tangent has no amplitude because it is unbounded. This is a good example of why applying sine/cosine vocabulary mechanically to every trig function is unsafe.

Symmetry

• sin(−x) = −sin x: sine is odd.
• cos(−x) = cos x: cosine is even.
• tan(−x) = −tan x: tangent is odd.

These symmetries are not merely graph facts; they become algebraic tools in simplification and equation solving.

---

7 · Transforming sine and cosine graphs

The SEC syllabus includes graphs of forms such as

• y = a sin(bx) + c;
• y = a sin(x/b) + c;
• y = a cos(bx) + c;
• y = a cos(x/b) + c;
• y = a tan(bx),

with the stated restrictions on a, b and c.

Amplitude

For sine and cosine, multiplying by a scales the vertical values. The amplitude is |a|. A negative a also reflects the graph in the x-axis.

Vertical shift

Adding c shifts the graph vertically. The midline becomes y = c. For y = 3 sin x + 2:

• amplitude = 3;
• midline = y = 2;
• maximum = 5;
• minimum = −1.

Horizontal scaling and period

For y = sin(bx) or cos(bx), the period is divided by b:

period = 360°/b or 2π/b.

For y = sin(x/b) or cos(x/b), the period is multiplied by b:

period = 360°b or 2πb.

The change happens because the inside argument reaches a complete cycle faster or slower.

Worked graph analysis

For y = −2 cos(3x) + 1, with x in degrees:

• amplitude = 2;
• period = 360°/3 = 120°;
• midline y = 1;
• maximum = 3;
• minimum = −1;
• the negative coefficient reflects the basic cosine graph vertically.

Tangent transformation

For y = a tan(bx), the period is 180°/b or π/b. Multiplication by a stretches or reflects the branches vertically but does not create a bounded amplitude.

---

8 · Fundamental identities: relationships that are always true

An identity is different from an equation to solve. An equation is true only for particular values. An identity is true for every value in its valid domain.

The fundamental identities include:

• tan A = sin A / cos A;
• cot A = cos A / sin A;
• sin²A + cos²A = 1;
• sec²A = 1 + tan²A;
• cosec²A = 1 + cot²A.

The last two are consequences of the Pythagorean identity. Divide sin²A + cos²A = 1 by cos²A to obtain tan²A + 1 = sec²A. Divide by sin²A to obtain 1 + cot²A = cosec²A.

Why identities matter

Identity work trains equivalence control. A complicated expression can be rewritten into another form without changing its value. The main challenge is choosing the transformation that moves the expression towards the target rather than away from it.

---

9 · Proving identities: do not attack both sides randomly

A reliable identity-proof strategy is:

1. Start from the more complicated side.
2. Convert sec, cosec, cot or tan into sine and cosine when useful.
3. Use one standard identity at a time.
4. Factorise, combine fractions or rationalise only when the algebra suggests it.
5. Move towards the form of the other side.
6. Do not treat the identity as an equation whose sides may be manipulated simultaneously without explanation.

Worked proof A

Prove

(1 − cos²A)/sin A = sin A.

Start with the left side:

(1 − cos²A)/sin A

= sin²A/sin A

= sin A.

We used 1 − cos²A = sin²A. The route is short because the target already contains sin A.

Worked proof B

Prove

(sec A − cos A)/tan A = sin A.

Left side:

[1/cos A − cos A]/[sin A/cos A]

= [(1 − cos²A)/cos A]·[cos A/sin A]

= sin²A/sin A

= sin A.

Converting reciprocal functions into sine and cosine exposed the Pythagorean identity.

Identity proof is not “showing both sides are equal by calculator”

Checking a few numerical angles can support confidence, but it does not prove an identity for every admissible angle. The proof must use exact algebraic relationships.

---

10 · Addition and subtraction formulae

The syllabus includes the expansions:

• sin(A ± B) = sinA cosB ± cosA sinB;
• cos(A ± B) = cosA cosB ∓ sinA sinB;
• tan(A ± B) = (tanA ± tanB)/(1 ∓ tanA tanB).

These formulae let us construct exact values for angles that are sums or differences of standard angles, and they become building blocks for double-angle formulae and identity manipulation.

Worked exact value · sin75°

75° = 45° + 30°.

sin75° = sin45°cos30° + cos45°sin30°

= (√2/2)(√3/2) + (√2/2)(1/2)

= (√6 + √2)/4.

Worked exact value · cos15°

15° = 45° − 30°.

cos15° = cos45°cos30° + sin45°sin30°

= (√6 + √2)/4.

The plus sign appears because cos(A − B) = cosAcosB + sinAsinB.

Sign discipline

The cosine formula changes the internal sign compared with the angle sign. This is a high-frequency error. Instead of relying on a vague memory, write the formula before substitution when accuracy matters.

---

11 · Double-angle formulae

Set B = A in the addition formulae:

• sin2A = 2sinAcosA;
• cos2A = cos²A − sin²A = 2cos²A − 1 = 1 − 2sin²A;
• tan2A = 2tanA/(1 − tan²A), where defined.

The three forms of cos2A are especially useful because different problems need different variables. If the expression contains only sin²A, use cos2A = 1 − 2sin²A. If it contains only cos²A, use cos2A = 2cos²A − 1.

Worked route · exact double angle

Given sinA = 3/5 and A is acute, find cos2A.

Use cos2A = 1 − 2sin²A:

cos2A = 1 − 2(9/25) = 1 − 18/25 = 7/25.

No square-root calculation for cosA was needed. Formula choice removed unnecessary work.

Worked route · solve using a double-angle identity

Solve 2sin x cos x = 1/2 for 0° ≤ x ≤ 360°.

Use sin2x = 2sinxcosx:

sin2x = 1/2.

Now 0° ≤ 2x ≤ 720°. Within this doubled interval, sin is 1/2 at 30°,150°,390°,510°.

Therefore

x = 15°, 75°, 195°, 255°.

When the argument becomes 2x, the interval must be transformed too. This is a major source of missing solutions.

---

12 · R-formula: compress two trig terms into one

The SEC syllabus includes expressing

a cos θ + b sin θ

in a form such as

R cos(θ − α) or R sin(θ + α),

with the sign chosen to match the coefficients.

The purpose is not cosmetic. One sinusoidal term makes amplitude, maximum, minimum and equations easier to read.

Deriving R cos(θ − α)

Expand:

R cos(θ − α) = R[cosθ cosα + sinθ sinα].

Match with a cosθ + b sinθ:

• R cosα = a;
• R sinα = b.

Square and add:

R²(cos²α + sin²α) = a² + b².

Therefore

R = √(a² + b²).

Also tanα = b/a when the chosen form and quadrant make that interpretation appropriate.

Worked route · 3cosθ + 4sinθ

Write 3cosθ + 4sinθ as R cos(θ − α).

R = √(3² + 4²) = 5.

5cosα = 3 and 5sinα = 4, so cosα = 3/5, sinα = 4/5 and tanα = 4/3.

Thus

3cosθ + 4sinθ = 5cos(θ − α), where α = tan⁻¹(4/3).

Maximum and minimum from R-form

Since −1 ≤ cos(θ − α) ≤ 1:

−5 ≤ 5cos(θ − α) ≤ 5.

Therefore the maximum of 3cosθ + 4sinθ is 5 and the minimum is −5.

A two-term expression has become one sinusoid whose amplitude is visible immediately.

R-form as an equation-solving tool

To solve 3cosθ + 4sinθ = 2, rewrite as 5cos(θ − α) = 2. Then cos(θ − α) = 2/5 and solve within the transformed interval. This is often cleaner than attempting to isolate sin or cos in the original expression.

---

13 · Solving trigonometric equations: interval first, not last

The current SEC requirement focuses on simple trigonometric equations in a given interval. That phrase should change how the solution is organised.

A reliable process is:

1. Write the original interval clearly.
2. Simplify the equation to one trig function where possible.
3. Find the reference or principal angle.
4. Use the function’s signs, graph and period to generate all solutions in the interval.
5. If the argument is bx or x/b, transform the interval for that argument.
6. Check endpoints and excluded values.
7. Return answers in the requested angular unit.

Worked equation A

Solve 2sinθ = √3 for 0° ≤ θ ≤ 360°.

sinθ = √3/2.

Reference angle 60°. Sine is positive in Quadrants I and II.

θ = 60°, 120°.

Worked equation B

Solve tanθ = −1 for 0 ≤ θ ≤ 2π.

Reference angle π/4. Tangent is negative in Quadrants II and IV.

θ = 3π/4, 7π/4.

Worked equation C · transformed argument

Solve cos2x = 1/2 for 0° ≤ x ≤ 180°.

Then 0° ≤ 2x ≤ 360°.

cos2x = 1/2 gives 2x = 60° or 300°.

Therefore x = 30° or 150°.

Students who solve only over the original x-interval for the doubled angle often lose solutions.

---

14 · Equations that need algebra before trigonometry

Many A-Math trig equations are disguised algebra problems. The trig function acts like a variable.

Worked route A · quadratic in sin x

Solve 2sin²x − 3sinx + 1 = 0 for 0° ≤ x ≤ 360°.

Let u = sinx:

2u² − 3u + 1 = 0

(2u − 1)(u − 1) = 0.

So sinx = 1/2 or sinx = 1.

For sinx = 1/2: x = 30°,150°.

For sinx = 1: x = 90°.

Thus x = 30°,90°,150°.

Worked route B · factorise after using double angle

Solve 3sin2x − 2cosx = 0 for 0° ≤ x ≤ 360°.

Use sin2x = 2sinxcosx:

6sinxcosx − 2cosx = 0.

Factor:

2cosx(3sinx − 1) = 0.

So cosx = 0 or sinx = 1/3.

cosx = 0 gives x = 90°,270°.

sinx = 1/3 gives two solutions in Quadrants I and II, which can be found with a calculator and rounded as requested.

Factoring preserved the branch cosx = 0. Dividing both sides by cosx would have lost those valid solutions.

Never divide by an expression that might be zero without considering that branch

This is a general algebra principle that becomes especially important in trigonometry. If an equation contains cosx as a factor, dividing by cosx assumes cosx ≠ 0. Solve the zero-factor branch separately before dividing.

---

15 · Equations requiring identities

Some equations contain more than one trig function. The first job is to convert them to a common function or useful factorisation.

Worked route A

Solve cos2x = sinx for 0° ≤ x ≤ 360°.

Use cos2x = 1 − 2sin²x:

1 − 2sin²x = sinx.

2sin²x + sinx − 1 = 0.

(2sinx − 1)(sinx + 1) = 0.

So sinx = 1/2 or sinx = −1.

Therefore x = 30°,150°,270°.

Worked route B

Solve 1 + tan²x = 4 for 0° ≤ x < 360°.

tan²x = 3, so tanx = ±√3.

Reference angle 60°. Tangent is positive in I and III, negative in II and IV.

x = 60°,120°,240°,300°.

The identity sec²x = 1 + tan²x is not even required here unless the original expression came in sec form. The point is to choose the identity that reduces the number of functions.

---

16 · Trigonometric models: amplitude and period acquire meaning

The syllabus includes using trigonometric functions as models. A model such as

h(t) = A sin(Bt) + C

or

h(t) = A cos(Bt) + C

can represent a repeating height, displacement, seasonal quantity or oscillation.

Parameter meaning

• |A| = amplitude: maximum deviation from the midline;
• C = vertical shift or midline;
• period = 2π/B in radians, or 360°/B in degrees;
• maximum = C + |A|;
• minimum = C − |A|.

Worked model · Ferris-wheel style height

A seat’s height is modelled by

h(t) = 12 − 10cos(πt/20),

where h is in metres and t in seconds.

• midline = 12 m;
• amplitude = 10 m;
• maximum height = 22 m;
• minimum height = 2 m;
• angular coefficient = π/20, so period T satisfies (π/20)T = 2π; hence T = 40 s.

The negative cosine means the model begins at the minimum when t = 0: h(0) = 12 − 10 = 2.

Model limits

A trigonometric model assumes repeating behaviour. Real systems may change amplitude, period or baseline over time. The mathematical model can be exact as a formula while still being an approximation of the physical world.

---

17 · G2 and G3: same trig core, different integration pressure

The listed trigonometric-functions, identities and equations content is shared across G2 K232 and G3 K341. The difference lies in the wider subject and the independence expected from the learner.

• G2: stabilise exact values, graph interpretation, core identities, interval solutions and simple proofs; connect them cleanly without unnecessary complexity.
• G3: do all of the above while handling denser algebraic manipulation, more mixed-topic questions, stronger transfer into calculus and less explicit method signalling.

The explanation of sin(A + B) does not need two versions. The performance ceiling changes through problem design, connection density and the amount of scaffolding removed.

---

18 · The trigonometry error map

“Weak at trigonometry” is too broad to repair. Common first weak links include:

• Angle-unit failure: degree/radian calculator mode does not match the question.
• Quadrant failure: reference-angle magnitude is correct but sign is wrong.
• Inverse-function confusion: sin⁻¹ is mistaken for cosec.
• Special-value fragility: exact values are replaced with decimals or remembered inconsistently.
• Graph-period failure: b is used to multiply rather than divide the base period in sin(bx).
• Amplitude failure: tangent is assigned an amplitude even though it is unbounded.
• Identity-law failure: a standard identity is recalled with a sign error.
• Proof-direction failure: both sides are manipulated randomly instead of simplifying one towards the other.
• Addition-formula sign failure: cos(A − B) is expanded with the wrong sign.
• R-form return failure: R is found but α or the correct ± form is not matched.
• Interval failure: the argument bx is solved over the original x-interval rather than the transformed interval.
• Lost branch: division by sinx or cosx removes zero-factor solutions.
• General-solution overreach: time is spent generating unrestricted families when the SEC question asks only for a stated interval.
• Model interpretation failure: amplitude and period are calculated but not connected to the real situation.

The first weak-link diagnostic

1. Can the learner give exact values for 30°,45°,60°?
2. Can the learner determine signs by quadrant?
3. Can the learner identify period and amplitude from a transformed graph?
4. Can the learner use sin²A + cos²A = 1 accurately?
5. Can the learner solve a simple equation over a stated interval?
6. Can the learner prove a one-line identity without altering both sides randomly?
7. Can the learner use one addition/double-angle formula correctly?
8. Can the learner form R and identify α?

Stop at the first unstable stage. A student who cannot control quadrants does not need a harder identity proof. A student who knows identities but misses interval solutions needs periodicity and graph repair, not more symbolic manipulation.

---

19 · Original guided practice

The following questions are original to this classroom guide. They move from function control into graph, identity, formula, equation and modelling work. Complete the first pass without the worked answers and record the first point at which the method becomes uncertain.

A · Angles, exact values and six functions

1. Convert 150° to radians.
2. Convert 7π/6 to degrees.
3. Find exact values of sin150°, cos150° and tan150°.
4. Find exact values of sin225°, cos225° and tan225°.
5. Given tanθ = 3/4 and θ is in Quadrant III, find all six trigonometric functions.
6. Given cosθ = −5/13 and θ is in Quadrant II, find sinθ and tanθ.
7. Evaluate exactly: sin60°cos30° + cos60°sin30°.
8. Explain why sin⁻¹(1/2) = 30° does not imply that sinθ = 1/2 has only one solution in 0° ≤ θ ≤ 360°.

B · Graph structure

9. State the amplitude and period of y = 4sin3x, where x is in degrees.
10. State the amplitude, period, midline, maximum and minimum of y = −2cos(2x) + 5.
11. State the period of y = 3tan4x in degrees.
12. State the period of y = sin(x/3) in degrees.
13. State the range of y = 5cosx − 2.
14. Explain why tangent has no amplitude.
15. Sketch one period of y = 2sinx + 1, marking maximum, minimum and midline.

C · Fundamental identities and simplification

16. Simplify (1 − cos²A)/sinA.
17. Simplify sec²A − tan²A.
18. Simplify (1 + cot²A)/cosec²A.
19. Express tanA + cotA as a single fraction in sinA and cosA, then simplify.
20. Prove (secA − cosA)/tanA = sinA.
21. Prove (1 − sinA)(1 + sinA) = cos²A.
22. Prove tanA + cotA = secA cosecA.

D · Addition and double-angle formulae

23. Find exact sin75°.
24. Find exact cos15°.
25. Find exact tan75°.
26. Given sinA = 3/5 and A is acute, find sin2A and cos2A.
27. Given tanA = 2 and A is acute, find tan2A.
28. Show that sin(A + B) + sin(A − B) = 2sinAcosB.
29. Express cos2A entirely in terms of sinA.
30. Express cos2A entirely in terms of cosA.

E · R-formula

31. Express 3cosθ + 4sinθ as Rcos(θ − α), where R > 0 and 0° < α < 90°.
32. Hence state the maximum and minimum values of 3cosθ + 4sinθ.
33. Express 5sinθ − 12cosθ as Rsin(θ − α) or an equivalent valid R-form.
34. Find the maximum value of 8sinθ + 15cosθ.
35. Use an R-form to solve 3cosθ + 4sinθ = 5 for 0° ≤ θ ≤ 360°.

F · Equations in stated intervals

36. Solve sinθ = √3/2 for 0° ≤ θ ≤ 360°.
37. Solve cosθ = −1/2 for 0 ≤ θ ≤ 2π.
38. Solve tanθ = 1 for 0° ≤ θ < 360°.
39. Solve cos2x = 1/2 for 0° ≤ x ≤ 180°.
40. Solve sin3x = 0 for 0° ≤ x ≤ 180°.
41. Solve 2sin²x − 3sinx + 1 = 0 for 0° ≤ x ≤ 360°.
42. Solve cos2x = sinx for 0° ≤ x ≤ 360°.
43. Solve 3sin2x − 2cosx = 0 for 0° ≤ x ≤ 360°.
44. Solve 1 + tan²x = 4 for 0° ≤ x < 360°.
45. Solve 2cos²x − 1 = 0 for 0° ≤ x ≤ 360°.

G · Modelling and mixed reasoning

46. For h(t) = 12 − 10cos(πt/20), state the amplitude, midline, maximum, minimum and period.
47. A seasonal quantity is modelled by Q(t) = 40 + 8sin(πt/6), with t in months. Find its period and range.
48. A model y = A sin(Bt) + C has maximum 17, minimum 5 and period 8. Find |A|, C and B if t is measured in units giving radians in the argument.
49. Explain why an R-form can be useful when modelling a quantity of the form a cosθ + b sinθ.
50. A student obtains x = 30° from sinx = 1/2 on 0° ≤ x ≤ 360° and stops. Diagnose the error.
51. A student divides 2cosx(3sinx − 1) = 0 by cosx and solves only sinx = 1/3. Diagnose the error.
52. A student enters sin(π/6) into a calculator in degree mode. Explain why this is a representation error.
53. A student proves an identity by showing both sides equal 0.75 for A = 30°. Explain why that is not a proof.
54. A graph y = 3sin2x + 4 is said to have amplitude 4. Correct the statement and explain the role of 4.
55. Explain why SEC interval solutions should be complete even though general solutions are not required.

20 · Worked answers and reasoning checkpoints

1. 150° × π/180 = 5π/6.
2. (7π/6)×180/π = 210°.
3. Reference angle 30°, Quadrant II: sin150° = 1/2, cos150° = −√3/2, tan150° = −√3/3.
4. Reference angle 45°, Quadrant III: sin225° = −√2/2, cos225° = −√2/2, tan225° = 1.
5. sinθ = −3/5, cosθ = −4/5, tanθ = 3/4, cosecθ = −5/3, secθ = −5/4, cotθ = 4/3.
6. Use a 5–12–13 triangle. Quadrant II gives sinθ = 12/13, tanθ = −12/5.
7. This is sin(60° + 30°) = sin90° = 1.
8. sin⁻¹(1/2) returns only the principal value 30°. On the full interval, sine is also 1/2 in Quadrant II, giving 150°.
9. Amplitude 4; period 120°.
10. Amplitude 2; period 180°; midline y = 5; maximum 7; minimum 3.
11. Period = 180°/4 = 45°.
12. Period = 360°×3 = 1080°.
13. Midline −2, amplitude 5, so range [−7,3].
14. Tangent is unbounded and has no maximum/minimum distance from a midline, so amplitude is not defined.
15. One period follows the usual sine shape shifted up 1 and stretched to amplitude 2: midline y = 1, max 3, min −1.
16. sinA.
17. 1.
18. 1.
19. tanA + cotA = sinA/cosA + cosA/sinA = (sin²A + cos²A)/(sinAcosA) = 1/(sinAcosA).
20. As shown in the worked proof: convert sec and tan to sine/cosine, use 1 − cos²A = sin²A, obtaining sinA.
21. (1 − sinA)(1 + sinA) = 1 − sin²A = cos²A.
22. tanA + cotA = 1/(sinAcosA), while secA cosecA = 1/(cosA sinA). Hence identical.
23. (√6 + √2)/4.
24. (√6 + √2)/4.
25. tan75° = tan(45° + 30°) = [1 + 1/√3]/[1 − 1/√3] = 2 + √3.
26. cosA = 4/5. sin2A = 2(3/5)(4/5) = 24/25. cos2A = 1 − 2(9/25) = 7/25.
27. tan2A = 2(2)/(1 − 4) = −4/3.
28. Expand both sine expressions: the cosA sinB terms cancel, leaving 2sinAcosB.
29. cos2A = 1 − 2sin²A.
30. cos2A = 2cos²A − 1.
31. R = 5 and tanα = 4/3. 3cosθ + 4sinθ = 5cos(θ − α), α = tan⁻¹(4/3).
32. Maximum 5, minimum −5.
33. R = 13. One valid form is 13sin(θ − α) with cosα = 5/13 and sinα = 12/13, so α = tan⁻¹(12/5).
34. R = √(8² + 15²) = 17.
35. 5cos(θ − α) = 5 gives cos(θ − α) = 1, so θ − α = 0° modulo 360°. In the stated interval the valid solution is θ = α = tan⁻¹(4/3) (with endpoint interpretation according to the chosen α range).
36. 60°,120°.
37. cosθ = −1/2 in Quadrants II and III: θ = 2π/3,4π/3.
38. 45°,225°.
39. 0° ≤ 2x ≤ 360°. 2x = 60°,300°, so x = 30°,150°.
40. 0° ≤ 3x ≤ 540°. sin3x = 0 at 0°,180°,360°,540°. Thus x = 0°,60°,120°,180°.
41. 30°,90°,150°.
42. 30°,150°,270°.
43. 2cosx(3sinx − 1) = 0. cosx = 0 gives 90°,270°. sinx = 1/3 gives x = sin⁻¹(1/3) and 180° − sin⁻¹(1/3). Include all four solutions.
44. tan²x = 3, so tanx = ±√3. 60°,120°,240°,300°.
45. 2cos²x − 1 = 0 gives cos²x = 1/2, so cosx = ±√2/2. 45°,135°,225°,315°.
46. Amplitude 10; midline 12; maximum 22; minimum 2; period 40.
47. Angular coefficient π/6 gives period 2π/(π/6) = 12 months. Range = 40 ± 8, so [32,48].
48. Amplitude |A| = (17 − 5)/2 = 6. Midline C = (17 + 5)/2 = 11. Period 8 gives B = 2π/8 = π/4.
49. R-form converts the sum of sine and cosine with the same angle into one sinusoid, making amplitude, extrema, phase and equation solving easier to read.
50. The student found the principal/reference solution but missed the Quadrant-II solution. Complete interval answer is 30°,150°.
51. Dividing by cosx assumes cosx ≠ 0 and loses the branch cosx = 0. That branch must be solved separately.
52. π/6 is a radian measure. Degree mode interprets the numerical input π/6 as degrees rather than radians, changing the angle itself.
53. Checking one value only shows the equality is true at that angle. An identity must be shown algebraically for every admissible angle.
54. In y = 3sin2x + 4, amplitude = 3. The 4 is the vertical shift, so the midline is y = 4.
55. Although unrestricted general-solution formulae are excluded, the stated interval may contain several repeated-cycle solutions. “No general solution required” does not mean “give only one angle”.

---

21 · A two-week revision architecture

Trigonometry is too broad to revise as one undifferentiated chapter. Its dependencies should return in layers.

1. Day 1: special angles, radians and quadrants.
2. Day 2: six functions and reciprocal relationships.
3. Day 4: sine/cosine/tangent graphs, amplitude and period.
4. Day 6: fundamental identities and one-step proofs.
5. Day 8: addition and double-angle formulae.
6. Day 10: R-formulae and extrema.
7. Day 12: interval equations, including transformed arguments and quadratic forms.
8. Day 14: modelling plus an unlabelled mixed set where the learner must decide whether the problem is graph, identity, equation or model.

Seven-minute retrieval test

• Write exact sin, cos and tan values for 30°,45°,60°.
• State the signs of sin, cos, tan in all four quadrants.
• State periods of sinx, cosx and tanx.
• Write the three Pythagorean-style identities.
• Write sin(A+B), cos(A+B), tan(A+B).
• Write sin2A, the three cos2A forms and tan2A.
• Explain how to find R in a cosθ + b sinθ.
• Explain why bx changes the interval in an equation.

22 · For teachers: teach the return path between representations

The old textbook’s strongest feature is its progression. It does not begin with identity proofs. It first gives the learner angle, function, graph and equation experience. That order is worth preserving.

A useful teaching routine is to show one mathematical object in four forms. For example, take sinθ = 1/2:

• triangle: special angle 30°;
• circle: y-coordinate/radius equals 1/2 in Quadrants I and II;
• graph: horizontal line y = 1/2 meets sine twice per 0°–360° cycle;
• equation: θ = 30°,150° over the stated interval.

The learner should not have separate memories for those four statements. They describe the same function.

Teach identity choice, not identity hunting

When a proof stalls, ask what the target contains. If the target is only sine and cosine, reciprocal functions are probably clutter. If the expression contains 1 − sin²A, the Pythagorean identity suggests cos²A. If the equation contains sin2A and cosA, the double-angle formula may expose a factor.

The point is to make transformations directional. Random identity substitution produces longer expressions and teaches the learner that proof is guesswork.

Do not over-train general solution

Periodicity should be understood deeply, but current SEC preparation should respect the specification. Train complete interval solutions, including intervals spanning more than one cycle. General solution can be enrichment or future-study preparation, not a substitute for mastering the actual examination task.

23 · For parents: what real progress sounds like

Early in the chapter, a student may ask, “Which trig formula is this?” As control grows, the language changes:

• “The angle is in Quadrant III, so sine and cosine must both be negative.”
• “This graph has amplitude 2 and period 120°, so the 3 inside the sine compressed the cycle.”
• “I want only sine and cosine, so I will replace sec and tan first.”
• “The argument is 2x, so I need to double the interval before listing solutions.”
• “I cannot divide by cosx yet because cosx might be zero.”
• “The two terms can be compressed into Rcos(θ − α), so the maximum is R.”
• “The calculator gave a principal angle; I still need every valid angle in the interval.”

That shift shows that the learner is reading the structure instead of searching memory for an isolated recipe.

24 · What carries forward

Trigonometry remains active throughout the rest of Additional Mathematics. Differentiation requires derivatives of sine, cosine and tangent at G3. Integration includes trigonometric functions. Kinematics may combine periodic models with rates. Coordinate geometry and plane geometry continue to depend on exact angle and ratio reasoning. Trigonometric identities become simplification tools inside calculus rather than disappearing when the chapter ends.

The old textbook’s next chapter is Proofs in Plane Geometry. That topic is still relevant in the SEC Additional Mathematics geometry strand, but the current G2 and G3 syllabuses should control exactly which proof relationships are core. The next classroom article will therefore crosswalk the old Chapter 10 against the 2027 syllabus before publication rather than assuming every historical theorem retains the same status.

---

Official syllabus references

• Singapore Examinations and Assessment Board · 2027 SEC G2 Additional Mathematics K232
• Singapore Examinations and Assessment Board · 2027 SEC G3 Additional Mathematics K341

Curriculum and assessment requirements can change. The official SEAB syllabuses remain the controlling sources for current subject codes, examinable content and examination structure.

Return to the Additional Mathematics Learning Hub