Trigonometry
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📋 Summary — What This Chapter Covers
- Trig functions (sin, cos, tan) give ratios of sides in right-angled triangles, and their reciprocals (sec, cosec, cot) are just "1 over" each one.
- The unit circle extends sin, cos and tan to ANY angle — not just angles inside a triangle (0°–90°).
- Graphs of sin, cos, tan are periodic (repeat forever) — sketching them is the most reliable way to find every solution to an equation.
- Transformations of trig graphs work exactly like transformations of any other function (translate, stretch, reflect).
- Trig identities (like sin²θ + cos²θ = 1) let you rewrite equations into solvable forms and prove new results.
- Solving trig equations — because trig functions repeat, equations have infinite solutions; you always work within a given interval, using the CAST diagram or a sketch to catch every one.
- Non-right-angled triangles need the sine rule, cosine rule, and area formula (SOHCAHTOA won't work there).
1️⃣ Trigonometric Functions & SOHCAHTOA
For an acute angle inside a right-angled triangle, the three trig ratios compare pairs of sides:
Think of it like this: the hypotenuse is always the longest side (opposite the right angle) and never changes name. But "opposite" and "adjacent" depend on which angle you're looking from — always ask "opposite to what?" before labelling a triangle.
The domain/range table (know this cold)
| Function | Domain | Range |
|---|---|---|
| sin(x) | x ∈ ℝ (all real numbers) | −1 ≤ sin(x) ≤ 1 |
| cos(x) | x ∈ ℝ | −1 ≤ cos(x) ≤ 1 |
| tan(x) | x ∈ ℝ, but x ≠ ±90°, ±270°... (undefined here) | tan(x) ∈ ℝ (unbounded) |
tan θ = sin θ / cos θ. At 90°, cos θ = 0 — and you can never divide by zero. That's exactly why the graph of tan x shoots off to infinity (an asymptote) at 90°, 270°, etc.
Non-right-angled triangles: Sine Rule, Cosine Rule & Area
SOHCAHTOA only works when there's a 90° angle. For any other triangle, label it with capital letters for angles (A, B, C) and lowercase for the side opposite that angle (a, b, c) — then use:
The sine rule always hands you the acute angle first. If the real angle is obtuse, you must take 180° minus that answer. This is a classic mark-scheme trap — always sketch the triangle to sanity-check whether the angle "looks" acute or obtuse before finalising your answer.
Reciprocal trig functions
Secant, cosecant and cotangent are simply "flipped" versions of cos, sin and tan. A great memory trick: look at the third letter of each name.
cosec x = 1/sin x (3rd letter "s" → reciprocal of sin)
cot x = 1/tan x (3rd letter "t" → reciprocal of tan)
sec x = 1/cos x, but sin⁻¹x ≠ 1/sin x. The "−1" superscript for an inverse function (which undoes sin to give you back an angle) is a completely different concept from the reciprocal (which is a fraction). This mix-up costs easy marks every year.
Without a calculator, find the exact value of sec(π/3).
In triangle ABC, angle A = 40°, side b = 8 cm, side c = 10 cm. Find the area of the triangle.
2️⃣ The Unit Circle — The Key to Everything
The unit circle is a circle of radius 1, centred at the origin (0, 0). Its entire purpose is to let you calculate sin, cos and tan for any angle — not just the acute angles that fit inside a right-angled triangle. This is the single most important idea in this whole chapter, so let's build it up slowly.
Imagine a line ("radius") from the origin, sweeping anticlockwise from the positive x-axis for a positive angle θ (and clockwise for a negative angle). Wherever that line touches the circle gives you a coordinate point (x, y). Because the radius is exactly 1, right-triangle trigonometry (SOHCAHTOA) tells us something beautiful:
In other words: every point on the unit circle IS (cos θ, sin θ). That's it — that's the whole trick. Once you see this, you understand exactly why sin and cos can be negative (the y or x coordinate is negative in certain quadrants) and why they repeat every 360° (you're just going round the circle again).
The CAST diagram — where each function is positive
Split the circle into four quadrants at every 90°. In each quadrant, only certain trig functions come out positive:
(90°–180°)
(0°–90°)
(180°–270°)
(270°–360°)
Reading anticlockwise starting from the bottom-right quadrant, the positive functions spell out C-A-S-T. A popular way to remember it starting from the top-right and going anticlockwise is "All Students Take Calculus."
Your calculator only ever gives you ONE answer (the "primary value") when you press sin⁻¹, cos⁻¹ or tan⁻¹. But trig equations almost always have more than one valid solution in a given range. The CAST diagram (or a graph sketch) is how you find all the others — called "secondary values."
Given: One solution of cos θ = 0.8 is θ = 0.6435 rad. Find all other solutions in −2π ≤ θ ≤ 2π.
Step 1: Cosine is positive in quadrants A and C (first and fourth), so sketch the angle 0.6435 into both of those quadrants.
Step 2: The four possible angles (measuring from the positive x-axis, both clockwise and anticlockwise, since our range includes negative values) are: 0.6435, −0.6435, and the "wrap-around" versions using 2π − 0.6435 = 5.6397 and −2π + 0.6435 = −5.6397.
Answer: θ = −5.64, −0.644, 0.644, 5.64 (3 s.f.)
A point on the unit circle has coordinates (0.5, 0.866) to 3 s.f. Find the angle θ in degrees.
3️⃣ Graphs of Trigonometric Functions
The unit circle is the "why" — the graphs are the "how" you'll actually solve most exam questions. Sketch these confidently and you can read off every solution just by drawing a horizontal line.
| Graph | Repeats every | Passes through | Range |
|---|---|---|---|
| y = sin x | 360° (2π rad) | the origin (0,0) | −1 ≤ y ≤ 1 |
| y = cos x | 360° (2π rad) | (0, 1) | −1 ≤ y ≤ 1 |
| y = tan x | 180° (π rad) | the origin (0,0) | all real y (unbounded) |
- sin(−x) = −sin(x) → sin has rotational symmetry about the origin (it's an "odd" function)
- cos(−x) = cos(x) → cos is symmetrical about the y-axis (it's an "even" function)
- tan x has vertical asymptotes at ±90°, ±270°, ±450°... (in radians: ±π/2, ±3π/2...)
How to sketch a trig graph, step by step
- Check units: if π appears in the given domain, work in radians; otherwise use degrees.
- Label the x-axis in multiples of 90° (or π/2 rad), covering the full given domain.
- Label the y-axis: −1 to 1 for sin/cos; for tan just draw in the asymptotes first.
- Draw the curve: mark max/min points for sin/cos (or asymptotes for tan) first, then join with a smooth curve, keeping the symmetry in mind.
Using the graph to find every solution
Say you need to solve sin x = −0.25 for −180° ≤ x ≤ 270°. Sketch y = sin x across that domain, then draw the horizontal line y = −0.25. Every point where the line crosses the curve is a valid solution — the graph does the "how many solutions are there" thinking for you visually.
Solve: sin x = −0.25 for −180° ≤ x ≤ 270°.
Step 1: Calculator gives the primary value: sin⁻¹(−0.25) = −14.5°.
Step 2: Using the graph's symmetry, the second solution within one period is 180 − (−14.5) = 194.5°... but we can also go the other way: −180 + 14.5 = −165.5°.
Answer: x = −165.5°, −14.5°, 194.5° (all fall inside −180° to 270°)
By sketching a graph, find all solutions to cos x = 0.5 for 0° ≤ x ≤ 360°.
4️⃣ Transformations of Trig Graphs
Trig graphs transform exactly the same way as any other function graph — the rules you already know for translating/stretching/reflecting y = f(x) apply directly here.
| Transformation | Effect on y = sin(x) |
|---|---|
| Vertical translation up/down by a | y = sin(x) + a / y = sin(x) − a |
| Horizontal translation right/left by a | y = sin(x − a) / y = sin(x + a) |
| Vertical stretch, scale factor a | y = a sin(x) |
| Horizontal stretch, scale factor a | y = sin(x/a) |
| Reflection in the x-axis | y = −sin(x) |
Combined transformations: a sin(bx) + c
This is the form you'll see most in exam questions — it combines a stretch, a horizontal squash, and a vertical shift all in one:
The principal axis is the horizontal line the wave oscillates around. Once you sketch that line, the maximum is exactly a above it (y = c + a) and the minimum is exactly a below it (y = c − a) — this makes sketching much faster than plotting point by point.
Given y = a sin(bx) + c, always apply transformations in this order: stretches first, then translations, then reflections last. Doing them out of order gives a completely different (wrong) graph.
Sketch: y = cos 3x for 0° ≤ x ≤ 360°, then find all values where cos 3x = 0.
Step 1: Original cos x has key points (0,1), (90,0), (180,−1), (270,0), (360,1). Since we're squashing horizontally by a factor of 3, the x-coordinates get divided by 3 — so the new key points are (0,1), (30,0), (60,−1), (90,0), (120,1)... and the pattern repeats three times faster.
Answer: cos 3x = 0 at x = 30°, 90°, 150°, 210°, 270°, 330° (adding 60° each time — read straight off where the curve crosses the x-axis).
State the amplitude, period, and principal axis of y = 4 sin(2x) − 1.
5️⃣ Trigonometric Identities
An identity (shown with the symbol ≡, meaning "identical to") is a statement that's true for every value of the angle — not just some. Identities are tools: you use them to rewrite a messy equation into a form you can actually solve.
The two identities you MUST know
Rearranged forms of identity 2 are hugely useful for turning a mix of sin and cos into just one function:
Two further Pythagorean identities
Divide sin²θ + cos²θ = 1 through by cos²θ (or sin²θ) and you get two more useful identities — you don't need to derive these in an exam, just know and apply them:
If an equation mixes sin and cos → use sin²θ + cos²θ = 1. If it mixes tan and sec → use 1 + tan²θ = sec²θ. If it mixes cot and cosec → use 1 + cot²θ = cosec²θ. Matching the "pair" in the equation to the identity that connects them is 90% of the battle.
Show that: 2sin²x − cos x = 0 can be written as a cos²x + b cos x + c = 0.
Step 1: Substitute sin²x = 1 − cos²x (rearranged Pythagorean identity): 2(1 − cos²x) − cos x = 0
Step 2: Expand: 2 − 2cos²x − cos x = 0
Answer: 2cos²x + cos x − 2 = 0 (i.e. a = 2, b = 1, c = −2)
Proving new identities
To prove that one expression equals another, start on ONE side only (usually the messier side) and simplify step-by-step using known identities until you land on the other side. Never work on both sides at once — that's circular logic and won't get marks.
Prove: sec²x (cot²x − cos²x) = cot²x
Step 1: Rewrite sec and cot using definitions: (1/cos²x)(cos²x/sin²x − cos²x)
Step 2: Expand (dividing by cos²x): 1/sin²x − 1
Step 3: Rewrite 1/sin²x as cosec²x: cosec²x − 1
Step 4: Use 1 + cot²x = cosec²x, so cosec²x − 1 = cot²x. Proven! ✓
Show that (sin θ + cos θ)² ≡ 1 + 2 sin θ cos θ.
6️⃣ Solving Trigonometric Equations
This is where everything comes together. Because sin/cos repeat every 360° and tan repeats every 180°, a trig equation almost always has infinitely many solutions — which is exactly why every question gives you a specific interval to search within.
The general method
- Solve as normal using your calculator (or exact values) to get the primary value.
- Use the CAST diagram or a graph sketch to find the secondary value(s) within one period.
- Add or subtract the period (360°/2π for sin & cos; 180°/π for tan) repeatedly until you've captured every solution inside the given interval.
- Double check: throw out any solution that falls outside the given range.
Forgetting to check whether solutions actually fall inside the given interval — especially negative solutions when the interval starts at 0°. Always list every candidate solution, then go back and cross out anything outside the range. Also remember: sin x = k and cos x = k only have solutions when −1 ≤ k ≤ 1, but tan x = k works for any value of k.
Equations with a transformed angle: sin(ax + b)
When the angle itself has been transformed (e.g. solve sin(2x − 30°) = 0.5), the cleanest method is substitution:
- Let u = ax + b (the "inside" of the function).
- Transform the interval the same way: if 0° ≤ x ≤ 360°, then the new interval for u is (a·0 + b) ≤ u ≤ (a·360 + b).
- Solve for u as a normal equation (primary value + secondary values via CAST/graph), within the new transformed interval.
- Undo the substitution to convert every u-solution back into an x-solution.
Quadratic trig equations
These involve sin²θ, cos²θ or tan²θ. Use a Pythagorean identity to get everything in terms of ONE trig function, then treat it exactly like a normal quadratic — factorise or use the quadratic formula. A great trick: substitute c = cos θ (or s = sin θ) to make the factorising visually easier.
Solve: 9 sec²θ − 11 = 3 tan θ for 0 ≤ θ ≤ 2π.
Step 1: Substitute sec²θ = 1 + tan²θ: 9(1 + tan²θ) − 11 = 3 tan θ → 9tan²θ − 3tan θ − 2 = 0
Step 2: Let x = tan θ, factorise: (3x − 2)(3x + 1) = 0, so x = 2/3 or x = −1/3
Step 3: tan θ = 2/3 → primary value 0.588 rad; secondary (add π) → 3.730 rad
tan θ = −1/3 → primary value −0.322 rad (outside range, discard); secondary (add π, then 2π) → 2.820 rad, 5.961 rad
Answer: θ = 0.588, 2.820, 3.730, 5.961 (3 d.p.)
Your calculator has NO inverse button for sec, cosec or cot. Always rewrite them using their definitions (sec x = 1/cos x, etc.) to convert the equation into sin, cos or tan first — THEN solve normally.
Solve tan²x − 2tan x = 0 for 0° ≤ x ≤ 360°.
Solve 3 + 5cos(2x) = 1 for −180° < x < 180°.
🧠 What to Memorise
✅ Concepts Checklist
🎯 Exam Tips & Common Mistakes
The most common way to lose marks in this topic is simply not finding all the solutions in the interval. Always sketch (even roughly) — it makes the number of solutions visually obvious and stops you stopping too early.
If the interval uses π (radians), your calculator MUST be in radian mode. If it uses ° symbols, it must be in degree mode. Mixing these up gives wildly wrong (but "confident-looking") answers — always check the mode first, every single question.
When solving sin(ax+b) = k using substitution, it's very easy to solve for u and forget to convert back to x at the very end. Always finish by asking: "have I answered in terms of the original variable?"
Remember sin θ = k and cos θ = k only have real solutions when −1 ≤ k ≤ 1. If a quadratic trig equation produces a root outside this range (e.g. cos θ = 2), that root is simply discarded — it contributes zero solutions, not an error in your working.
Since calculators have no dedicated sec/cosec/cot buttons, always rewrite these in terms of sin, cos, tan FIRST before attempting to solve — never try to find "sec⁻¹" directly.
What examiners are looking for
- Clear, labelled working — especially showing which identity was used and why.
- Exact values (√2/2, √3/2, etc.) where the question says "without a calculator" or "exact value."
- Correct rounding as specified (check: decimal places vs significant figures — they're not the same!).
- All solutions listed, with any outside the range clearly excluded (not just silently dropped).
- For proofs: working from one side only, with each step justified by an identity — never algebra performed on both sides at once.
- 1️⃣ Trigonometric Functions & SOHCAHTOA
- 🎯 Exam Tips & Common Mistakes
- Non-right-angled triangles: Sine Rule, Cosine Rule & Area
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