MathematicsCore18 min read

Trigonometry

One circle explains every wave you will ever meet

This topic appears in:

01

Forget triangles for a moment

School teaches SOH-CAH-TOA first, and it works — for angles under 90°. But then you are asked for sin(210°) and the triangle story falls apart, because there is no triangle with a 210° angle.

The unit circle fixes this permanently. Draw a circle of radius 1. Sweep a line out from the centre at angle θ. Wherever the line meets the circle, that point's height is sin θ and its horizontal position is cos θ. That definition works for any angle at all, including negative ones and angles past 360°.

Press Play. The green sine wave on the right is literally the height of the rotating point, plotted against angle. The wave is not an analogy for the circle — it is the circle, unrolled.

02

Why sine goes negative

Between 180° and 360° the point on the circle sits below the centre. Its height is negative, so sin θ is negative. Nothing mysterious happened; the height simply went below zero.

The same reading explains why tan θ blows up at 90°. Since tan θ = sin θ / cos θ, and cos θ is the horizontal position, at 90° the point is directly above the centre — horizontal position 0 — and you are dividing by zero.

03

The identities worth memorising

IdentityWhere it comes from
sin²θ + cos²θ = 1Pythagoras on the radius-1 triangle
tan θ = sin θ / cos θDefinition
sin(−θ) = −sin θReflecting below the axis flips the height
cos(−θ) = cos θReflecting keeps the horizontal position
sin(θ + 360°) = sin θA full lap returns you to the same point
sin(2θ) = 2 sin θ cos θDouble-angle formula

The first identity is free

The point on the unit circle is at (cos θ, sin θ), and it is exactly 1 unit from the centre. Pythagoras on that right triangle gives cos²θ + sin²θ = 1² immediately. You never need to memorise it — you can rebuild it in five seconds from the picture.

04

Radians, and why they exist

Degrees are arbitrary — the number 360 comes from ancient Babylonian astronomy, not from mathematics. Radians measure angle by arc length: an angle of 1 radian sweeps out an arc exactly 1 radius long.

This matters because calculus only behaves nicely in radians. The clean result d/dx sin x = cos x is true in radians and false in degrees (you would pick up an ugly factor of π/180).

180° = π rad1 rad ≈ 57.3°full turn = 2πalways switch your calculator to RAD for calculus
05

The right-triangle ratios, which is where the paper starts

Before the unit circle, trigonometry is three ratios in a right-angled triangle. Label the sides relative to the angle you are working with: the hypotenuse is always opposite the right angle, the opposite is across from θ, and the adjacent is the remaining side touching θ.

The word "relative" is the whole difficulty. Move to the other acute angle of the same triangle and opposite and adjacent swap over, while the hypotenuse stays put.

sin θ = opposite / hypotenusecos θ = adjacent / hypotenusetan θ = opposite / adjacentcosec θ = 1/sin θsec θ = 1/cos θcot θ = 1/tan θthe three reciprocal ratios are examined by name — note that sec pairs with cos, not with sin

The two sides used by the chosen ratio are highlighted. Change θ and watch which side is called "opposite" — it is decided by the angle you are standing at, never by the orientation of the page.

06

Angles of elevation and depression

These two words turn a height-and-distance problem into a right-angled triangle, and they are the standard application in the Punjab paper.

The angle of elevation is measured upward from the horizontal to your line of sight; the angle of depression is measured downward from the horizontal. Because the two horizontals are parallel, the angle of elevation from the bottom always equals the angle of depression from the top — which is how you transfer a given angle into the triangle you actually want.

Worked example

From a point 40 m from the foot of a tower, the angle of elevation of the top is 55°. Find the height of the tower.

  1. Sketch the right-angled triangle: the tower is the opposite side, the 40 m is adjacent to the 55° angle.The sketch is what tells you which ratio to use; drawing it is worth the time.
  2. Opposite and adjacent together mean tan: tan 55° = h / 40.The hypotenuse is neither given nor wanted, which rules out sine and cosine.
  3. h = 40 tan 55° = 40 × 1.428.
  4. h = 57.1 m.Sensible: the angle exceeds 45°, so the height should exceed the 40 m base distance, and it does.

57.1 m

Add the observer's height if the question gives it

If the angle is measured from an instrument 1.5 m above the ground, your triangle gives the height above the instrument. The tower is 1.5 m taller than your answer. This final addition is a mark of its own and is the most commonly dropped mark in the whole topic.

07

Triangles that are not right-angled

SOH-CAH-TOA needs a right angle. When there is not one, two rules take over, and choosing between them is a matter of counting what you have been given.

Use the sine rule when you have a side opposite a known angle — that is, when the given information pairs a side with the angle facing it. Use the cosine rule when you have three sides, or two sides and the angle between them, because in those cases no such pair exists.

a / sin A = b / sin B = c / sin C(sine rule)a² = b² + c² − 2bc cos A(cosine rule)area = ½ ab sin Cthe cosine rule reduces to Pythagoras when A = 90°, because cos 90° = 0
a, b, c
the three sideseach named after the angle opposite it
A, B, C
the three angles
C in the area formula
the angle between the two sides usedit must be the included angle
Worked example

In triangle ABC, b = 8 cm, c = 5 cm and ∠A = 60°. Find a and the area.

  1. Two sides and the included angle, and no side-angle pair — so this is the cosine rule.Deciding which rule applies before substituting is what the first mark is for.
  2. a² = 64 + 25 − 2(8)(5) cos 60° = 89 − 80(0.5) = 49.cos 60° = 0.5 exactly, which is why the examiner chose that angle.
  3. a = 7 cm.Sensible: a lies between c and b + c, as any third side must.
  4. Area = ½(8)(5) sin 60° = 20 × 0.866 = 17.3 cm².The angle used must be the one between the two sides, which 60° is.

a = 7 cm; area ≈ 17.3 cm²

Set the calculator to degrees

A calculator left in radian mode gives answers that look plausible and are completely wrong. Check the mode indicator before the first calculation of every trigonometry question — and switch it back to radians for calculus.

Practice questions

6 questions · 25 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

3 · 7 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
State the sine rule and say when it is used in preference to the cosine rule.
Model answer

a/sin A = b/sin B = c/sin C. Use it when you have a matched pair — a side and the angle opposite it — plus one more piece of information.

Examiner tip. The phrase "opposite pair" is what identifies the rule. Without a matched pair the sine rule cannot start.

SQ2[2 marks]
Write down the exact values of sin 30°, cos 60° and tan 45°.
Model answer

sin 30° = 1/2, cos 60° = 1/2, tan 45° = 1

Examiner tip. These come up in non-calculator papers constantly. Learn the 30–60–90 and 45–45–90 triangles once and you can rebuild all six.

SQ3[3 marks]
Prove that (1 − cos²θ)/(sin θ cos θ) = tan θ.
Model answer

1 − cos²θ = sin²θ, so the expression is sin²θ/(sin θ cos θ) = sin θ/cos θ = tan θ.

Examiner tip. Start from the messier side and simplify towards the other. Working on both sides at once is not a proof.

Long questions

2 · 13 marks

Theory and numerical together, as they appear in the long-question section.

LQ1[6 marks]
In triangle ABC, AB = 7.0 cm, AC = 9.0 cm and angle BAC = 52°.
  1. Calculate BC. [3]
  2. Calculate the area of the triangle. [2]
  3. Calculate angle ABC. [1]
Mark scheme
  1. Uses the cosine rule a² = b² + c² − 2bc cos Atwo sides and the included angle[1]
  2. BC² = 49 + 81 − 2(7)(9)cos 52°[1]
  3. BC = 7.29 cm[1]
  4. Uses ½ab sin C = ½(7)(9)sin 52°[1]
  5. = 24.8 cm²[1]
  6. Sine rule: sin B = 9 sin 52° / 7.29 → B = 76.6°[1]

(a) 7.29 cm (b) 24.8 cm² (c) 76.6°

Examiner tip. Two sides and the angle between them means cosine rule. A matched opposite pair means sine rule. Deciding which before you write anything saves the whole question.

LQ2[7 marks]
A vertical mast stands on level ground. From a point P the angle of elevation of the top is 34°. From a point Q, 45 m closer to the mast and in line with P, the angle of elevation is 58°.
  1. Draw a labelled diagram of the situation. [1]
  2. Calculate the height of the mast. [5]
  3. Calculate the distance of Q from the foot of the mast. [1]
Mark scheme
  1. Diagram with the mast vertical, both angles marked at ground level and PQ = 45 m[1]
  2. Lets the height be h and the distance from Q be d[1]
  3. From Q: h = d tan 58°[1]
  4. From P: h = (d + 45) tan 34°[1]
  5. Equates: d tan 58° = (d + 45) tan 34°[1]
  6. Solves: d(1.600 − 0.6745) = 30.35 → d = 32.8 m[1]
  7. h = 32.8 × tan 58° = 52.5 m[1]

height ≈ 52.5 m, Q is ≈ 32.8 m from the foot

Examiner tip. Two unknowns need two equations. Name both distances before you touch the calculator — the algebra is easy once the naming is done.

Exam questions

1 · 5 marks

Multi-part questions with a full mark scheme.

Q1[5 marks]
Solve 2 sin θ = 1 for 0° ≤ θ ≤ 360°.
  1. Find the principal value. [2]
  2. Find all solutions in the given range and justify how you found the second. [3]
Mark scheme
  1. sin θ = 1/2[1]
  2. θ = 30°[1]
  3. Sine is also positive in the second quadrant[1]
  4. θ = 180° − 30° = 150°[1]
  5. No further solutions in the range, so θ = 30° and 150°[1]

θ = 30° and 150°

Examiner tip. A calculator gives you one angle. The range in the question tells you how many there should be — sketch the sine curve across that range and count the crossings.