Pearson Edexcel International Advanced Level Mathematics: Triangle trig, radians and trig graphs (YMA01)

This guide follows the Pearson Edexcel International Advanced Subsidiary/Advanced Level Mathematics, Further Mathematics and Pure Mathematics specification (Issue 3, April 2019). It covers Unit P1 topics 3.1, 3.2 and 3.3: the sine and cosine rules, radian measure, and the graphs of sine, cosine and tangent. The specification lists P1 as a compulsory unit for both IAS Mathematics and IAL Mathematics (and Pure Mathematics).

Use it with the revision notes and the practice set. For the rest of the unit, see Pure Mathematics 1 and the P1 revision notes. Course hub: Edexcel A Level Mathematics. Printable checklist. Not sure where you stand? Try a free 10-minute diagnostic.

What this unit covers

Spec	What you must be able to do
3.1	Use the sine and cosine rules, and area = ½ab sin C, including the ambiguous case of the sine rule
3.2	Use radian measure, with s = rθ for arc length and A = ½r²θ for sector area
3.3	Know the graphs, symmetries and periodicity of sine, cosine and tangent, including curves such as y = 3 sin x, y = sin(x + π/6) and y = sin 2x

Solving trigonometric equations, identities and the transformation of general graphs are separate specification topics. This guide only uses graph transformations where they apply to the three trig curves.

3.1 Sine rule, cosine rule and area

Label a triangle ABC with side a opposite angle A, and so on.

Sine rule:    a / sin A = b / sin B = c / sin C
              (use the upside-down form sin A / a = ... when finding an angle)
Cosine rule:  a² = b² + c² − 2bc cos A
              cos A = (b² + c² − a²) / (2bc)
Area:         ½ ab sin C   (the angle must be between the two sides used)

Which rule?

You know	Use
Two sides and the angle between them (SAS)	Cosine rule for the third side
Three sides (SSS)	Cosine rule for an angle
Two angles and a side (AAS or ASA)	Sine rule (the third angle is 180° minus the other two)
Two sides and an angle not between them (SSA)	Sine rule, and watch for two triangles

Worked example 1: SAS then an angle

In triangle ABC, b = 11.6, c = 8.4 and A = 52°. Find a, then angle C.

a² = 11.6² + 8.4² − 2(11.6)(8.4)cos 52°
   = 134.56 + 70.56 − 194.88 × 0.61566
   = 85.14
a = 9.23   (3 s.f.)

sin C / 8.4 = sin 52° / 9.227
sin C = 8.4 × 0.78801 / 9.227 = 0.7174
C = 45.8°

C is opposite the shortest side, so it must be acute and the sine rule gives one answer. If you tried angle B (opposite the longest side) with the sine rule, it could be 82.2° or 97.8°, and you could not choose. B = 180° − 52° − 45.8° = 82.2°. Finding the angle opposite the shortest side first avoids the doubt.

Worked example 2: SSS and an obtuse angle

A triangle has sides 7, 9 and 12. Find its largest angle.

Put the longest side, 12, opposite the angle you want, since the biggest angle faces it.

cos θ = (7² + 9² − 12²) / (2 × 7 × 9)
      = (49 + 81 − 144) / 126
      = −14/126 = −1/9
θ = 96.4°

A negative cosine means an obtuse angle. The cosine rule never gives a false second answer, which is why it is the better tool for a possibly obtuse angle.

Worked example 3: two angles and a side

In triangle ABC, A = 59°, C = 63° and a = 22. Find c.

B = 180° − 59° − 63° = 58°
c / sin 63° = 22 / sin 59°
c = 22 × 0.89101 / 0.85717 = 22.9   (3 s.f.)

The ambiguous case

You are given two sides and a non-included angle: say a, b and A. Since sin B = sin(180° − B), the equation sin B = k can give two angles, B and 180° − B. Both are valid if the second leaves room for angle A in the triangle: A + (180° − B) < 180°, which simplifies to A < B.

Worked example 4: two triangles

In triangle ABC, A = 34°, a = 9 and b = 13. Find all possible values of c.

sin B = 13 sin 34° / 9 = 0.80772
B = 53.9°   or   B = 180° − 53.9° = 126.1°

Check each against A = 34°: 34° + 126.1° = 160.1°, which is below 180°, so both triangles exist.

Case 1: B = 53.9°  → C = 180 − 34 − 53.9 = 92.1°
        c = 9 sin 92.1° / sin 34° = 16.1
Case 2: B = 126.1° → C = 180 − 34 − 126.1 = 19.9°
        c = 9 sin 19.9° / sin 34° = 5.47

Answer: c = 16.1 or c = 5.47.

Two triangles are only possible when the side opposite the given angle is shorter than the other given side but longer than b sin A. Here 13 sin 34° = 7.27 < 9 < 13. If a ≥ b, the second angle is too big and only one triangle exists.

Worked example 5: area and an unknown angle

A triangle has two sides of length 10 and 9, and area 40. Find the possible sizes of the angle between them, and the third side in each case.

½ × 10 × 9 × sin C = 40
sin C = 80/90 = 0.8889
C = 62.7°   or   C = 117.3°

Both values give a triangle, because the sine of an acute angle and of its supplement are equal. The cosine rule gives the third side:

C = 62.7°:  c² = 100 + 81 − 180 cos 62.7° = 98.5  → c = 9.93
C = 117.3°: c² = 100 + 81 − 180 cos 117.3° = 263.5 → c = 16.2

Examiner insight: when a question says “find the angle” from an area, a single answer can lose the second solution. Mention that sin C = k gives C and 180° − C, then test each against the context.

3.2 Radian measure

One radian is the angle at the centre of a circle that cuts off an arc equal in length to the radius. A full turn is 2π radians, so

180° = π radians
degrees → radians: multiply by π/180
radians → degrees: multiply by 180/π

Common angles you should know without working:

Degrees	30°	45°	60°	90°	120°	135°	150°	180°	270°	360°
Radians	π/6	π/4	π/3	π/2	2π/3	3π/4	5π/6	π	3π/2	2π

Leave answers in terms of π unless the question asks for a decimal. For example 72° = 72π/180 = 2π/5 ≈ 1.26 rad, and 2 rad = 2 × 180/π ≈ 114.6°.

Arc length and sector area

With θ in radians and radius r:

arc length   s = rθ
sector area  A = ½ r² θ

Both formulae are only valid when θ is in radians. In degrees the equivalents carry a factor of θ/360, so work in radians throughout.

Worked example 6: a sector

A sector has radius 9 cm and angle 1.2 radians. Find the arc length, area and perimeter.

s = 9 × 1.2 = 10.8 cm
A = ½ × 81 × 1.2 = 48.6 cm²
perimeter = 2 × 9 + 10.8 = 28.8 cm

The perimeter of a sector includes both radii. Leaving them out is a common slip.

Worked example 7: working backwards

A sector has arc length 15 cm and area 75 cm². Find r and θ.

s = rθ = 15       A = ½ r² θ = ½ r (rθ) = ½ r × 15 = 75
r = 150/15 = 10 cm
θ = 15/10 = 1.5 radians

The substitution A = ½ r s is quick whenever arc length and area are both known.

Worked example 8: a segment

A chord AB subtends an angle of 2 radians at the centre O of a circle of radius 6 cm. Find the area of the minor segment and its perimeter.

A segment is the sector OAB minus triangle OAB.

sector      = ½ × 36 × 2         = 36
triangle    = ½ × 6 × 6 × sin 2  = 18 × 0.90930 = 16.37
segment     = 36 − 16.37         = 19.6 cm²   (3 s.f.)

arc AB      = 6 × 2 = 12
chord AB    : isosceles triangle, apex 2 rad
            = 2 × 6 × sin(1) = 10.10
perimeter   = 12 + 10.10 = 22.1 cm

Evaluate sin 2 in radians: sin 2 = 0.9093. A value of 0.0349 means you have used 2 degrees.

3.3 Graphs of sine, cosine and tangent

The tables below use radians.

Function	Period	Range	Key features on 0 to 2π
y = sin x	2π	−1 to 1	zeros at 0, π, 2π; max at π/2; min at 3π/2
y = cos x	2π	−1 to 1	max at 0 and 2π; zeros at π/2, 3π/2; min at π
y = tan x	π	all real	zeros at 0, π, 2π; asymptotes at x = π/2 and 3π/2

Symmetries and periodicity

sin(x + 2π) = sin x     cos(x + 2π) = cos x     tan(x + π) = tan x
sin(−x) = −sin x        (odd: rotational symmetry about the origin)
cos(−x) = cos x         (even: reflection in the y-axis)
tan(−x) = −tan x
sin(π − x) = sin x      cos(π − x) = −cos x     tan(π − x) = −tan x
sin(x + π) = −sin x     cos(x + π) = −cos x
sin(π/2 − x) = cos x

You can read each one off the graph, so sketch a quick curve rather than memorise them.

Stretches, shifts and changes of period

Equation	Effect on y = sin x	Period
y = 3 sin x	stretch parallel to the y-axis, scale factor 3; range −3 to 3	2π
y = sin(x + π/6)	translation of π/6 to the left	2π
y = sin 2x	stretch parallel to the x-axis, scale factor ½	π
y = cos(x/2)	stretch parallel to the x-axis, scale factor 2	4π

For y = sin(ax), the period is 2π/a. For y = tan(ax), it is π/a.

Worked example 9: sketching y = sin(x + π/6)

Sketch y = sin(x + π/6) for 0 ≤ x ≤ 2π, labelling intercepts and turning points.

y-intercept: x = 0 → sin(π/6) = 1/2        point (0, 1/2)
maximum:     x + π/6 = π/2 → x = π/3       point (π/3, 1)
zeros:       x + π/6 = π   → x = 5π/6
             x + π/6 = 2π  → x = 11π/6
minimum:     x + π/6 = 3π/2 → x = 4π/3    point (4π/3, −1)
right end:   x = 2π → sin(13π/6) = 1/2     point (2π, 1/2)

The curve starts at (0, 1/2), rises to (π/3, 1), falls through (5π/6, 0) to (4π/3, −1), then rises through (11π/6, 0) to (2π, 1/2). It is the sine curve moved π/6 to the left, so the curve begins part-way up the first rise.

Worked example 10: y = sin 2x

Sketch y = sin 2x for 0 ≤ x ≤ 2π. The period is π, so there are two full waves.

zeros:   x = 0, π/2, π, 3π/2, 2π
maxima:  (π/4, 1) and (5π/4, 1)
minima:  (3π/4, −1) and (7π/4, −1)

Reading the tangent graph

y = tan x climbs from −∞ to +∞ between each pair of asymptotes, x = π/2 + nπ, passing through zero midway. It has no maximum or minimum. In y = tan 2x the asymptotes move to x = π/4 + nπ/2.

Common errors

• Using ½ab sin C with an angle that is not between the two sides.
• Using the sine rule for an angle opposite the longest side and ignoring the obtuse alternative.
• Forgetting the second ambiguous-case triangle, or keeping one that makes the angle sum exceed 180°.
• Applying s = rθ or A = ½r²θ with θ in degrees.
• Leaving the two radii out of a sector perimeter.
• Stretching y = sin 2x vertically; the 2 changes the period, not the height.
• Sketching y = sin(x + π/6) shifted to the right.
• Rounding intermediate values early. Keep full figures until the last line.

Next steps

Work through the revision notes, then test yourself with the practice questions. Tick off topics on the checklist and take a diagnostic to find gaps.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics, Specification, Issue 3, April 2019 (Pearson Education Limited), Unit P1 topics 3.1 to 3.3.