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IGCSE Mathematics 0580 · Chapter 9 notes · ~34 min read

Pythagoras & trigonometry

Right-angled triangles first (Pythagoras, SOH CAH TOA, bearings), then the sine rule, cosine rule, 3D problems and trig graphs. Every method is shown on a labelled diagram, so you can see which rule to pick.

PythagorasSOHCAHTOABearingsAngles of elevationSine & cosine rulesArea ½ab sin C3DTrig graphs
RSby Riyaz Sir
mentorriyaz.in
No tag = Core (everyone)EXTENDED Extended onlyhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • Pythagoras: in a right-angled triangle c² = a² + b². Add the squares to find the hypotenuse and subtract to find a shorter side.
  • SOH CAH TOA links an angle to two sides. Label the sides (opposite, adjacent, hypotenuse), pick the ratio, then solve. Use sin⁻¹, cos⁻¹, tan⁻¹ to find angles.
  • Not right-angled? Use the sine rule for a side and its opposite angle (a pair), and the cosine rule for two sides and the angle between them, or three sides. EXT

By the end you can

  • Use Pythagoras' theorem to find a missing side and to test for a right angle.
  • Use sine, cosine and tangent to find sides and angles in right-angled triangles.
  • Solve problems with three-figure bearings, and angles of elevation and depression.
  • Use the sine rule, the cosine rule and area = ½ab sin C in any triangle. EXT
  • Solve 3D problems by finding right-angled triangles inside solids. EXT
  • Know the exact values of sin, cos and tan for 0°, 30°, 45°, 60° and 90°. EXT
  • Sketch the graphs of sin x, cos x and tan x, and solve trigonometric equations. EXT

1Pythagoras' theorem

The theorem

16925435Pythagorasc² = a² + b²25 = 9 + 165² = 3² + 4²the square on thehypotenuse equals thesum of the squares onthe other two sides
Build a square on each side. The two small squares (9 + 16) have exactly the same area as the big one (25).

Hypotenuse — the longest side of a right-angled triangle. It is always opposite the right angle.

Pythagoras' theorem — in a right-angled triangle, c² = a² + b², where c is the hypotenuse. It works only for right-angled triangles.

  • Pythagorean triples are whole numbers that fit: 3, 4, 5 and 5, 12, 13 and 8, 15, 17 and 7, 24, 25. Multiples work too, so 6, 8, 10 and 9, 12, 15 are right-angled.
  • Test for a right angle: if the longest side squared equals the other two squared and added, the triangle is right-angled. This is called the converse.
Worked example

Do the sides 8 cm, 15 cm and 17 cm make a right-angled triangle?

  1. longest side = 17, so test 17² = 289
  2. 8² + 15² = 64 + 225 = 289
  3. 289 = 289, so yes, it is right-angled
Worked example

Do the sides 6 cm, 7 cm and 10 cm make a right-angled triangle?

  1. 6² + 7² = 36 + 49 = 85
  2. 10² = 100
  3. 85 ≠ 100, so no

Finding a missing side

Longest side: ADDShorter side: SUBTRACT96cc² = 6² + 9²= 36 + 81 = 117c = √117 = 10.85a13a² = 13² − 5²= 169 − 25 = 144a = √144 = 12Always check: the hypotenuse is the longest side.
Look for the hypotenuse first. If you are finding it, add the squares. If you are finding a shorter side, subtract.
  • Step 1: sketch the triangle and mark the right angle. The side opposite it is the hypotenuse c.
  • Step 2: write c² = a² + b² and put the numbers in.
  • Step 3: if you want a shorter side, rearrange to a² = c² − b².
  • Step 4: take the square root at the end. Keep full calculator accuracy until the final answer.
Worked example

Find the hypotenuse of a right-angled triangle with shorter sides 6 cm and 9 cm.

  1. c² = 6² + 9²
  2. = 36 + 81 = 117
  3. c = √117 = 10.8 cm (3 s.f.)
Worked example

A right-angled triangle has hypotenuse 13 cm and one shorter side 5 cm. Find the other side.

  1. a² = 13² − 5²
  2. = 169 − 25 = 144
  3. a = √144 = 12 cm
Worked example

Find the hypotenuse when the shorter sides are 7.5 cm and 4 cm.

  1. c² = 7.5² + 4² = 56.25 + 16 = 72.25
  2. c = √72.25 = 8.5 cm
Common mistakeAdding when you should subtract, or the other way round. Quick check: the hypotenuse must be the longest side. If your answer for a shorter side is bigger than the hypotenuse, you added.
Common mistakeForgetting the final square root and giving c² as the answer.

Pythagoras in real problems

LadderRectangleIsosceles triangle1.5 m5 mhh² = 5² − 1.5²h = 4.77 m12915diagonal= √(12² + 9²) = 15h101066h² = 10² − 6², h = 8area = ½ × 12 × 8 = 48
Most word problems hide a right-angled triangle: a ladder against a wall, the diagonal of a rectangle, or half of an isosceles triangle.
  • Ladders and walls: the wall, the ground and the ladder make a right-angled triangle. The ladder is the hypotenuse.
  • Diagonal of a rectangle: it splits the rectangle into two right-angled triangles. For a 12 cm by 9 cm rectangle the diagonal is √(12² + 9²) = 15 cm.
  • Isosceles triangle: drop the perpendicular height from the top. It splits the base in half and makes two right-angled triangles.
  • Distance between two points: the horizontal and vertical differences are the two shorter sides, so d = √((x₂ − x₁)² + (y₂ − y₁)²). This is the length formula from coordinate geometry.
Worked example

A 5 m ladder leans against a wall with its foot 1.5 m from the wall. How high up the wall does it reach?

  1. the ladder is the hypotenuse, so h² = 5² − 1.5²
  2. = 25 − 2.25 = 22.75
  3. h = √22.75 = 4.77 m (3 s.f.)
Worked example

An isosceles triangle has equal sides of 10 cm and a base of 12 cm. Find its height and its area.

  1. half the base = 6 cm
  2. h² = 10² − 6² = 64, so h = 8 cm
  3. area = ½ × 12 × 8 = 48 cm²
Worked example

Find the distance between the points (1, 2) and (7, 10).

  1. horizontal difference = 7 − 1 = 6
  2. vertical difference = 10 − 2 = 8
  3. d = √(6² + 8²) = √100 = 10
Exam tipIn "show that" questions give the exact form first (for example √117) and then the rounded value. Never round in the middle of a calculation.

2Trigonometry in right-angled triangles

Naming the sides

θ at the bottom rightθ at the topθoppositeadjacenthypotenuseθadjacentoppositehypotenuseThe opposite is across from θ. The adjacent is next to θ.The hypotenuse is opposite the right angle (the longest side).
The hypotenuse never changes. The opposite and adjacent sides swap over when you move to the other angle.

Opposite — the side across from the angle θ. It does not touch θ.

Adjacent — the side next to θ that is not the hypotenuse.

  • Label in this order: hypotenuse first (opposite the right angle), then the opposite (across from θ), and the adjacent is what is left.
  • The angle decides the labels. The same side can be "opposite" for one angle and "adjacent" for the other.

SOH CAH TOA

OSHSOHsin θ = O ÷ HO = H × sin θH = O ÷ sin θACHCAHcos θ = A ÷ HA = H × cos θH = A ÷ cos θOTATOAtan θ = O ÷ AO = A × tan θA = O ÷ tan θCover the one you want: side by side means × and one above the other means ÷.O = opposite, A = adjacent, H = hypotenuse, S = sin, C = cos, T = tan
Cover up the quantity you want. If the other two are side by side, multiply. If one is above the other, divide.
sin θ = oppositehypotenuse
SOH
cos θ = adjacenthypotenuse
CAH
tan θ = oppositeadjacent
TOA
  • Pick the ratio by looking at the two sides involved: the one you have and the one you want. If they are O and H use sin, A and H use cos, O and A use tan.
  • The ratios depend only on the angle, not on the size of the triangle. A bigger triangle with the same angles has the same sin, cos and tan.
  • Sine and cosine of an angle are never more than 1, because the hypotenuse is the longest side.
Memory trickSome Old Hens Can't Always Hide Their Old Age. The first letters spell SOH CAH TOA.

Finding sides and angles

Find a sideFind an angle35°x12sin 35° = x ÷ 12x = 12 × sin 35°x = 6.88θ513sin θ = 5 ÷ 13θ = sin⁻¹(5 ÷ 13)θ = 22.6°Calculator in DEG mode. Use sin⁻¹, cos⁻¹, tan⁻¹ to find an angle.
To find a side, multiply (or divide) by the ratio. To find an angle, use the inverse button on your calculator.
  • Finding a side: write the ratio with numbers, then rearrange. If the unknown is on top of the fraction, multiply. If it is on the bottom, divide.
  • Finding an angle: work out the ratio as a number, then use sin⁻¹, cos⁻¹ or tan⁻¹. For example, if sin θ = 5/13, then θ = sin⁻¹(5 ÷ 13).
  • Check the mode. Your calculator must be in degrees (DEG). Radians give wrong answers.
Worked example

Find x when the hypotenuse is 12 cm and the angle opposite x is 35°.

  1. x is opposite and 12 is the hypotenuse, so use sin
  2. sin 35° = x ÷ 12
  3. x = 12 × sin 35° = 6.88 cm
Worked example

Find x when the opposite side is 9 cm and the angle is 52°, with x being the adjacent side.

  1. O and A, so use tan
  2. tan 52° = 9 ÷ x
  3. x = 9 ÷ tan 52° = 7.03 cm (x is on the bottom, so divide)
Worked example

Find the hypotenuse h when the adjacent side is 8 cm and the angle is 40°.

  1. A and H, so use cos
  2. cos 40° = 8 ÷ h
  3. h = 8 ÷ cos 40° = 10.4 cm
Worked example

Find angle θ when the opposite side is 5 cm and the hypotenuse is 13 cm.

  1. sin θ = 5 ÷ 13
  2. θ = sin⁻¹(5 ÷ 13) = 22.6°
Worked example

Find angle θ when the opposite side is 7 cm and the adjacent side is 4 cm.

  1. tan θ = 7 ÷ 4 = 1.75
  2. θ = tan⁻¹(1.75) = 60.3°
Worked example

An isosceles triangle has equal sides of 10 cm and base angles of 70°. Find the base and the area.

  1. drop the height to split the base in two. Half the base = 10 × cos 70° = 3.420
  2. base = 2 × 3.420 = 6.84 cm
  3. height = 10 × sin 70° = 9.397
  4. area = ½ × 6.840 × 9.397 = 32.1 cm²
Common mistakeCalculator in radian mode. Check that it shows D or DEG at the top.
Common mistakeRounding in the middle. Round only at the end, to 3 significant figures (or 1 decimal place for angles).
Common mistakeMixing up opposite and adjacent. Always label first, from the angle you are using.
Exam tipPythagoras or trig? If you have two sides and want the third, use Pythagoras. If an angle is involved (given or wanted), use SOH CAH TOA.

3Bearings, elevation and depression

Three-figure bearings

NESW000°090°180°270°Three figures, measuredclockwise from NorthNNAB065°245°bearing of Bfrom A = 065°bearing of Afrom B = 245°back bearing= ± 180°
Start at North, turn clockwise to the direction of travel, and write the angle with three figures.

Bearing — the direction of one point from another, measured clockwise from North and written with three figures. East is 090°, South is 180° and West is 270°.

Back bearing — the bearing for the return journey. Add or subtract 180°.

  • Draw a North line at the point you are going FROM. The word "of B from A" means start at A.
  • Three figures always: write 065°, never 65°.
  • Back bearing: if the bearing is less than 180°, add 180°. If it is more than 180°, subtract 180°.
  • Using trigonometry: draw the triangle with the North line as one side. The angle from North is usually the angle in the triangle.
Worked example

B is 6 km east and 8 km north of A. Find the distance AB and the bearing of B from A.

  1. AB = √(6² + 8²) = 10 km
  2. angle from North: tan θ = 6 ÷ 8 = 0.75
  3. θ = tan⁻¹(0.75) = 36.87°
  4. bearing = 037°
Worked example

The bearing of B from A is 065°. Find the bearing of A from B.

  1. 065 + 180 = 245°
Worked example

A ship sails 15 km on a bearing of 120°. How far east and how far south of its start is it?

  1. 120° is 60° past East (090°), so the angle from South is 60°
  2. east = 15 × sin 60° = 13.0 km
  3. south = 15 × cos 60° = 7.5 km
Common mistakeMeasuring from the wrong North line. The bearing of A from B uses the North line at B.
Common mistakeMeasuring anticlockwise, or giving only two figures. Bearings are always clockwise and three digits.

Angles of elevation and depression

Angle of elevationAngle of depression34°looking UP from the horizontalthe line of sight is above it36°36°looking DOWN from the horizontalangle of depression = angle of elevation(alternate angles)
Both angles are measured from the horizontal. Elevation looks up and depression looks down. They are equal because they are alternate angles.

Angle of elevation — the angle up from the horizontal to a line of sight.

Angle of depression — the angle down from the horizontal to a line of sight.

  • Always draw a right-angled triangle with a horizontal line and a vertical line.
  • The angle of depression is not inside the triangle at the top. The matching angle inside the triangle is at the bottom and is equal to it (alternate angles).
Worked example

From a point 40 m from the foot of a tower, the angle of elevation of the top is 32°. Find the height of the tower.

  1. O and A, so tan 32° = h ÷ 40
  2. h = 40 × tan 32° = 25.0 m
Worked example

From the top of a 50 m cliff the angle of depression of a boat is 18°. How far is the boat from the foot of the cliff?

  1. the angle inside the triangle at the boat is also 18°
  2. tan 18° = 50 ÷ d
  3. d = 50 ÷ tan 18° = 154 m
Common mistakeMeasuring the angle of depression from the vertical. It is always from the horizontal.

4Sine and cosine rules EXT

Which rule?EXTENDED

Right-angled triangle?YESNOUse thesePythagorastwo sides known,find the thirdSOH CAH TOAone side + one angleWhat are you given?a side and the angleopposite it (a pair)SINE RULEside or angletwo sides and theangle between (SAS)COSINE RULEfind a sideall three sides(SSS)COSINE RULEfind an angleAlways sketch and label the triangle first.
Ask whether the triangle has a right angle. If not, look at what you are given: a pair, two sides and the angle between, or three sides.
  • Labelling: the angles are A, B, C and the sides opposite them are a, b, c. So side a is opposite angle A.
  • A pair means a side and its opposite angle, such as a and A. If you have a pair, use the sine rule.
  • SAS means two sides and the angle between them. Use the cosine rule to find the third side.
  • SSS means all three sides and no angle. Use the cosine rule to find an angle.
  • Two angles given? Find the third using angles add up to 180°, then you have a pair.

The sine ruleEXTENDED

ABCabcEach side is matched with theangle opposite it (same colour)The sine ruleasin A=bsin B=csin CUse it when you know a PAIR:a side and the angle opposite it.Only two of the fractionsare used in any one question.Side? Unknown on top:x ÷ sin 40° = 8 ÷ sin 65° → x = 5.67Angle? sin on top:sin θ ÷ 6 = sin 50° ÷ 9 → θ = 30.7°
Each side is divided by the sine of the angle opposite it. All three fractions are equal.
asin A = bsin B = csin C
use to find a side
sin Aa = sin Bb = sin Cc
use to find an angle
  • Use only two of the three fractions in any question. Choose the two that contain the pair you know and the thing you want.
  • To find a side: put the unknown on top. To find an angle: put sin on top.
  • Obtuse angles: sin⁻¹ gives an acute angle. If the diagram shows an obtuse angle, the answer is 180° minus your calculator value.
Worked example

Find x if x ÷ sin 40° = 8 ÷ sin 65°.

  1. x = 8 × sin 40° ÷ sin 65°
  2. = 8 × 0.6428 ÷ 0.9063 = 5.67 cm
Worked example

In triangle ABC, angle A = 48°, angle B = 71° and a = 12 cm. Find b.

  1. pair is a and A, and b is opposite B
  2. b ÷ sin 71° = 12 ÷ sin 48°
  3. b = 12 × sin 71° ÷ sin 48° = 15.3 cm
Worked example

In the same triangle find c.

  1. angle C = 180° − 48° − 71° = 61°
  2. c = 12 × sin 61° ÷ sin 48° = 14.1 cm
Worked example

Find angle θ if sin θ ÷ 6 = sin 50° ÷ 9, and θ is acute.

  1. sin θ = 6 × sin 50° ÷ 9 = 0.5107
  2. θ = sin⁻¹(0.5107) = 30.7°
Common mistakeUsing a side and an angle that are not opposite each other. The side and angle in a pair must be across from each other.

The cosine ruleEXTENDED

c = 9b = 7a = ?52°ABCThe cosine rulea² = b² + c² − 2bc cos Afind a SIDE (two sides and the angle between)cos A =b² + c² − a²2bcfind an ANGLE (all three sides)The side on the left (a) and theangle inside the cos (A) arealways opposite each other.Here: a² = 7² + 9² − 2 × 7 × 9 × cos 52° = 52.4so a = 7.24
Use it when the angle is between the two known sides (SAS), or when you know all three sides (SSS).
a² = b² + c² − 2bc cos A
to find a side
cos A = b² + c² − a²2bc
to find an angle
  • The side on the left (a) and the angle in the cos (A) are opposite each other. The other two sides, b and c, are the ones around angle A.
  • Order of working: do 2 × b × c × cos A first, then subtract from b² + c². Do not subtract before multiplying by cos A.
  • A negative cosine means the angle is obtuse (bigger than 90°). The calculator gives it correctly.
Worked example

Find a when b = 7 cm, c = 9 cm and A = 52°.

  1. a² = 7² + 9² − 2 × 7 × 9 × cos 52°
  2. = 49 + 81 − 126 × 0.6157
  3. = 130 − 77.57 = 52.43
  4. a = 7.24 cm
Worked example

Triangle with sides 5 cm, 7 cm and 10 cm. Find the largest angle.

  1. largest angle is opposite 10, so a = 10, b = 5, c = 7
  2. cos A = (5² + 7² − 10²) ÷ (2 × 5 × 7) = −26 ÷ 70 = −0.3714
  3. A = cos⁻¹(−0.3714) = 111.8° (obtuse, as the cosine is negative)
Worked example

Triangle with sides 6 cm, 7 cm and 8 cm. Find the largest angle.

  1. the angle is opposite 8
  2. cos A = (6² + 7² − 8²) ÷ (2 × 6 × 7) = 21 ÷ 84 = 0.25
  3. A = cos⁻¹(0.25) = 75.5°
Common mistakeWriting (b² + c² − 2bc) cos A. The cos only multiplies the 2bc part.
Exam tipLargest angle first is a good check. It is opposite the longest side. If it is obtuse the cosine is negative.

Area of a triangle with sineEXTENDED

a = 11b = 8h40°CBAArea = ½ ab sin Cbase a, height h = b sin C,so area = ½ × a × b sin CThe angle C must be BETWEENthe two sides a and b.a = 11, b = 8, C = 40°Area = ½ × 11 × 8 × sin 40°= 44 × 0.6428= 28.3 cm²sin C turns a slanted side into a height
The height is b sin C, so the area is half the base times that height. The angle must be between the two sides.
Area = ½ ab sin C
a and b are two sides, C is the angle between them
  • Use this when you do not have the perpendicular height. The sine rule gives the height for you.
  • Works for obtuse angles too. For C = 150°, sin 150° = 0.5.
  • Finding an angle: rearrange to sin C = 2 × area ÷ (ab). Use the diagram to decide if C is acute or obtuse.
Worked example

Find the area of a triangle with sides 11 cm and 8 cm and an angle of 40° between them.

  1. area = ½ × 11 × 8 × sin 40°
  2. = 44 × 0.6428 = 28.3 cm²
Worked example

Triangle ABC has AB = 9 cm, AC = 11 cm and angle A = 64°. Find its area and the length BC.

  1. area = ½ × 9 × 11 × sin 64° = 49.5 × 0.8988 = 44.5 cm²
  2. BC² = 9² + 11² − 2 × 9 × 11 × cos 64° = 202 − 198 × 0.4384 = 115.2
  3. BC = 10.7 cm
Worked example

A triangle has area 30 cm², sides 10 cm and 8 cm and an acute angle C between them. Find C.

  1. ½ × 10 × 8 × sin C = 30
  2. 40 sin C = 30, so sin C = 0.75
  3. C = sin⁻¹(0.75) = 48.6°

5Trigonometry in 3D EXT

Cuboids and the space diagonalEXTENDED

865ABCDEFGHspace diagonal EC (red)Step 1: base diagonal ACAC² = 8² + 6² = 100, AC = 10Step 2: space diagonal ECEC² = 10² + 5² = 125, EC = 11.2Step 3: angle between EC and the basetan θ = 5 ÷ 10, so θ = 26.6°AC = 10AE = 5EC26.6°
Find a right-angled triangle inside the solid. Base diagonal first, then the diagonal through the solid.

Angle between a line and a plane — the angle between the line and its shadow (projection) on the plane. For a diagonal of a cuboid, the shadow on the base is the base diagonal.

  • Method: (1) pick the right-angled triangle that contains your line, (2) redraw it flat with its true lengths, (3) use Pythagoras or SOH CAH TOA.
  • Base diagonal: √(length² + width²). Space diagonal: √(length² + width² + height²).
  • The angle between the diagonal and the base is found using the height (opposite) and the base diagonal (adjacent).
Worked example

A cuboid is 8 cm by 6 cm by 5 cm. Find the space diagonal and the angle it makes with the base.

  1. base diagonal = √(8² + 6²) = √100 = 10 cm
  2. space diagonal = √(10² + 5²) = √125 = 11.2 cm
  3. tan θ = 5 ÷ 10
  4. θ = 26.6°
Worked example

A cuboid has AB = 9 cm, BC = 12 cm and CG = 8 cm. Find AG.

  1. AC = √(9² + 12²) = 15 cm
  2. AG = √(15² + 8²) = √289 = 17 cm

PyramidsEXTENDED

68edge VAABCDVMsquare base 8 × 8, height 6Angle of an EDGE with the base65.668.2546.7°tan θ = 6 ÷ 5.66θ = 46.7°Angle of a FACE with the base647.2156.3°tan θ = 6 ÷ 4θ = 56.3°
The apex sits directly above the centre of the base. Two different right-angled triangles give the angle of an edge and the angle of a face.
  • Angle of an edge with the base: use the triangle with the height, the half-diagonal of the base and the edge.
  • Angle of a face with the base: use the triangle with the height, half a side of the base and the slant height of the face.
  • For a square base of side s: the half-diagonal is s√2 ÷ 2, and half a side is s ÷ 2.
Worked example

A square-based pyramid has base 8 cm by 8 cm and height 6 cm. Find the length of a sloping edge and the angle it makes with the base.

  1. half-diagonal = ½ × 8√2 = 4√2 = 5.657 cm
  2. edge² = 5.657² + 6² = 32 + 36 = 68
  3. edge = 8.25 cm
  4. tan θ = 6 ÷ 5.657, so θ = 46.7°
Worked example

For the same pyramid find the angle between a triangular face and the base.

  1. half a side = 4 cm, height = 6 cm
  2. tan θ = 6 ÷ 4 = 1.5
  3. θ = 56.3°
Common mistakeUsing the edge where you need the half-diagonal, or the side where you need the half-diagonal of the base. Draw the flat triangle every time.

6Trigonometric graphs and equations EXT

Exact valuesEXTENDED

1√3260°30°half an equilateraltriangle of side 211√245°45°a square cutalong its diagonalLearn the twotriangles and youcan rebuild thewhole tableθ0°30°45°60°90°sin θ01/2√2/2√3/21cos θ1√3/2√2/21/20tan θ0√3/31√3—tan 90° is undefined. √3/3 is the same as 1/√3.
Two special triangles give every exact value you need. Half an equilateral triangle gives 30° and 60°. A square cut along its diagonal gives 45°.
  • Pattern for sin: the values for 0°, 30°, 45°, 60°, 90° are √0 ÷ 2, √1 ÷ 2, √2 ÷ 2, √3 ÷ 2 and √4 ÷ 2. The cosines are the same list in reverse.
  • For tan: divide sin by cos. So tan 30° = (1/2) ÷ (√3/2) = 1/√3 and tan 60° = √3.
  • Exact means no decimals. Leave surds and fractions in your answer when the question says "exact".
Worked example

Find the exact value of sin 60° × cos 30°.

  1. sin 60° = √3/2 and cos 30° = √3/2
  2. product = (√3 × √3) ÷ 4 = 3/4
Worked example

An equilateral triangle has sides 6 cm. Find its exact height.

  1. the height splits it into a 30°-60°-90° triangle with hypotenuse 6
  2. h = 6 × sin 60° = 6 × √3/2 = 3√3 cm
Worked example

Find the exact length x if the hypotenuse is 10 cm and the angle opposite x is 30°.

  1. x = 10 × sin 30° = 10 × 1/2 = 5 cm

Graphs of sin, cos and tanEXTENDED

y901802703601−10x°y = sin xy = cos xsin xstarts at 0, peak 1 at 90°zero at 0°, 180° and 360°minimum −1 at 270°cos xstarts at 1, zero at 90°minimum −1 at 180°zero again at 270°, back to 1 at 360°Both repeat every 360° and stay between −1 and 1.
Both waves repeat every 360° and stay between −1 and 1. The cosine curve is the sine curve shifted 90° to the left.
  • Key points of y = sin x: (0°, 0), (90°, 1), (180°, 0), (270°, −1), (360°, 0).
  • Key points of y = cos x: (0°, 1), (90°, 0), (180°, −1), (270°, 0), (360°, 1).
  • Period is how far you go before the shape repeats. For sin and cos it is 360°. Amplitude is the maximum height above the middle, which is 1 for both.
  • The curves cross where sin x = cos x, at 45° and 225°.

The tangent graphEXTENDED

y90180270360321−1−2−30x°(45°, 1)asymptoteasymptotey = tan xtan x repeats every 180° and passes through 0° and 180°.It has asymptotes (lines it never touches) at 90° and 270°.Unlike sin and cos, it is not limited to between −1 and 1.
The tangent curve has vertical asymptotes at 90° and 270°. It repeats every 180° and has no maximum or minimum.
  • Asymptotes are lines the curve gets closer and closer to, but never touches. For y = tan x they are x = 90° and x = 270°.
  • Key points: tan 0° = 0, tan 45° = 1, tan 135° = −1, tan 180° = 0.
  • Between the asymptotes the curve rises from −∞ to +∞, so every value of tan x happens exactly once in each cycle.

Transformed trig graphsEXTENDED

y = 2 sin xamplitude 218036021−1−20y = sin 2xperiod 180°18036021−1−20y = sin x + 1moved up 118036021−1−20y = a sin bx + ca = amplitude(the height)b = cycles in 360°period = 360° ÷ bc = slideup or down
In y = a sin bx + c, the number a stretches the height, b squashes the width, and c slides the whole graph up or down.
  • a is the amplitude. y = 2 sin x goes up to 2 and down to −2.
  • b changes the period. The period is 360° ÷ b. So y = sin 2x repeats every 180°.
  • c moves the graph up (c positive) or down. y = sin x + 1 goes from 0 to 2.
Worked example

State the amplitude and the period of y = 3 cos 2x.

  1. amplitude = 3
  2. period = 360° ÷ 2 = 180°
Worked example

Find the maximum and minimum values of y = 2 sin x − 1.

  1. sin x lies between −1 and 1
  2. 2 sin x lies between −2 and 2
  3. subtract 1: maximum = 1 and minimum = −3

Solving trigonometric equationsEXTENDED

Aall +Ssin +Ttan +Ccos +90°180°270°0°sin: x and 180° − xcos: x and 360° − xtan: x and 180° + xy901802703601−10x°30°150°sin x = 0.5Solve sin x = 0.5 for 0° ≤ x ≤ 360°1. calculator: sin⁻¹ 0.5 = 30°2. sin is + in quadrants 1 and 23. second answer: 180° − 30° = 150°x = 30° or x = 150°
Use the calculator for the first answer, then use the CAST diagram or the graph to find the second answer in the range 0° to 360°.
  • Step 1: get the trig function alone: for example 2 sin x = 1.4 becomes sin x = 0.7.
  • Step 2: use the inverse on the calculator to find the first answer.
  • Step 3: find the second answer using the graph or CAST. Sin: x and 180° − x. Cos: x and 360° − x. Tan: x and 180° + x.
  • Negative answer from the calculator? Add 360° (or 180° for tan) to make it fit the range.
  • How many solutions? For sin x = k or cos x = k with k between −1 and 1 there are two in 0° to 360°. If k = 1 or −1 there is one. If k is bigger than 1 there are none.
Worked example

Solve sin x = 0.5 for 0° ≤ x ≤ 360°.

  1. sin⁻¹(0.5) = 30°
  2. second answer: 180° − 30° = 150°
  3. x = 30° or 150°
Worked example

Solve cos x = −0.5 for 0° ≤ x ≤ 360°.

  1. cos⁻¹(−0.5) = 120°
  2. second answer: 360° − 120° = 240°
  3. x = 120° or 240°
Worked example

Solve tan x = −1 for 0° ≤ x ≤ 360°.

  1. tan⁻¹(−1) = −45°
  2. add 180° to get 135°
  3. add 360° to get 315°
  4. x = 135° or 315°
Worked example

Solve 2 sin x = 1.4 for 0° ≤ x ≤ 360°.

  1. sin x = 0.7
  2. first answer = sin⁻¹(0.7) = 44.4°
  3. second = 180° − 44.4° = 135.6°
  4. x = 44.4° or 135.6°
Worked example

Solve sin x = −0.5 for 0° ≤ x ≤ 360°.

  1. sin⁻¹(−0.5) = −30°
  2. sin is negative in quadrants 3 and 4
  3. x = 180° + 30° = 210° and x = 360° − 30° = 330°
  4. x = 210° or 330°
Common mistakeGiving only one answer. Always check for a second one in the range.
Exam tipCheck on the graph. Draw a horizontal line at the value on the sine or cosine curve. It cuts the curve at your solutions.

7Test yourself

Test yourself — cover the answers first

Quick check

  1. Find the hypotenuse of a right-angled triangle with shorter sides 9 cm and 12 cm.
  2. A right-angled triangle has hypotenuse 17 cm and one shorter side 8 cm. Find the other side.
  3. Is a triangle with sides 7 cm, 24 cm and 25 cm right-angled?
  4. Find the length of the diagonal of a rectangle 5 cm by 12 cm.
  5. Find the distance between the points (2, 1) and (8, 9).
  6. A 6 m ladder rests against a wall with its foot 2 m from the wall. How high up the wall does it reach? Give your answer to 3 significant figures.
  7. Name the longest side of a right-angled triangle.
  8. Find angle θ if sin θ = 0.5.
  9. Find x when the hypotenuse is 20 cm and the angle opposite x is 30°.
  10. Find the adjacent side when the hypotenuse is 15 cm and the angle is 60°.
  11. Find angle θ when the opposite and adjacent sides are both 8 cm.
  12. Find angle θ when the opposite side is 5 cm and the hypotenuse is 9 cm.
  13. A tree casts a shadow 12 m long when the angle of elevation of the Sun is 40°. Find the height of the tree.
  14. Write down the three-figure bearing for due West.
  15. The bearing of B from A is 050°. Find the bearing of A from B.
  16. The bearing of Q from P is 285°. Find the bearing of P from Q.
  17. The angle of elevation of A from B is 25°. What is the angle of depression of B from A?
  18. EXT Write down the exact value of cos 60°.
  19. EXT Write down the exact value of tan 45°.
  20. EXT In triangle ABC, a = 10 cm, A = 30° and B = 45°. Find b.
  21. EXT Write down the cosine rule for finding side a.
  22. EXT Find the area of a triangle with sides 6 cm and 9 cm and an angle of 30° between them.
  23. EXT State the amplitude and the period of y = 2 sin 3x.
  24. EXT Solve cos x = 0.5 for 0° ≤ x ≤ 360°.
  25. EXT Find the length of the space diagonal of a cuboid 3 cm by 4 cm by 12 cm.

Exam-style

  1. The foot of a ladder is 1.8 m from a wall and its top reaches 6 m up the wall. Find the length of the ladder and the angle it makes with the ground. [4]
  2. An isosceles triangle has equal sides of 13 cm and a base of 10 cm. Find its height, its area and a base angle. [5]
  3. From the top of a 45 m building the angle of depression of a car is 28°. Find how far the car is from the foot of the building. [3]
  4. A ship sails 12 km on a bearing of 040° from P to Q, then 9 km on a bearing of 130° to R. Show that angle PQR = 90°, then find PR and the bearing of R from P. [6]
  5. EXT In triangle ABC, AB = 8 cm, BC = 11 cm and angle ABC = 70°. Find AC and the area of the triangle. [5]
  6. EXT In triangle PQR, PQ = 7 cm, QR = 9 cm and PR = 12 cm. Find the largest angle. [3]
  7. EXT In triangle ABC, A = 35°, B = 80° and AB = 14 cm. Find BC. [4]
  8. EXT A pyramid has a rectangular base 12 cm by 5 cm. The apex is 8 cm directly above the centre of the base. Find the length of a sloping edge and the angle it makes with the base. [5]
  9. EXT A cuboid ABCDEFGH has AB = 9 cm, BC = 12 cm and CG = 8 cm. Find AG and the angle between AG and the base. [4]
  10. EXT Solve 3 sin x + 1 = 0 for 0° ≤ x ≤ 360°. [3]
  11. EXT Sketch y = 2 sin x for 0° ≤ x ≤ 360°, marking the maximum, minimum and where the curve crosses the x-axis. [3]
Show answers
  1. 15 cm
  2. 15 cm
  3. Yes: 49 + 576 = 625 = 25²
  4. 13 cm
  5. 10
  6. 5.66 m
  7. hypotenuse
  8. 30°
  9. 10 cm
  10. 7.5 cm
  11. 45°
  12. 33.7°
  13. 10.1 m
  14. 270°
  15. 230°
  16. 105°
  17. 25°
  18. 1/2
  19. 1
  20. 14.1 cm
  21. a² = b² + c² − 2bc cos A
  22. 13.5 cm²
  23. amplitude 2, period 120°
  24. 60° and 300°
  25. 13 cm
  26. length = √(1.8² + 6²) = √39.24 = 6.26 m
    tan θ = 6 ÷ 1.8
    θ = tan⁻¹(3.333) = 73.3°
  27. half the base = 5 cm
    h = √(13² − 5²) = √144 = 12 cm
    area = ½ × 10 × 12 = 60 cm²
    cos θ = 5 ÷ 13, so θ = 67.4°
  28. angle inside the triangle at the car = 28°
    tan 28° = 45 ÷ d
    d = 45 ÷ tan 28° = 84.6 m
  29. back bearing of PQ = 040° + 180° = 220°
    angle PQR = 220° − 130° = 90°
    PR = √(12² + 9²) = 15 km
    angle QPR: tan = 9 ÷ 12, so 36.9°
    bearing of R from P = 040° + 36.9° = 077°
  30. AC² = 8² + 11² − 2 × 8 × 11 × cos 70° = 185 − 60.20 = 124.8
    AC = 11.2 cm
    area = ½ × 8 × 11 × sin 70° = 41.3 cm²
  31. largest angle is opposite 12, so it is angle Q
    cos Q = (7² + 9² − 12²) ÷ (2 × 7 × 9) = −14 ÷ 126 = −0.1111
    Q = 96.4°
  32. angle C = 180° − 35° − 80° = 65°
    pair: AB is opposite C, and BC is opposite A
    BC ÷ sin 35° = 14 ÷ sin 65°
    BC = 14 × sin 35° ÷ sin 65° = 8.86 cm
  33. base diagonal = √(12² + 5²) = 13 cm, half = 6.5 cm
    edge = √(6.5² + 8²) = √106.25 = 10.3 cm
    tan θ = 8 ÷ 6.5, so θ = 50.9°
  34. AC = 15 cm
    AG = √(15² + 8²) = 17 cm
    tan θ = 8 ÷ 15
    θ = 28.1°
  35. sin x = −1/3
    sin⁻¹(−1/3) = −19.5°
    x = 180° + 19.5° = 199.5° and x = 360° − 19.5° = 340.5°
  36. wave starting at the origin
    maximum (90°, 2) and minimum (270°, −2)
    crosses the x-axis at 0°, 180° and 360°
Practise this chapter
Pythagoras' theoremRight-angled trianglesExact trigonometric valuesSine & cosine rules
Next chapterTransformations & vectors →
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