Mini-Lesson

Geometry & Trigonometry

6 missions. Guess, learn, then check. Your place is saved.

Mission 1What's the longest pencil that fits in a 20 × 12 × 9 cm box?
Mission 2How can you find a mountain's height without climbing it?
Mission 330 cm pizza cut in 8, 24 cm cut in 6: which slice has more crust?
Mission 4Why is $sin 15 0^{\circ}$ sin 150° exactly equal to $sin 3 0^{\circ}$ sin 30°?
Mission 5A Ferris wheel turns for 10 minutes. How often is your seat at 20 m?
Mission 6How could you find $sin 7 5^{\circ}$ sin 75° exactly without a calculator?

The specification points and sources are listed under the lesson: see Sources.

Mission 1

What's the longest pencil that fits in a 20 × 12 × 9 cm box?

Make a guess: What's the longest pencil that fits in a 20 × 12 × 9 cm box?

Geometry · 3-D shapes

Distance and midpoint in 3-D

Points in space have three coordinates $(x, y, z)$ (x, y, z).
Distance between $(x_{1}, y_{1}, z_{1})$ (x₁, y₁, z₁) and $(x_{2}, y_{2}, z_{2})$ (x₂, y₂, z₂):

d = (x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2} + (z_{1} - z_{2})^{2}

Midpoint: $(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2}, \frac{z _{1} + z _{2}}{2})$ ((x₁+x₂)/2, (y₁+y₂)/2, (z₁+z₂)/2)
In a cuboid $a \times b \times c$ a × b × c, Pythagoras on the base gives $a^{2} + b^{2}$ √(a²+b²); Pythagoras again with the height gives the space diagonal $a^{2} + b^{2} + c^{2}$ √(a²+b²+c²).

Worked example Geometry · 3-D shapes

Space diagonal and its angle with the base

Question

A cuboid measures 8 cm by 6 cm by 5 cm. Find the length of a space diagonal and the angle it makes with the base.

1
Base diagonal by Pythagoras: $8^{2} + 6^{2} = 10$ √(8²+6²)=10 cm
2
Space diagonal in the vertical right-angled triangle: $1 0^{2} + 5^{2} = 125 = 11.2$ √(10²+5²)=√125=11.2 cm (3 s.f.)
3
The angle with the base is between the diagonal and its projection, the base diagonal: $tan θ = \frac{5}{10}$ tan θ=5/10

Answer

11.2

cm and

θ = 26. 6^{\circ}

(3 s.f.)

⚠️ Watch out

the angle with the base uses the base diagonal (10 cm), not an edge;

tan^{- 1} \frac{5}{8}

gives the wrong

32. 0^{\circ}

.

Calculate

?

A

is

(2, - 1, 5)

and

B

is

(- 4, 1, 1)

.

M

is the midpoint of

A B

. Find the distance from

M

to

C (1, 6, 0)

. Give an exact answer.

Geometry · 3-D shapes

Angles in 3-D

The angle between a line and a plane is the angle between the line and its projection on the plane (the "shadow" straight below it).
Drop a perpendicular from a point on the line to the plane: this makes a right-angled triangle containing the angle.
The angle between two intersecting lines is found in the triangle that both lines lie in.
At SL, 3-D angle questions use right-angled trigonometry: sketch the triangle flat, label it, then use SOH CAH TOA.

Quick check

?

V A BC D

is a right pyramid on a square base

A BC D

.

M

is the centre of the base, directly below

V

, and

N

is the midpoint of

A B

. Which angle is the angle between the edge

V A

and the base?

Geometry · 3-D shapes

Volume and surface area

Solid	Volume	Surface area
Right pyramid	$\frac{1}{3} \times$ 1/3× base area $\times$ × height	base + triangular faces
Right cone	$\frac{1}{3} π r^{2} h$ 1/3π r²h	curved: $π r l$ π rl ( $l$ l = slant height)
Sphere	$\frac{4}{3} π r^{3}$ 4/3π r³	$4 π r^{2}$ 4π r²
Hemisphere	$\frac{2}{3} π r^{3}$ 2/3π r³	curved: $2 π r^{2}$ 2π r²

These formulas are in the IB formula booklet. For a combined solid, add the volumes, but count only the surfaces on the outside.
Slant height of a cone: $l = r^{2} + h^{2}$ l=√(r²+h²).

Worked example Geometry · 3-D shapes

A cone on a hemisphere

Question

A toy is a cone of radius 3 cm and height 4 cm fixed to a hemisphere of radius 3 cm. Find its volume and its total surface area.

1
Cone volume: $\frac{1}{3} π (3^{2}) (4) = 12 π$ 1/3π(3²)(4)=12π
2
Hemisphere volume: $\frac{2}{3} π (3^{3}) = 18 π$ 2/3π(3³)=18π, so total volume $= 30 π = 94.2$ =30π=94.2 cm³ (3 s.f.)
3
Slant height: $l = 3^{2} + 4^{2} = 5$ l=√(3²+4²)=5, so the curved cone surface is $π (3) (5) = 15 π$ π(3)(5)=15π
4
Curved hemisphere surface: $2 π (3^{2}) = 18 π$ 2π(3²)=18π; the flat circles are joined inside, so they don't count

Answer

volume

94.2

cm³, surface area

33 π = 104

cm² (3 s.f.)

⚠️ Watch out

using the whole sphere's

4 π r^{2}

gives

51 π

; only half the sphere is there.

Calculate

?A right pyramid has a square base of side 10 cm and vertical height 12 cm. Find the angle between a sloping edge and the base. Give your answer in degrees to 3 significant figures.

Calculate

?A grain silo is a cylinder of radius 4 m and height 10 m with a hemisphere of radius 4 m on top. Find the volume of the silo to 3 significant figures.

m³

Mission check

?What's the longest pencil that fits in a 20 × 12 × 9 cm box?

Mission 2

How can you find a mountain's height without climbing it?

Make a guess: How can you find a mountain's height without climbing it?

Geometry · trigonometry

Right-angled trigonometry, elevation and bearings

In a right-angled triangle: $sin θ = \frac{opp}{hyp}$ sin θ=opp/hyp, $cos θ = \frac{adj}{hyp}$ cos θ=adj/hyp, $tan θ = \frac{opp}{adj}$ tan θ=opp/adj. Use $sin^{- 1}$ sin⁻¹, $cos^{- 1}$ cos⁻¹, $tan^{- 1}$ tan⁻¹ to find angles.
Angle of elevation: measured up from the horizontal. Angle of depression: measured down from the horizontal.
Bearings are measured clockwise from north, written with three figures (e.g. 040°).
Always sketch and label a diagram from the words first.

Quick check

?From the top of a vertical cliff 40 m high, the angle of depression of a boat is 25°. How far is the boat from the foot of the cliff, to 3 significant figures?

Geometry · trigonometry

Sine rule, cosine rule and area

Label the side opposite angle $A$ A as $a$ a, and so on.
Sine rule: $\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}$ a/(sin A)=b/(sin B)=c/(sin C). Use it with a side and its opposite angle known.
Cosine rule: $c^{2} = a^{2} + b^{2} - 2 ab cos C$ c²=a²+b²−2abcos C, or $cos C = \frac{a ^{2} + b ^{2} - c ^{2}}{2 ab}$ cos C=(a²+b²−c²)/2ab. Use it with two sides and the angle between them, or with all three sides.
Area $= \frac{1}{2} ab sin C$ =1/2absin C ( $C$ C is the angle between sides $a$ a and $b$ b).
Finding an angle with the sine rule can give two possible triangles (the ambiguous case); you meet it in the unit circle mission.

Worked example Geometry · trigonometry

Cosine rule, then area

Question

In triangle

PQR

,

PQ = 8

cm,

PR = 11

cm and angle

QPR = 4 0^{\circ}

. Find

QR

and the area of the triangle.

1
Two sides and the included angle, so use the cosine rule: $Q R^{2} = 8^{2} + 1 1^{2} - 2 (8) (11) cos 4 0^{\circ}$ QR²=8²+11²−2(8)(11)cos 40°
2
Evaluate: $Q R^{2} = 50.18 \dots$ QR²=50.18…, so $QR = 7.08$ QR=7.08 cm (3 s.f.)
3
Area with the same two sides and included angle: $\frac{1}{2} (8) (11) sin 4 0^{\circ}$ 1/2(8)(11)sin 40°

Answer

QR = 7.08

cm, area

= 28.3

cm² (3 s.f.)

⚠️ Watch out

with your GDC in radian mode,

cos 40

gives

QR = 17.4

cm; set degree mode for degree questions.

Sort it

?Which rule do you use first?

Tap a statement, then tap the group it belongs in. Then press Check.

Worked example Geometry · trigonometry

Height from two angles of elevation

Question

From

A

, the angle of elevation of the top

T

of a tower is 20°. From

B

, 500 m closer on level ground, it is 32°. Find the height of the tower.

1
In triangle $A BT$ ABT: angle $A BT = 18 0^{\circ} - 3 2^{\circ} = 14 8^{\circ}$ ABT=180°−32°=148°, so angle $A TB = 18 0^{\circ} - 2 0^{\circ} - 14 8^{\circ} = 1 2^{\circ}$ ATB=180°−20°−148°=12°
2
Sine rule: $\frac{BT}{sin 2 0 ^{\circ}} = \frac{500}{sin 1 2 ^{\circ}}$ BT/(sin 20°)=500/(sin 12°), so $BT = 822.5$ BT=822.5 m
3
Right-angled triangle below $T$ T: height $= BT sin 3 2^{\circ}$ =BTsin 32°

Answer

height

= 436

m (3 s.f.)

⚠️ Watch out

the angle at

T

is the difference of the two elevations,

3 2^{\circ} - 2 0^{\circ}

, not their sum.

Calculate

?A ship sails 15 km from port

P

on a bearing of 040° to

Q

, then 9 km on a bearing of 110° to

R

. Find the distance

PR

. Give your answer to 3 significant figures.

km

Calculate

?A triangular field has sides 50 m, 70 m and 90 m. Find its area. Give your answer to the nearest square metre.

m²

Mission check

?How can you find a mountain's height without climbing it?

Mission 3

30 cm pizza cut in 8, 24 cm cut in 6: which slice has more crust?

Make a guess: 30 cm pizza cut in 8, 24 cm cut in 6: which slice has more crust?

Geometry · circles

Arc length and sector area

A sector is a "pizza slice" of a circle; its curved edge is an arc.
A sector with angle $θ^{\circ}$ θ° is the fraction $\frac{θ}{360}$ θ/360 of the whole circle, so

arc length = \frac{θ}{360} \times 2 π r sector area = \frac{θ}{360} \times π r^{2}

Perimeter of a sector $=$ = arc $+ 2 r$ + 2r (two radii).

Worked example Geometry · circles

A sector in degrees

Question

A sector has radius 9 cm and angle 140°. Find its arc length, area and perimeter.

1
Fraction of the circle: $\frac{140}{360} = \frac{7}{18}$ 140/360=7/18
2
Arc length: $\frac{7}{18} \times 2 π (9) = 7 π = 22.0$ 7/18× 2π(9)=7π=22.0 cm (3 s.f.)
3
Area: $\frac{7}{18} \times π (9^{2}) = 31.5 π = 99.0$ 7/18× π(9²)=31.5π=99.0 cm² (3 s.f.)
4
Perimeter: $7 π + 2 (9) = 40.0$ 7π+2(9)=40.0 cm (3 s.f.)

Answer

arc

22.0

cm, area

99.0

cm², perimeter

40.0

cm

⚠️ Watch out

the perimeter includes both straight radii, not just the arc.

Geometry · circles

Radians

One radian is the angle at the centre when the arc length equals the radius.
A full turn is $2 π$ 2π radians, so $π rad = 18 0^{\circ}$ π rad=180°.
Degrees to radians: multiply by $\frac{π}{180}$ π/180. Radians to degrees: multiply by $\frac{180}{π}$ 180/π.
Learn: $3 0^{\circ} = \frac{π}{6}$ 30°=π/6, $4 5^{\circ} = \frac{π}{4}$ 45°=π/4, $6 0^{\circ} = \frac{π}{3}$ 60°=π/3, $9 0^{\circ} = \frac{π}{2}$ 90°=π/2.
With $θ$ θ in radians: arc $l = r θ$ l=rθ and sector area $A = \frac{1}{2} r^{2} θ$ A=1/2r²θ. Set your GDC to radian mode.

Worked example Geometry · circles

A sector in radians

Question

A sector has radius 8 cm and angle 1.2 radians. Find its arc length, area and perimeter.

1
Arc length: $l = r θ = 8 (1.2) = 9.6$ l=rθ=8(1.2)=9.6 cm
2
Area: $A = \frac{1}{2} r^{2} θ = \frac{1}{2} (8^{2}) (1.2) = 38.4$ A=1/2r²θ=1/2(8²)(1.2)=38.4 cm²
3
Perimeter: $9.6 + 2 (8) = 25.6$ 9.6+2(8)=25.6 cm

Answer

arc

9.6

cm, area

38.4

cm², perimeter

25.6

cm

⚠️ Watch out

l = r θ

only works with

θ

in radians; 1.2 rad is about

68. 8^{\circ}

, not

1. 2^{\circ}

.

Quick check

?Write

13 5^{\circ}

in radians.

Calculate

?A sector of a circle of radius 6 cm has area 30 cm². Find the angle of the sector in radians, to 3 significant figures.

Geometry · circles

Segments and compound shapes

A segment is the region between a chord and its arc.
Segment area $=$ = sector area $-$ − triangle area, with the triangle area $\frac{1}{2} r^{2} sin θ$ 1/2r²sin θ (two radii and the angle between them).
For a shape made of sectors, rectangles and triangles: find each piece, then add or subtract.
Work with unrounded values and round only the final answer to 3 significant figures.

Calculate

?A circle has centre

O

and radius 10 cm. Points

A

and

B

lie on the circle with angle

A OB = 1.4

radians. Find the area of the segment between chord

A B

and the minor arc

A B

. Give your answer to 3 significant figures.

cm²

Calculate

?A fan-shaped badge is a sector of radius 12 cm. Its perimeter is 40 cm. Find its area.

cm²

Mission check

?30 cm pizza cut in 8, 24 cm cut in 6: which slice has more crust?

Mission 4

Why is $sin 15 0^{\circ}$ sin 150° exactly equal to $sin 3 0^{\circ}$ sin 30°?

Make a guess: Why is $sin 15 0^{\circ}$ sin 150° exactly equal to $sin 3 0^{\circ}$ sin 30°?

Trigonometry · unit circle

Sine and cosine on the unit circle

The unit circle has centre $O$ O and radius 1. Turn a radius $θ$ θ anticlockwise from the positive $x$ x-axis.
Its end point is $(cos θ, sin θ)$ (cos θ, sin θ): cosine is the $x$ x-coordinate, sine is the height.
$tan θ = \frac{sin θ}{cos θ}$ tan θ=(sin θ)/(cos θ), the gradient of the radius; the line is $y = x tan θ$ y=xtan θ.
This works for any angle, so signs follow the quadrant: in the 2nd quadrant $sin θ > 0$ sin θ>0 and $cos θ < 0$ cos θ<0.

Trigonometry · unit circle

Angles in different quadrants

Reflect in the $y$ y-axis: $sin (18 0^{\circ} - θ) = sin θ$ sin (180°−θ)=sin θ, $cos (18 0^{\circ} - θ) = - cos θ$ cos (180°−θ)=−cos θ.
Reflect in the $x$ x-axis: $sin (- θ) = - sin θ$ sin (−θ)=−sin θ, $cos (- θ) = cos θ$ cos (−θ)=cos θ.
Half a turn: $sin (18 0^{\circ} + θ) = - sin θ$ sin (180°+θ)=−sin θ, $cos (18 0^{\circ} + θ) = - cos θ$ cos (180°+θ)=−cos θ.
In radians, $18 0^{\circ}$ 180° is $π$ π: for example $sin (π - x) = sin x$ sin (π−x)=sin x.

Calculate

?The angle

θ

is obtuse and

sin θ = sin 5 0^{\circ}

. Find

θ

in degrees.

Trigonometry · unit circle

Exact values

$θ$ θ	$0$ 0	$\frac{π}{6}$ π/6	$\frac{π}{4}$ π/4	$\frac{π}{3}$ π/3	$\frac{π}{2}$ π/2
$sin θ$ sin θ	$0$ 0	$\frac{1}{2}$ 1/2	$\frac{2}{2}$ (√2)/2	$\frac{3}{2}$ (√3)/2	$1$ 1
$cos θ$ cos θ	$1$ 1	$\frac{3}{2}$ (√3)/2	$\frac{2}{2}$ (√2)/2	$\frac{1}{2}$ 1/2	$0$ 0
$tan θ$ tan θ	$0$ 0	$\frac{3}{3}$ (√3)/3	$1$ 1	$3$ √3	undefined

For a multiple such as $\frac{3 π}{4}$ 3π/4: find the reference angle to the $x$ x-axis ( $\frac{π}{4}$ π/4), take its value, then fix the sign from the quadrant.

Quick check

?Without a calculator, find the exact value of

cos \frac{5 π}{6}

.

Trigonometry · unit circle

Identities

From the unit circle and Pythagoras: $cos^{2} θ + sin^{2} θ = 1$ cos²θ+sin²θ=1.
Use it to find one ratio from another; the quadrant decides the sign of the square root.
Double angle: $sin 2 θ = 2 sin θ cos θ$ sin 2θ=2sin θcos θ.
$cos 2 θ = cos^{2} θ - sin^{2} θ = 2 cos^{2} θ - 1 = 1 - 2 sin^{2} θ$ cos 2θ=cos²θ−sin²θ=2cos²θ−1=1−2sin²θ.

Worked example Trigonometry · unit circle

One ratio from another

Question

sin θ = \frac{3}{5}

and

θ

is obtuse. Find

cos θ

and

tan θ

.

1
Pythagorean identity: $cos^{2} θ = 1 - (\frac{3}{5})^{2} = \frac{16}{25}$ cos²θ=1−(3/5)²=16/25
2
$θ$ θ is in the 2nd quadrant, so cosine is negative: $cos θ = - \frac{4}{5}$ cos θ=−4/5
3
$tan θ = \frac{sin θ}{cos θ} = \frac{3/5}{- 4/5}$ tan θ=(sin θ)/(cos θ)=(3/5)/(−4/5)

Answer

cos θ = - \frac{4}{5}

,

tan θ = - \frac{3}{4}

⚠️ Watch out

cos^{2} θ = \frac{16}{25}

gives

\pm \frac{4}{5}

; taking

+ \frac{4}{5}

ignores that

θ

is obtuse.

Calculate

?Given that

cos x = \frac{3}{4}

and

x

is acute, find

sin 2 x

. Give your answer to 3 significant figures.

Worked example Trigonometry · unit circle

The ambiguous case of the sine rule

Question

In triangle

A BC

,

A B = 12

cm,

BC = 8

cm and angle

A = 3 0^{\circ}

. Find the possible sizes of angle

C

.

1
Sine rule: $\frac{sin C}{12} = \frac{sin 3 0 ^{\circ}}{8}$ (sin C)/12=(sin 30°)/8, so $sin C = 0.75$ sin C=0.75
2
Inverse sine: $C = 48. 6^{\circ}$ C=48.6°; the unit circle gives a second angle, $18 0^{\circ} - 48. 6^{\circ} = 131. 4^{\circ}$ 180°−48.6°=131.4°
3
Test each: $3 0^{\circ} + 131. 4^{\circ} < 18 0^{\circ}$ 30°+131.4°<180°, so both triangles exist

Answer

C = 48. 6^{\circ}

or

C = 131. 4^{\circ}

(1 d.p.)

⚠️ Watch out

when the known angle is opposite the shorter given side, check

18 0^{\circ}

minus your answer; a GDC shows only the acute one.

Mission check

?Why is

sin 15 0^{\circ}

exactly equal to

sin 3 0^{\circ}

?

Mission 5

A Ferris wheel turns for 10 minutes. How often is your seat at 20 m?

Make a guess: A Ferris wheel turns for 10 minutes. How often is your seat at 20 m?

Trigonometry · graphs and equations

The graphs of sine, cosine and tangent

$y = sin x$ y=sin x and $y = cos x$ y=cos x repeat every $2 π$ 2π (360°): period $2 π$ 2π, amplitude 1, range $- 1 \leq y \leq 1$ −1≤ y≤ 1.
$y = cos x$ y=cos x is $y = sin x$ y=sin x translated $\frac{π}{2}$ π/2 to the left.
$y = tan x$ y=tan x has period $π$ π, no amplitude, and vertical asymptotes at $x = \frac{π}{2}, \frac{3 π}{2}, \dots$ x=π/2, 3π/2, …
Radians are assumed unless degrees are stated.

Trigonometry · graphs and equations

$y = a sin (b (x + c)) + d$ y=asin (b(x+c))+d

$∣ a ∣$ |a| is the amplitude: a vertical stretch, scale factor $a$ a.
Period $= \frac{2 π}{b}$ =2π/b: a horizontal stretch, scale factor $\frac{1}{b}$ 1/b.
$c$ c translates the graph $c$ c units to the left (so $x - \frac{π}{4}$ x−π/4 moves it right).
$y = d$ y=d is the principal axis: maximum $d + ∣ a ∣$ d+|a|, minimum $d - ∣ a ∣$ d−|a|.
The same rules hold for $a cos (b (x + c)) + d$ acos (b(x+c))+d.

Quick check

?What is the period of

f (x) = 3 cos (2 (x - \frac{π}{4})) + 1

?

Worked example Trigonometry · graphs and equations

Building a tide model

Question

At a harbour, the water depth is 8 m at high tide (

t = 0

) and 2 m at low tide, and the pattern repeats every 12 hours. Model the depth

d

metres at

t

hours in the form

d = a cos (b t) + k

.

1
Principal axis: $k = \frac{8 + 2}{2} = 5$ k=(8+2)/2=5
2
Amplitude: $a = \frac{8 - 2}{2} = 3$ a=(8−2)/2=3 (positive, since the model starts at a maximum)
3
Period: $\frac{2 π}{b} = 12$ 2π/b=12, so $b = \frac{π}{6}$ b=π/6

Answer

d = 3 cos (\frac{π t}{6}) + 5

⚠️ Watch out

b

is not the period;

b = 12

would make the tide repeat every

\frac{π}{6}

hours.

Calculate

?The depth of the water in the harbour is

d = 3 cos (\frac{π t}{6}) + 5

metres,

t

hours after high tide. Find the depth 2.5 hours after high tide, to 3 significant figures.

m

Trigonometry · graphs and equations

Solving trig equations in an interval

Rearrange to $sin (\dots) = k$ sin (…)=k, $cos (\dots) = k$ cos (…)=k or $tan (\dots) = k$ tan (…)=k. Find the first value with the inverse function.
Use the graph or unit circle for the others: $sin$ sin repeats as $π - α$ π−α, $cos$ cos as $2 π - α$ 2π−α, $tan$ tan as $α + π$ α+π, then add periods.
For $sin (2 x)$ sin (2x), change the interval first: $0 \leq x \leq 2 π$ 0≤ x≤ 2π means $0 \leq 2 x \leq 4 π$ 0≤ 2x≤ 4π.
A quadratic in $sin x$ sin x or $cos x$ cos x: use $sin^{2} x + cos^{2} x = 1$ sin²x+cos²x=1 if needed, factorise, then reject values outside $[- 1, 1]$ [−1, 1].
Graphically: count every intersection of the curve with the line in the interval.

Worked example Trigonometry · graphs and equations

An equation with $2 x$ 2x

Question

Solve

2 sin (2 x) = 3

for

0 \leq x \leq 2 π

.

1
Rearrange: $sin (2 x) = \frac{3}{2}$ sin (2x)=(√3)/2, and widen the interval: $0 \leq 2 x \leq 4 π$ 0≤ 2x≤ 4π
2
First quadrant value and its partner: $2 x = \frac{π}{3}$ 2x=π/3 or $π - \frac{π}{3} = \frac{2 π}{3}$ π−π/3=2π/3
3
Add $2 π$ 2π to each while still within $4 π$ 4π: $2 x = \frac{7 π}{3}$ 2x=7π/3 or $\frac{8 π}{3}$ 8π/3
4
Halve every value

Answer

x = \frac{π}{6}, \frac{π}{3}, \frac{7 π}{6}, \frac{4 π}{3}

⚠️ Watch out

halving the interval too early, or forgetting to widen it, loses the last two solutions.

Quick check

?Solve

2 sin^{2} x + 5 cos x + 1 = 0

for

0 \leq x \leq 2 π

.

Calculate

?The harbour model is

d = 3 cos (\frac{π t}{6}) + 5

for

0 \leq t \leq 12

. A boat needs a depth of more than 6.5 m. For how many hours of this 12-hour cycle can it use the harbour?

hours

Mission check

?Seat height

h (t) = 16 - 15 cos (\frac{π t}{2})

m after

t

minutes. For

0 \leq t \leq 10

, how often is

h = 20

?

Mission 6

How could you find $sin 7 5^{\circ}$ sin 75° exactly without a calculator?

Make a guess: How could you find $sin 7 5^{\circ}$ sin 75° exactly without a calculator?

Geometry and trigonometry · trig identities

Reciprocal trig ratios

$sec θ = \frac{1}{cos θ}$ sec θ = 1/(cos θ), $cosec θ = \frac{1}{sin θ}$ cosec θ = 1/(sin θ), $cot θ = \frac{1}{tan θ} = \frac{cos θ}{sin θ}$ cot θ = 1/(tan θ) = (cos θ)/(sin θ)
Each is undefined where its denominator is 0: $sec θ$ sec θ has vertical asymptotes where $cos θ = 0$ cos θ = 0.
Divide $sin^{2} θ + cos^{2} θ = 1$ sin²θ + cos²θ = 1 by $cos^{2} θ$ cos²θ to get $tan^{2} θ + 1 = sec^{2} θ$ tan²θ + 1 = sec²θ.
Divide it by $sin^{2} θ$ sin²θ instead to get $1 + cot^{2} θ = cosec^{2} θ$ 1 + cot²θ = cosec²θ.
Signs follow the parent ratio: $sec θ$ sec θ is negative exactly where $cos θ$ cos θ is.

Quick check

?Given that

sec θ = 3

and

\frac{3 π}{2} < θ < 2 π

, find the exact value of

tan θ

.

Geometry and trigonometry · trig identities

Inverse trig functions

$sin x$ sin x is not one-to-one, so restrict its domain to $- \frac{π}{2} \leq x \leq \frac{π}{2}$ −π/2 ≤ x ≤ π/2. Its inverse is $arcsin x$ arcsin x.

Function	Domain	Range
$arcsin x$ arcsin x	$- 1 \leq x \leq 1$ −1 ≤ x ≤ 1	$- \frac{π}{2} \leq y \leq \frac{π}{2}$ −π/2 ≤ y ≤ π/2
$arccos x$ arccos x	$- 1 \leq x \leq 1$ −1 ≤ x ≤ 1	$0 \leq y \leq π$ 0 ≤ y ≤ π
$arctan x$ arctan x	$x \in R$ x ∈ ℝ	$- \frac{π}{2} < y < \frac{π}{2}$ −π/2 < y < π/2

Each graph is the reflection of the restricted graph in $y = x$ y = x. The graph of $y = arctan x$ y = arctan x has horizontal asymptotes $y = \pm \frac{π}{2}$ y = ±π/2.
The output must lie in the range, so $arcsin (sin x) = x$ arcsin (sin x) = x only when $x$ x is in $[- \frac{π}{2}, \frac{π}{2}]$ [−π/2, π/2].

Quick check

?Find the exact value of

arcsin (sin \frac{5 π}{6})

.

Geometry and trigonometry · trig identities

Compound angle identities

These are in the formula booklet:

$sin (A \pm B) = sin A cos B \pm cos A sin B$ sin (A ± B) = sin Acos B ± cos Asin B $cos (A \pm B) = cos A cos B \mp sin A sin B$ cos (A ± B) = cos Acos B ∓ sin Asin B $tan (A \pm B) = \frac{tan A \pm tan B}{1 \mp tan A tan B}$ tan (A ± B) = (tan A ± tan B)/(1 ∓ tan Atan B)

Warning: $sin (A + B) \neq = sin A + sin B$ sin (A + B) ≠ sin A + sin B. Try $A = 3 0^{\circ}$ A = 30°, $B = 6 0^{\circ}$ B = 60°: the left side is 1, the right side is about 1.37.
To find an exact value, split the angle into two angles whose exact ratios you know ( $3 0^{\circ}$ 30°, $4 5^{\circ}$ 45°, $6 0^{\circ}$ 60°, or $\frac{π}{6}$ π/6, $\frac{π}{4}$ π/4, $\frac{π}{3}$ π/3).
Note the sign flip in the cosine identity: $cos (A + B)$ cos (A + B) has a minus.

Worked example Geometry and trigonometry · trig identities

An exact value from a compound angle

Question

Find the exact value of

cos 10 5^{\circ}

.

1
Split the angle: $10 5^{\circ} = 6 0^{\circ} + 4 5^{\circ}$ 105° = 60° + 45°
2
Use the identity: $cos 10 5^{\circ} = cos 6 0^{\circ} cos 4 5^{\circ} - sin 6 0^{\circ} sin 4 5^{\circ}$ cos 105° = cos 60°cos 45° − sin 60°sin 45°
3
Substitute exact values: $\frac{1}{2} \cdot \frac{2}{2} - \frac{3}{2} \cdot \frac{2}{2} = \frac{2}{4} - \frac{6}{4}$ 1/2·(√2)/2 − (√3)/2·(√2)/2 = (√2)/4 − (√6)/4
4
Sense-check the sign: $10 5^{\circ}$ 105° is in the second quadrant, so the cosine is negative

Answer

cos 10 5^{\circ} = \frac{2 - 6}{4}

⚠️ Watch out

cos 6 0^{\circ} + cos 4 5^{\circ} = \frac{1 + 2}{2}

is bigger than 1, which no cosine can be.

Calculate

?Angles

A

and

B

are acute, with

sin A = \frac{3}{5}

and

cos B = \frac{5}{13}

. Then

sin (A + B) = \frac{p}{65}

. Find

p

.

Geometry and trigonometry · trig identities

Double angles and symmetry

Put $B = A$ B = A in the compound identities to derive the double angle identities:

sin 2 A = 2 sin A cos A, cos 2 A = cos^{2} A - sin^{2} A, tan 2 A = \frac{2 tan A}{1 - tan ^{2} A}

Compound identities also explain the symmetry of the graphs. With $A = π$ A = π: $sin (π - θ) = sin π cos θ - cos π sin θ = sin θ$ sin (π − θ) = sin πcos θ − cos πsin θ = sin θ.
In the same way, $cos (π - θ) = - cos θ$ cos (π − θ) = −cos θ and $tan (π - θ) = - tan θ$ tan (π − θ) = −tan θ.

Calculate

?Given that

tan θ = \frac{1}{3}

, find the exact value of

tan 2 θ

. Give your answer as a decimal.

Mission check

?How could you find

sin 7 5^{\circ}

exactly without a calculator?

Recap

Big ideas

3-D distance: $(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2} + (z_{1} - z_{2})^{2}$ √((x₁−x₂)²+(y₁−y₂)²+(z₁−z₂)²); a cuboid's space diagonal is $a^{2} + b^{2} + c^{2}$ √(a²+b²+c²).
The angle between a line and a plane is the angle with its projection: find the right-angled triangle and use trigonometry.
Combined solids: add volumes; count only the outside surfaces.
Right-angled triangles: SOH CAH TOA. Elevation is up from the horizontal; depression is down; bearings are clockwise from north.
Sine rule needs a side with its opposite angle; cosine rule needs two sides and the included angle, or three sides. Area $= \frac{1}{2} ab sin C$ =1/2absin C.
Break a real problem into triangles and sketch a labelled diagram first.
Arc $= \frac{θ}{360} \times 2 π r$ =θ/360× 2π r and sector $= \frac{θ}{360} \times π r^{2}$ =θ/360×π r²; perimeter adds two radii.
In radians: $π = 18 0^{\circ}$ π=180°, arc $l = r θ$ l=rθ, sector $A = \frac{1}{2} r^{2} θ$ A=1/2r²θ.
Segment $=$ = sector $-$ − triangle, with the triangle $\frac{1}{2} r^{2} sin θ$ 1/2r²sin θ.
On the unit circle, the point at angle $θ$ θ is $(cos θ, sin θ)$ (cos θ, sin θ) and $tan θ = \frac{s i n θ}{c o s θ}$ tan θ=(sin θ)/(cos θ); the quadrant fixes the signs.
$sin (18 0^{\circ} - θ) = sin θ$ sin (180°−θ)=sin θ, so the sine rule can give two angles: test both.
$cos^{2} θ + sin^{2} θ = 1$ cos²θ+sin²θ=1, $sin 2 θ = 2 sin θ cos θ$ sin 2θ=2sin θcos θ, $cos 2 θ = 2 cos^{2} θ - 1$ cos 2θ=2cos²θ−1; learn the exact values.
$y = a sin (b (x + c)) + d$ y=asin (b(x+c))+d: amplitude $∣ a ∣$ |a|, period $\frac{2 π}{b}$ 2π/b, translation $c$ c left, principal axis $y = d$ y=d.
Solve in an interval: first value, then the symmetry partner, then add periods; widen the interval for $sin (b x)$ sin (bx).
Quadratic trig equations: factorise, then reject values outside $[- 1, 1]$ [−1, 1].
$sec θ = \frac{1}{c o s θ}$ sec θ = 1/(cos θ), $cosec θ = \frac{1}{s i n θ}$ cosec θ = 1/(sin θ), $cot θ = \frac{1}{t a n θ}$ cot θ = 1/(tan θ), with $1 + tan^{2} θ = sec^{2} θ$ 1 + tan²θ = sec²θ and $1 + cot^{2} θ = cosec^{2} θ$ 1 + cot²θ = cosec²θ.
$arcsin$ arcsin, $arccos$ arccos, $arctan$ arctan undo restricted trig functions; their outputs lie in $[- \frac{π}{2}, \frac{π}{2}]$ [−π/2, π/2], $[0, π]$ [0, π] and $(- \frac{π}{2}, \frac{π}{2})$ (−π/2, π/2).
Compound angle identities give exact values and the double angle identities (set $B = A$ B = A); $sin (A + B) \neq = sin A + sin B$ sin (A + B) ≠ sin A + sin B.

🏆

Mini-lesson complete!

⭐⭐⭐

You've worked through Geometry & Trigonometry for IB Diploma Mathematics: Analysis & Approaches HL. 🎉

Mission 1What's the longest pencil that fits in a 20 × 12 × 9 cm box?
Mission 2How can you find a mountain's height without climbing it?
Mission 330 cm pizza cut in 8, 24 cm cut in 6: which slice has more crust?
Mission 4Why is $sin 15 0^{\circ}$ sin 150° exactly equal to $sin 3 0^{\circ}$ sin 30°?
Mission 5A Ferris wheel turns for 10 minutes. How often is your seat at 20 m?
Mission 6How could you find $sin 7 5^{\circ}$ sin 75° exactly without a calculator?

Your stars: 0 / 0

Next: more IB Mathematics: Analysis & Approaches · HL lessons and games on the IB Mathematics: Analysis & Approaches · HL page.

📣 Smashed it? Share your score

Challenge a mate to beat your stars, or show a parent how you got on.

→ Back to IB Mathematics: Analysis & Approaches · HL

Specification points and sources for this lesson are listed at the bottom of the page.

Geometry & Trigonometry

What's the longest pencil that fits in a 20 × 12 × 9 cm box?

Distance and midpoint in 3-D

Space diagonal and its angle with the base

Angles in 3-D

Volume and surface area

A cone on a hemisphere

How can you find a mountain's height without climbing it?

Right-angled trigonometry, elevation and bearings

Sine rule, cosine rule and area

Cosine rule, then area

Height from two angles of elevation

30 cm pizza cut in 8, 24 cm cut in 6: which slice has more crust?

Arc length and sector area

A sector in degrees

Radians

A sector in radians

Segments and compound shapes

Why is sin150∘sin 150° exactly equal to sin30∘sin 30°?

Sine and cosine on the unit circle

Angles in different quadrants

Exact values

Identities

One ratio from another

The ambiguous case of the sine rule

A Ferris wheel turns for 10 minutes. How often is your seat at 20 m?

The graphs of sine, cosine and tangent

y=asin(b(x+c))+dy=asin (b(x+c))+d

Building a tide model

Solving trig equations in an interval

An equation with 2x2x

How could you find sin75∘sin 75° exactly without a calculator?

Reciprocal trig ratios

Inverse trig functions

Compound angle identities

An exact value from a compound angle

Double angles and symmetry

Big ideas

Mini-lesson complete!

📣 Smashed it? Share your score

Why is $sin 15 0^{\circ}$ sin 150° exactly equal to $sin 3 0^{\circ}$ sin 30°?

$y = a sin (b (x + c)) + d$ y=asin (b(x+c))+d

An equation with $2 x$ 2x

How could you find $sin 7 5^{\circ}$ sin 75° exactly without a calculator?