DP Maths AI SL 3.2 Trigonometric Ratios and Identities

DP Maths AI SL 3.2 Trigonometric Ratios and Identities - Full notes

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DP Maths AI SL | 3.2 Trigonometric Ratios and Identities

DP MATHEMATICS AI · TOPIC 3: GEOMETRY AND TRIGONOMETRY

3.2

Trigonometric Ratios and Identities

Full notes for Applications and Interpretation SL

Contents

1	Trigonometric ratios in right-angled triangles
2	Exact values for special angles
3	Angles beyond 90°: the unit circle
4	Signs of the ratios in the four quadrants
5	Related angles and symmetry
6	Supplementary angles and the sine rule
7	The Pythagorean identity
8	The tangent identity
9	Finding one ratio from another
10	Simplifying trigonometric expressions
11	Proving simple identities
12	Solving basic trigonometric equations
13	Solving equations with technology
14	Inverse trigonometric functions on the GDC
15	Application: gradients, slopes and ramps
16	Application: resolving forces and velocities
17	Sensitivity and accuracy
18	Mixed review: worked problems
19	Investigation: designing an accessible ramp
20	Exam-style questions with markscheme
21	Common misconceptions and links
22	Check your understanding
23	Summary and topic map
24	Answers to practice exercises

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and work through every worked example before attempting the practice exercise.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
use sine, cosine and tangent in right-angled triangles
recall exact values for 30°, 45° and 60°
define sine and cosine for any angle using the unit circle
determine signs and related angles in each quadrant
use cos2θ + sin2θ = 1 and tan θ = sin θ/cos θ
find one ratio from another, simplify expressions and prove simple identities
solve trigonometric equations in an interval
apply trigonometry to slopes, forces and velocities

SL booklet: this booklet contains the content for Applications and Interpretation SL. Sections marked for HL in the HL edition are not included.

Guiding questions

How can trigonometric ratios be defined for any angle? How do identities simplify problems and equations?

How this topic is examined

Ratios and identities are tested in short Paper 1 questions (exact values, finding a ratio given another and a quadrant, simple equations) and underpin all triangle and periodic modelling on Paper 2.

Draw the unit circle or a quick sketch to decide signs and the number of solutions, and give exact values when the question asks for them.

Mathematics: Applications and Interpretation places strong emphasis on technology, modelling and interpretation. Most papers allow a GDC, so you are expected to use it efficiently, record the method you used, and interpret every answer in context.

Command term	What the examiner wants
Write down	an answer with little or no working; it should be obtainable by inspection
Find / Calculate / Determine	obtain the answer showing relevant working, including what you entered into a GDC
Show that	obtain the given result; every step must be shown and the result must not be assumed
Estimate	obtain an approximate value
Sketch	a diagram showing the general shape and key features, labelled
Interpret / Comment	explain the meaning of a result in context, or give a judgement
Justify / Explain	give valid reasons or evidence
Hence	use the previous result

How to use these notes

Box	What it's for
Think about it	a question to get you thinking before a section
Why it works	a derivation or explanation of a key result
Key point	a result or idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
GDC tip	how to use your graphic display calculator efficiently
In the real world	where the mathematics is used outside the classroom
Worked example	a full solution with IB-style marks (M1 method, A1 answer, R1 reasoning, AG answer given), graded basic, standard and exam level
Practice exercise	graded questions for each section; full answers are at the end of the booklet

1 Trigonometric ratios in right-angled triangles

Opposite, adjacent, hypotenuse.

Think about it

Two different-sized right-angled triangles both have a 35° angle. Why must the ratio opposite/hypotenuse be the same in both?

For an acute angle θ in a right-angled triangle, the ratios of sides depend only on θ (similar triangles have equal ratios). These ratios are sine, cosine and tangent:

WORKED EXAMPLE 1 Finding a side and an angle (basic)

A right triangle has hypotenuse 13 and an angle of 27°. Find the side opposite 27°. In another, opposite 5 and adjacent 12: find the angle.

Solution

13 sin 27° = 5.90A1

tan−1(5/12) = 22.6°A1

Practice exercise 1

1. Find the adjacent side if the hypotenuse is 20 and θ = 64°.

2. Find θ if cos θ = 0.35 (acute).

2 Exact values for special angles

From two special triangles.

Half of an equilateral triangle of side 2 has sides 1, √3, 2 and angles 30°, 60°, 90°. Half of a square of side 1 has sides 1, 1, √2 and angles 45°, 45°, 90°. These give exact values:

θ	0°	30°	45°	60°	90°
sin θ	0	1/2	√2/2	√3/2	1
cos θ	1	√3/2	√2/2	1/2	0
tan θ	0	√3/3	1	√3	undefined

WORKED EXAMPLE 2 Exact calculation (standard)

A ladder 8 m long makes 60° with the ground. Find exactly how high it reaches.

Solution

8 sin 60°M1

= 4√3 mA1

Practice exercise 2

1. Evaluate sin 30° + cos 60°.

2. Evaluate tan 45° × sin 60°.

3. Evaluate 2 sin 45° cos 45°.

3 Angles beyond 90°: the unit circle

Coordinates define the ratios.

On a circle of radius 1 centred at the origin, the point at angle θ (anticlockwise from the positive x-axis) has coordinates (cos θ, sin θ). This defines sine and cosine for every angle; tan θ = sin θ/cos θ is the gradient of the radius.

Figure 1. 30°, 150°, 210° and 330° share the same reference angle.

4 Signs of the ratios in the four quadrants

All Students Take Calculus.

Quadrant	Angles	Positive
1st	0° to 90°	sin, cos, tan
2nd	90° to 180°	sin
3rd	180° to 270°	tan
4th	270° to 360°	cos

Practice exercise 3

1. State the sign of cos 200° and tan 300°.

2. In which quadrant is θ if sin θ < 0 and cos θ > 0?

5 Related angles and symmetry

Reference angles.

WORKED EXAMPLE 3 Exact values beyond 90° (standard)

Find exactly: (a) sin 150°; (b) cos 225°; (c) tan 300°.

Solution

(a) 1/2A1

(b) −√2/2A1

Practice exercise 4

1. Find exactly cos 120° and sin 240°.

2. Find all θ in [0°, 360°] with the same sine as 40°.

6 Supplementary angles and the sine rule

Why the ambiguous case exists.

Because sin θ = sin(180° − θ), the equation sin B = k (0 < k < 1) has two solutions between 0° and 180°: an acute B and an obtuse 180° − B. This is why the sine rule can give two triangles. The cosine rule avoids this because cos distinguishes acute (positive) from obtuse (negative) angles.

7 The Pythagorean identity

From the unit circle.

Why it works

The point (cos θ, sin θ) is on the circle x2 + y2 = 1, so cos2θ + sin2θ = 1 for every angle θ.

Practice exercise 5

1. Simplify 1 − sin250°.

2. Simplify 3 sin2x + 3 cos2x.

8 The tangent identity

tan = sin/cos.

This holds for all θ with cos θ ≠ 0. It lets equations such as sin x = 2 cos x be rewritten as tan x = 2.

9 Finding one ratio from another

Use the identity and the quadrant.

WORKED EXAMPLE 4 Given sine, find cosine (standard)

sin θ = 5/13 and θ is obtuse. Find cos θ and tan θ.

Solution

cos2θ = 1 − 25/169 = 144/169M1

Obtuse: cos θ = −12/13A1

tan θ = −5/12A1

Practice exercise 6

1. cos θ = 0.6 with 270° < θ < 360°. Find sin θ and tan θ.

2. tan θ = 3/4 with θ in the 3rd quadrant. Find sin θ.

10 Simplifying trigonometric expressions

Replace and cancel.

WORKED EXAMPLE 5 Simplifying (standard)

Simplify (1 − cos2x)/(sin x cos x).

Solution

= sin2x/(sin x cos x)M1

= sin x/cos x = tan xA1

Practice exercise 7

1. Simplify sin x/tan x.

2. Simplify (sin x + cos x)2 − 2 sin x cos x.

11 Proving simple identities

Start from one side.

WORKED EXAMPLE 6 Proof (standard)

Prove that (1 + sin x)(1 − sin x) ≡ cos2x.

Solution

LHS = 1 − sin2xM1

= cos2x = RHSA1 AG

WORKED EXAMPLE 7 Proof with tan (exam level)

Prove that tan x + 1/tan x ≡ 1/(sin x cos x).

Solution

LHS = sin x/cos x + cos x/sin x = (sin2x + cos2x)/(sin x cos x)M1 A1

= 1/(sin x cos x)AG

Practice exercise 8

1. Prove that sin4x − cos4x ≡ sin2x − cos2x.

2. Prove that (cos x)/(1 − sin x) ≡ (1 + sin x)/cos x.

12 Solving basic trigonometric equations

All solutions in the interval.

For sin x = k: principal value x1 = sin−1 k, second solution 180° − x1. For cos x = k: x1 = cos−1 k and 360° − x1. For tan x = k: x1 = tan−1 k and x1 + 180°. Then add or subtract 360° to stay in the interval.

WORKED EXAMPLE 8 Exact solutions (standard)

Solve 2 sin x = √3 for 0° ≤ x ≤ 360°.

Solution

sin x = √3/2M1

x = 60°, 120°A1

WORKED EXAMPLE 9 Using an identity (exam level)

Solve 2cos2x + sin x = 1 for 0° ≤ x ≤ 360°.

Solution

2(1 − sin2x) + sin x − 1 = 0: 2sin2x − sin x − 1 = 0M1 A1

(2 sin x + 1)(sin x − 1) = 0A1

x = 90°, 210°, 330°A1

Practice exercise 9

1. Solve cos x = −0.5 for 0° ≤ x ≤ 360°.

2. Solve tan x = 2 for 0° ≤ x ≤ 360°.

3. Solve sin x = 2 cos x for 0° ≤ x ≤ 360°.

13 Solving equations with technology

Graphs and intersections.

Graph each side and find the intersections in the interval. This handles equations that are hard to solve algebraically, such as 3 sin x = x/20 + 1 (degrees).

GDC tip

Set the window to the required interval (for example 0 ≤ x ≤ 360) and use 'intersect' for each solution. Count the solutions on the graph first.

Practice exercise 10

1. Solve 4 cos x = 3 sin x + 1 for 0° ≤ x ≤ 360°.

14 Inverse trigonometric functions on the GDC

Principal values.

sin−1 and tan−1 give angles between −90° and 90°; cos−1 gives angles between 0° and 180°. These are only one solution: use symmetry to find the others in the required interval.

Practice exercise 11

1. The GDC gives sin−1(−0.3) = −17.5°. Find all solutions of sin x = −0.3 in [0°, 360°].

15 Application: gradients, slopes and ramps

tan as a gradient.

The gradient of a line equals tan θ, where θ is the angle the line makes with the positive x-axis. Road signs give gradients as percentages: a 12% gradient rises 12 m per 100 m horizontally, so θ = tan−1(0.12) = 6.84°.

WORKED EXAMPLE 10 Road gradient (standard)

A road rises at 8°. Find its percentage gradient and the height gained over 2 km of road (along the slope).

Solution

tan 8° = 0.1405: 14.1%A1

Height = 2000 sin 8° = 278 mA1

Practice exercise 12

1. Find the angle of a line with gradient −1.5.

2. A 1 : 12 ramp rises 1 unit per 12 horizontally. Find its angle.

16 Application: resolving forces and velocities

Components.

A force or velocity of magnitude F at angle θ to the horizontal has horizontal component F cos θ and vertical component F sin θ. Components are added separately, then recombined with Pythagoras and tan.

WORKED EXAMPLE 11 Resolving (standard)

A ball is kicked at 22 m/s at 35° above the horizontal. Find the components of its velocity.

Solution

Horizontal 22 cos 35° = 18.0 m/sA1

Vertical 22 sin 35° = 12.6 m/sA1

Practice exercise 13

1. A boat heads north at 5 m/s while the current flows east at 2 m/s. Find the resultant speed and direction.

17 Sensitivity and accuracy

Small errors in angles.

For small angles, small errors in the angle can produce large errors in lengths found with tan, because tan changes rapidly near 90°. Always consider measurement accuracy when interpreting results.

WORKED EXAMPLE 12 Effect of a 1° error (standard)

A building's height is estimated as 50 tan θ with θ = 72° ± 1°. Find the range of heights.

Solution

50 tan 71° = 145.2; 50 tan 73° = 163.5M1 A1

Height between 145 m and 164 m: a large uncertaintyR1

18 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 13 Mixed problem (exam level)

Given cos θ = −1/3 with 180° < θ < 270°, find sin θ and tan θ exactly.

Solution

sin2θ = 8/9; third quadrant: sin θ = −2√2/3M1 A1

tan θ = 2√2A1

Practice exercise 14

1. Solve 3 tan2x = 1 for 0° ≤ x ≤ 360°.

2. Show that sin2x + cos2x + tan2x ≡ 1/cos2x.

19 Investigation: designing an accessible ramp

A Paper 3 style exploration.

Building regulations require wheelchair ramps to have a gradient of at most 1 : 12, with a level landing for every 760 mm of rise. 1. Find the maximum angle of the ramp. 2. A doorway is 1.1 m above the ground. Find the minimum horizontal length of ramp and the number of landings needed. 3. Find the length of ramp surface. 4. A designer proposes 1 : 15 to make the ramp easier. Find the extra horizontal space required. 5. Discuss the trade-off between steepness and space.

Answers. 1. tan−1(1/12) = 4.76°. 2. 13.2 m; 1.1/0.76 = 1.45, so the ramp is split into 2 sections with 1 intermediate landing. 3. √(13.22 + 1.12) = 13.25 m. 4. 16.5 − 13.2 = 3.3 m. 5. Gentler slopes are easier to use but need more space and cost more.

20 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

Question 1 (Paper 1 style) (about 8 minutes)

Let sin θ = 2/3, where 90° < θ < 180°.

(a) Find the exact value of cos θ. [3]

(b) Find the exact value of tan θ. [2]

Markscheme

Part	Answer	Marks
(a)	cos2θ = 1 − 4/9	M1
	= 5/9	A1
	cos θ = −√5/3	A1
(b)	(2/3)/(−√5/3)	M1
	−2/√5	A1

Examiner's comments. The sign follows from the quadrant.

Question 2 (Paper 1 style) (about 10 minutes)

Solve 2sin2x = 3 cos x for 0° ≤ x ≤ 360°. [6]

Markscheme

Part	Answer	Marks
	2(1 − cos2x) = 3 cos x	M1
	2cos2x + 3 cos x − 2 = 0	A1
	(2 cos x − 1)(cos x + 2) = 0	A1
	cos x = −2 impossible	R1
	x = 60°, 300°	A1 A1

Examiner's comments. Reject impossible values with a reason.

Question 3 (Paper 2 style) (about 8 minutes)

A hill road has a sign showing a 15% gradient.

(a) Find the angle of the road with the horizontal. [2]

(b) A cyclist rides 1.5 km along the road. Find the height gained. [2]

Markscheme

Part	Answer	Marks
(a)	tan−1(0.15)	M1
	8.53°	A1
(b)	1500 sin 8.53°	M1
	223 m	A1

Examiner's comments. Distance along the road is the hypotenuse.

21 Common misconceptions and links

Students often think…	Actually…
cos 120° = cos 60°	cos 120° = −cos 60°
sin−1 gives all solutions	only the principal value
sin2x means sin(x2)	it means (sin x)2
a 100% gradient is vertical	it is 45°

How 3.2 connects to the rest of the course

Topic	Connection
3.2 Laws of sines and cosines	non-right triangles
3.7 / 2.5 Trigonometric functions	graphs
3.10 Vectors (HL)	components
5.3 Derivatives	rates of change

22 Check your understanding

Section A: multiple choice

1. cos 60° =

A. 1/2 B. √3/2 C. √2/2 D. 1

2. sin(180° − θ) =

A. −sin θ B. cos θ C. sin θ D. −cos θ

3. In the 2nd quadrant, positive is

A. cos B. sin C. tan D. all

4. sin2x + cos2x =

A. 0 B. tan x C. 2 D. 1

5. tan 45° =

A. 0 B. 1 C. √3 D. undefined

Section B: structured

6. Solve 5 sin x = 2 for 0° ≤ x ≤ 360°. [2]

7. Simplify (1 − sin2x) tan x. [2]

Answers

1. A 2. C 3. B 4. D 5. B

6. 23.6°, 156.4°

7. sin x cos x

23 Summary and topic map

Idea	Remember
Right triangles	SOH CAH TOA
Exact values	30°, 45°, 60°
Unit circle	(cos θ, sin θ)
Quadrants	All, Sin, Tan, Cos
Identities	cos2 + sin2 = 1; tan = sin/cos
Equations	principal value, symmetry, period
Applications	gradients, components

Before the exam, make sure you…

•decide the quadrant first
•find all solutions in the interval
•reject impossible values
•consider measurement accuracy

Trigonometric Ratios and Identities on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 2. Topic map for Trigonometric Ratios and Identities.

24 Answers to practice exercises

Check your working against these answers. Full methods are expected in the exam.

Practice exercise 1

1. 8.77

2. 69.5°

Practice exercise 2

1. 1

2. √3/2

3. 1

Practice exercise 3

1. negative; negative

2. 4th

Practice exercise 4

1. −1/2; −√3/2

2. 40°, 140°

Practice exercise 5

1. cos250°

2. 3

Practice exercise 6

1. −0.8; −4/3

2. −3/5

Practice exercise 7

1. cos x

2. 1

Practice exercise 8

1. difference of squares and sin2 + cos2 = 1

2. multiply numerator and denominator by 1 + sin x

Practice exercise 9

1. 120°, 240°

2. 63.4°, 243.4°

3. 63.4°, 243.4°

Practice exercise 10

1. x = 41.6°, 244.7°

Practice exercise 11

1. 197.5°, 342.5°

Practice exercise 12

1. 123.7° (or 56.3° below the horizontal)

2. 4.76°

Practice exercise 13

1. 5.39 m/s on a bearing of 021.8°

Practice exercise 14

1. 30°, 150°, 210°, 330°

2. 1 + tan2x = (cos2x + sin2x)/cos2x

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