DP Maths AA HL - Geometry and Trigonometry - Trigonometric Ratios and Identities

DP Maths AA HL - Geometry and Trigonometry - Trigonometric Ratios and Identities - Full notes

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DP Maths AA HL | Trigonometric Ratios and Identities

Trigonometric Ratios and Identities

This topic contains 2 notes:

3.6 Trig Identities and Equations
3.9 Further Trigonometry

DP MATHEMATICS AA · TOPIC 3: GEOMETRY AND TRIGONOMETRY

3.6

Trigonometric Identities and Equations

Full notes for HL · Syllabus SL 3.6, 3.8

Contents

1	The Pythagorean identity
2	The tangent identity
3	Double angle identities
4	Proving trigonometric identities
5	Solving basic trigonometric equations
6	Equations with multiple angles
7	Quadratic trigonometric equations
8	Equations using identities
9	Solving graphically
10	Using the Pythagorean identity to simplify
11	Double angle identities in practice
12	Proofs of identities: strategies
13	Equations with multiple angles
14	Equations reducing to quadratics
15	Equations with tan
16	Graphical solutions and counting solutions
17	Trigonometric equations in context
18	Mixed review: worked problems
19	Additional exam-style practice
20	Exact values and identities together
21	Checking with technology
22	Exam-style questions with markscheme
23	Common misconceptions and links
24	Check your understanding
25	Summary and topic map
26	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 Quick check.

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 the identity cos2θ + sin2θ = 1
use tan θ = sin θ / cos θ
use the double angle identities for sin 2θ and cos 2θ (three forms)
prove simple trigonometric identities
solve trigonometric equations in a finite interval, finding all solutions
solve equations involving multiple angles such as sin 2x
solve quadratic equations in sin, cos or tan
use identities to rewrite and solve equations, and solve graphically with technology

This booklet covers SL 3.6, 3.8, which is common to SL and HL. HL students extend this topic in booklet 3.9.

Guiding questions

How can one trigonometric expression be rewritten in many equivalent forms? Why does a trigonometric equation usually have many solutions?

How this topic is examined

Identities and equations are central to Paper 1, often in 'show that' and 'hence solve' form: rewrite an equation using an identity to obtain a quadratic, then solve in a given interval. Paper 2 uses technology for equations that cannot be solved exactly.

Always find all solutions in the given interval, give exact values on Paper 1, and check the units (degrees or radians) of the interval.

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
Solve	find the values of the unknowns, using algebra or technology as allowed
Show that	obtain the given result; every step must be shown and the result must not be assumed
Hence / Hence or otherwise	use the previous result (or another valid method)
Sketch	a diagram showing the general shape and key features, labelled
Prove	a formal, rigorous argument from accepted facts
Justify / Explain	give valid reasons or evidence

How to use these notes

Each section teaches one part of the syllabus. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Why it works	a derivation or proof of a key result, so you understand where formulas come from
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 on Paper 2
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); examples are graded basic, standard and exam level
Practice exercise	graded questions for each section; full answers are at the end of the booklet
Quick check	short questions to check the key ideas

1 The Pythagorean identity

From the unit circle.

Think about it

A point on the unit circle has coordinates (cos θ, sin θ). What equation must those coordinates satisfy?

Because (cos θ, sin θ) lies on x2 + y2 = 1, for every θ:

Here cos2θ means (cos θ)2. Rearranged forms are just as useful: sin2θ = 1 − cos2θ and cos2θ = 1 − sin2θ.

WORKED EXAMPLE 1 Finding another ratio (standard)

Given cos θ = −12/13 and π/2 < θ < π, find sin θ.

Solution

sin2θ = 1 − 144/169 = 25/169M1

2nd quadrant, sin positive: sin θ = 5/13A1

Practice exercise 1

1. Given sin θ = 0.6 with θ obtuse, find cos θ.

2. Simplify 1 − cos22x.

3. Simplify (sin θ + cos θ)2 + (sin θ − cos θ)2.

2 The tangent identity

tan = sin/cos.

This is the gradient of the radius on the unit circle. It lets you convert equations such as sin x = √3 cos x into tan x = √3.

Practice exercise 2

1. Solve sin x = cos x for 0 ≤ x ≤ 2π.

2. Given tan θ = 2 with θ acute, find sin θ and cos θ exactly.

3 Double angle identities

Sine and cosine of twice an angle.

The three forms of cos 2θ come from replacing sin2θ or cos2θ using the Pythagorean identity. Choose the form that leaves only one trigonometric function in your equation. These identities are proved from the compound angle formulas (3.9, HL) or from De Moivre's theorem (1.12, HL); at SL they are given in the formula booklet.

WORKED EXAMPLE 2 Using a double angle (standard)

Given sin θ = 1/3 and 0 < θ < π/2, find the exact values of sin 2θ and cos 2θ.

Solution

cos θ = √(1 − 1/9) = 2√2/3A1

sin 2θ = 2(1/3)(2√2/3) = 4√2/9A1

cos 2θ = 1 − 2(1/9) = 7/9A1

Practice exercise 3

1. Given cos x = 3/5, find cos 2x.

2. Write sin 4x in terms of sin 2x and cos 2x.

3. Simplify 2 sin 3x cos 3x.

4. Write cos2x in terms of cos 2x.

4 Proving trigonometric identities

One side to the other.

Start from the more complicated side and use known identities, algebra (common denominators, factorizing, difference of squares) and the Pythagorean identity until you reach the other side. Write each step on a new line.

WORKED EXAMPLE 3 An identity (standard)

Prove that (1 − cos 2θ)/sin 2θ ≡ tan θ.

Solution

LHS = (1 − (1 − 2sin2θ))/(2 sin θ cos θ)M1

= 2sin2θ/(2 sin θ cos θ) = sin θ/cos θA1

= tan θ = RHSAG

WORKED EXAMPLE 4 Using Pythagoras (exam level)

Prove that sin4θ − cos4θ ≡ −cos 2θ.

Solution

LHS = (sin2θ − cos2θ)(sin2θ + cos2θ)M1

= (sin2θ − cos2θ)(1)A1

= −(cos2θ − sin2θ) = −cos 2θ = RHSAG

Practice exercise 4

1. Prove that (sin θ + cos θ)2 ≡ 1 + sin 2θ.

2. Prove that tan θ + 1/tan θ ≡ 2/sin 2θ.

3. Prove that cos4x − sin4x ≡ cos 2x.

5 Solving basic trigonometric equations

Find every solution in the interval.

To solve, for example, sin x = k in an interval: (1) find the principal value from the inverse function (or an exact value); (2) use symmetry to find the second solution in one period (sin x = sin(π − x); cos x = cos(−x), so the second cosine solution is 2π − x; tan repeats every π); (3) add or subtract multiples of the period to find all solutions in the interval. A sketch of the graph or the unit circle prevents missed solutions.

WORKED EXAMPLE 5 Exact solutions (basic)

Solve 2 cos x = −1 for 0 ≤ x ≤ 2π.

Solution

cos x = −½: reference angle π/3M1

cos negative in the 2nd and 3rd quadrantsR1

x = 2π/3, 4π/3A1

WORKED EXAMPLE 6 Degrees with a calculator (standard)

Solve 3 sin x = 2 for 0° ≤ x ≤ 360°.

Solution

sin x = 2/3: x = 41.8°A1

second solution 180° − 41.8° = 138.2°A1

Practice exercise 5

1. Solve tan x = −1 for 0 ≤ x ≤ 2π.

2. Solve sin x = −√3/2 for −π ≤ x ≤ π.

3. Solve 5 cos x = 2 for 0° ≤ x ≤ 360°.

6 Equations with multiple angles

Stretch the interval first.

For sin 2x = k with 0 ≤ x ≤ 2π, let u = 2x: then 0 ≤ u ≤ 4π, which covers two full periods. Find all u in the extended interval, then divide by 2. A multiple angle gives proportionally more solutions. For sin(x + π/6) = k, shift the interval in the same way.

WORKED EXAMPLE 7 Double angle equation (standard)

Solve sin 2x = √3/2 for 0 ≤ x ≤ 2π.

Solution

u = 2x, 0 ≤ u ≤ 4π: u = π/3, 2π/3, 7π/3, 8π/3M1 A1

x = π/6, π/3, 7π/6, 4π/3A1

WORKED EXAMPLE 8 Shifted angle (exam level)

Solve cos(x − π/4) = ½ for 0 ≤ x ≤ 2π.

Solution

u = x − π/4, −π/4 ≤ u ≤ 7π/4: u = π/3, 5π/3M1 A1

x = 7π/12, 23π/12A1

Common mistake

Solving sin 2x = k for x in [0, 2π] by only finding two solutions. The interval for 2x is [0, 4π], giving four.

Practice exercise 6

1. Solve cos 3x = 0 for 0 ≤ x ≤ π.

2. Solve tan(x/2) = 1 for 0 ≤ x ≤ 4π.

3. Solve 2 sin(2x + π/3) = 1 for 0 ≤ x ≤ π.

7 Quadratic trigonometric equations

Factorize as a quadratic.

Equations such as 2sin2x − sin x − 1 = 0 are quadratics in sin x. Let s = sin x, factorize or use the formula, then solve each resulting basic equation. Reject any value outside the range: sin x and cos x must lie between −1 and 1.

WORKED EXAMPLE 9 Quadratic in sine (standard)

Solve 2sin2x − sin x − 1 = 0 for 0 ≤ x ≤ 2π.

Solution

(2 sin x + 1)(sin x − 1) = 0M1

sin x = 1: x = π/2A1

sin x = −½: x = 7π/6, 11π/6A1

Practice exercise 7

1. Solve tan2x = 3 for 0 ≤ x ≤ 2π.

2. Solve 2cos2x + 3 cos x − 2 = 0 for 0° ≤ x ≤ 360°.

3. Solve sin2x = sin x for 0 ≤ x ≤ 2π.

8 Equations using identities

Reduce to a single function.

If an equation contains more than one trigonometric function or angle, use an identity to rewrite it in terms of a single function of a single angle. Typical moves: replace cos2x by 1 − sin2x; replace cos 2x by the form that matches the other term; replace sin 2x by 2 sin x cos x and factorize. Never divide by sin x or cos x, since this loses the solutions where they are zero; factorize instead.

WORKED EXAMPLE 10 Using cos 2x (exam level)

Solve cos 2x = sin x for 0 ≤ x ≤ 2π.

Solution

1 − 2sin2x = sin x, so 2sin2x + sin x − 1 = 0M1 A1

(2 sin x − 1)(sin x + 1) = 0A1

x = π/6, 5π/6, 3π/2A1

WORKED EXAMPLE 11 Using sin 2x and factorizing (exam level)

Solve sin 2x = sin x for 0 ≤ x ≤ 2π.

Solution

2 sin x cos x − sin x = 0, so sin x(2 cos x − 1) = 0M1 A1

sin x = 0: x = 0, π, 2πA1

cos x = ½: x = π/3, 5π/3A1

Common mistake

Dividing both sides of sin 2x = sin x by sin x, which loses x = 0, π and 2π.

Practice exercise 8

1. Solve 2cos2x = 3 sin x for 0 ≤ x ≤ 2π.

2. Solve cos 2x + cos x = 0 for 0 ≤ x ≤ 2π.

3. Show that 3 sin2x + cos 2x = 2 can be written as sin2x = 1, and solve for 0 ≤ x ≤ 2π.

9 Solving graphically

When there is no exact method.

Equations that mix trigonometric and other functions, such as sin x = x/2 or 3 cos x = x2, have no exact solution method. Graph both sides on a GDC (in the correct angle mode) and find the intersections. The number of solutions in an interval is the number of intersections.

WORKED EXAMPLE 12 Graphical solution (standard)

Solve 2 sin x = x − 1 for 0 ≤ x ≤ 2π.

Solution

Graph y = 2 sin x and y = x − 1M1

x = 2.38A1

Figure 1. One intersection in [0, 2π].

Practice exercise 9

1. Solve cos x = x for x ∈ ℝ.

2. How many solutions does sin x = x/5 have for x ≥ 0?

10 Using the Pythagorean identity to simplify

Replace sin2 or cos2.

Expressions containing both sin2θ and cos2θ can often be simplified with cos2θ + sin2θ = 1, or rewritten entirely in terms of one function. Factorizing (difference of squares, common factors) is often the key.

WORKED EXAMPLE 13 Simplifying (standard)

Simplify (1 − cos2x)/(1 − sin2x).

Solution

sin2x/cos2xM1

= tan2xA1

WORKED EXAMPLE 14 Factorizing (standard)

Simplify cos4x − sin4x.

Solution

(cos2x − sin2x)(cos2x + sin2x)M1

= cos 2xA1

Practice exercise 10

1. Simplify sin3x + sin x cos2x.

2. Simplify (sin x + cos x)2 − 1.

3. Simplify (1 − sin x)(1 + sin x)/cos x.

11 Double angle identities in practice

Choosing the right form of cos 2x.

cos 2x = cos2x − sin2x = 2cos2x − 1 = 1 − 2sin2x. Choose the form that matches the other terms in the expression or equation. sin 2x = 2 sin x cos x often produces a common factor.

WORKED EXAMPLE 15 Exact values (standard)

Given cos x = −3/5 with π/2 < x < π, find sin 2x and cos 2x.

Solution

sin x = 4/5A1

sin 2x = 2(4/5)(−3/5) = −24/25A1

cos 2x = 2(9/25) − 1 = −7/25A1

WORKED EXAMPLE 16 Half-angle (exam level)

Find the exact value of cos(π/8).

Solution

cos(π/4) = 2cos2(π/8) − 1M1

cos2(π/8) = (2 + √2)/4A1

cos(π/8) = √(2 + √2)/2A1

Practice exercise 11

1. Write sin2x in terms of cos 2x.

2. Find the exact value of sin2(π/12).

3. Simplify (sin 2x)/(1 + cos 2x).

12 Proofs of identities: strategies

A toolkit.

Useful moves: write everything in sin and cos; use a common denominator; factorize; use cos2 + sin2 = 1; multiply by a conjugate such as (1 + sin x). Work on one side only, and write each step clearly.

WORKED EXAMPLE 17 Conjugate (exam level)

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

Solution

Multiply by (1 + sin x)/(1 + sin x): cos x(1 + sin x)/(1 − sin2x)M1 A1

= cos x(1 + sin x)/cos2x = (1 + sin x)/cos xA1 AG

WORKED EXAMPLE 18 Double angle proof (exam level)

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

Solution

(2cos2x)/(2 sin x cos x)M1 A1

= cos x/sin x = 1/tan xAG

Practice exercise 12

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

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

13 Equations with multiple angles

Extend the interval.

For sin(kx) = c with x in [a, b], solve for u = kx in [ka, kb], then divide by k. The number of solutions increases by a factor of about k.

WORKED EXAMPLE 19 Triple angle (standard)

Solve sin 3x = ½ for 0 ≤ x ≤ π.

Solution

u = 3x ∈ [0, 3π]: u = π/6, 5π/6, 13π/6, 17π/6M1 A1

x = π/18, 5π/18, 13π/18, 17π/18A1

Practice exercise 13

1. Solve cos 2x = −√3/2 for 0 ≤ x ≤ 2π.

2. Solve tan(x/2) = √3 for 0 ≤ x ≤ 4π.

14 Equations reducing to quadratics

Use an identity, then factorize.

Replace cos2x by 1 − sin2x (or vice versa), or cos 2x by the appropriate form, to obtain a quadratic in a single trigonometric function. Factorize, reject impossible values (|sin x| > 1), and solve the basic equations.

WORKED EXAMPLE 20 Quadratic in cosine (exam level)

Solve 2sin2x + 3cos x = 3 for 0 ≤ x ≤ 2π.

Solution

2(1 − cos2x) + 3cos x − 3 = 0: 2cos2x − 3cos x + 1 = 0M1 A1

cos x = 1 or ½A1

x = 0, π/3, 5π/3, 2πA1

Practice exercise 14

1. Solve cos 2x = 3 sin x + 2 for 0 ≤ x ≤ 2π.

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

15 Equations with tan

Divide by cos only when safe.

Equations of the form a sin x = b cos x become tan x = b/a, provided cos x ≠ 0 (check whether cos x = 0 could be a solution). tan has period π, so solutions repeat every π.

WORKED EXAMPLE 21 Tan equation (standard)

Solve √3 sin x = cos x for 0 ≤ x ≤ 2π.

Solution

tan x = 1/√3 (cos x = 0 is not a solution)M1

x = π/6, 7π/6A1

Practice exercise 15

1. Solve sin 2x = cos 2x for 0 ≤ x ≤ π.

2. Solve tan2x = 3 for 0 ≤ x ≤ π.

16 Graphical solutions and counting solutions

Intersections.

Sketch both sides to count the solutions before solving. For equations like sin x = x/3 or cos x = 0.5x, technology gives the numerical solutions.

Figure 2. sin x = x/3 has three solutions.

WORKED EXAMPLE 22 Counting (standard)

Solve sin x = x/3.

Solution

Three intersectionsR1

x = 0, ±2.28A1 A1

Practice exercise 16

1. How many solutions has cos x = x/10 for x ≥ 0?

2. Solve 2 cos x = x.

17 Trigonometric equations in context

Interpreting solutions.

Modelling questions often require solving a trigonometric equation and interpreting the solutions as times. Convert to clock times or dates, and keep only solutions in the stated interval.

WORKED EXAMPLE 23 Tide (exam level)

Depth d(t) = 5 + 2 sin(πt/6), t in hours after midnight. Find when the depth is 6 m in the first 12 hours.

Solution

sin(πt/6) = ½: πt/6 = π/6, 5π/6M1 A1

t = 1 and t = 5 (01:00 and 05:00)A1

Practice exercise 17

1. For the tide, find when the depth is below 4 m in [0, 12].

18 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 24 Mixed problem (exam level)

(a) Show that sin 3x ≡ 3 sin x − 4sin3x (given sin 3x = sin(2x + x)). (b) Hence solve sin 3x = 2 sin x for 0 ≤ x ≤ π.

Solution

(a) sin 2x cos x + cos 2x sin x = 2 sin x cos2x + sin x − 2sin3x = 3 sin x − 4sin3xM1 A1

(b) sin x(1 − 4sin2x) = 0M1

x = 0, π, π/6, 5π/6A1

Practice exercise 18

1. Solve sin x + sin 2x = 0 for 0 ≤ x ≤ 2π.

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

19 Additional exam-style practice

Paper 1 questions.

WORKED EXAMPLE 25 Paper 1 question (exam level)

(a) Show that 3 cos 2x + 7 sin x = 5 can be written as 6sin2x − 7 sin x + 2 = 0. (b) Hence solve for 0 ≤ x ≤ π, giving answers to 3 s.f.

Solution

(a) 3(1 − 2sin2x) + 7 sin x − 5 = 0M1 AG

(b) (2 sin x − 1)(3 sin x − 2) = 0A1

x = π/6, 5π/6, 0.730, 2.41A1 A1

Practice exercise 19

1. Solve 4sin2x = 3 for 0 ≤ x ≤ 2π.

20 Exact values and identities together

Combining skills.

Exam questions often require using one identity to find an exact value, then substituting into another: for example, find cos x from sin x, then sin 2x and tan 2x.

WORKED EXAMPLE 26 Combined (exam level)

Given tan x = 2 with 0 < x < π/2, find sin 2x and cos 2x exactly.

Solution

sin x = 2/√5, cos x = 1/√5A1

sin 2x = 4/5; cos 2x = 1/5 − 4/5 = −3/5A1 A1

Practice exercise 20

1. Given sin x = 1/3 (x acute), find tan 2x exactly.

21 Checking with technology

Graph both sides.

An identity is true for all x, so the graphs of both sides coincide exactly. An equation is true only at the intersections. Graphing both sides is a quick check of either.

WORKED EXAMPLE 27 Check (standard)

Use technology to check that sin 2x = 2 sin x cos x.

Solution

The graphs coincide for all xR1

GDC tip

Graph y = (1 − cos 2x)/sin 2x and y = tan x on the same axes: if they overlap everywhere they are identical (except where undefined).

22 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

Question 1 (Paper 1 style) (about 15 minutes)

(a) Show that cos 2x + 3 sin x = 2 can be written as 2sin2x − 3 sin x + 1 = 0. [2]

(b) Hence solve cos 2x + 3 sin x = 2 for 0 ≤ x ≤ 2π. [5]

Markscheme

Part	Answer	Marks
(a)	1 − 2sin2x + 3 sin x = 2	M1
	2sin2x − 3 sin x + 1 = 0	AG A1
(b)	(2 sin x − 1)(sin x − 1) = 0	M1 A1
	sin x = ½: π/6, 5π/6	A1 A1
	sin x = 1: π/2	A1

Examiner's comments. Finding all three solutions was essential for full marks.

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

Let θ be an angle with cos θ = 2/3 and 3π/2 < θ < 2π.

(a) Find the exact value of sin θ. [2]

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

Markscheme

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

Examiner's comments. The fourth-quadrant sign of sin θ carries through to sin 2θ.

23 Common misconceptions and links

Students often think…	Actually…
sin 2x = 2 sin x	sin 2x = 2 sin x cos x
divide by sin x to simplify	factorize to keep solutions
sin x = k has one solution	usually two per period
cos2x = cos x2	cos2x = (cos x)2

How 3.6 connects to the rest of the course

Topic	Connection
3.4 Unit circle	exact values and quadrants
3.7 Trig functions	graphs and periods
3.9 Further trigonometry (HL)	compound angles
5.6 Calculus	differentiating trig functions

24 Check your understanding

Section A: multiple choice

1. sin240° + cos240° =

A. 1 B. 0 C. 80 D. 2

2. cos 2θ can be written as

A. 2 sin θ cos θ B. 1 + 2sin2θ C. 1 − 2sin2θ D. 2 cos θ

3. sin x = 0.5 in [0, 2π] has

A. 1 solution B. 2 solutions C. 4 solutions D. no solutions

4. sin 2x = 0.5 in [0, 2π] has

A. 2 solutions B. 1 solution C. 8 solutions D. 4 solutions

5. tan θ =

A. cos θ/sin θ B. sin θ/cos θ C. sin θ cos θ D. 1/sin θ

Section B: structured

6. Solve 2 cos2x − cos x = 0 for 0 ≤ x ≤ 2π. [4]

7. Prove that sin 2x/(1 + cos 2x) ≡ tan x. [3]

Answers

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

6. cos x(2 cos x − 1) = 0: π/2, 3π/2, π/3, 5π/3.

7. 2 sin x cos x/(2cos2x) = tan x.

25 Summary and topic map

Idea	Remember
Pythagorean	cos2θ + sin2θ = 1
Tangent	tan θ = sin θ/cos θ
Double angle	sin 2θ = 2 sin θ cos θ; cos 2θ three forms
Basic equations	principal value, symmetry, period
Multiple angles	extend the interval
Quadratics	factorize in sin or cos
Identities in equations	reduce to one function; factorize

Before the exam, make sure you…

•find every solution in the interval
•never divide by a trig function
•choose the right form of cos 2θ
•check calculator mode

Trigonometric Identities and Equations 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 3. Topic map for Trigonometric Identities and Equations.

26 Answers to practice exercises

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

Practice exercise 1

1. −0.8

2. sin22x

3. 2

Practice exercise 2

1. π/4, 5π/4

2. sin θ = 2/√5, cos θ = 1/√5

Practice exercise 3

1. −7/25

2. 2 sin 2x cos 2x

3. sin 6x

4. (1 + cos 2x)/2

Practice exercise 4

1. expand: sin2 + cos2 + 2 sin cos

2. (sin2 + cos2)/(sin cos) = 1/(½ sin 2θ)

3. difference of squares, then cos2 − sin2

Practice exercise 5

1. 3π/4, 7π/4

2. −π/3, −2π/3

3. 66.4°, 293.6°

Practice exercise 6

1. π/6, π/2, 5π/6

2. π/2, 5π/2

3. u = 2x + π/3 ∈ [π/3, 7π/3]: u = 5π/6, 13π/6: x = π/4, 11π/12

Practice exercise 7

1. π/3, 2π/3, 4π/3, 5π/3

2. cos x = ½ (cos x = −2 rejected): 60°, 300°

3. 0, π/2, π, 2π

Practice exercise 8

1. 2 − 2sin2x = 3 sin x: sin x = ½: π/6, 5π/6

2. 2cos2x + cos x − 1 = 0: cos x = ½ or −1: π/3, π, 5π/3

3. 3s2 + 1 − 2s2 = 2 gives s2 = 1: π/2, 3π/2

Practice exercise 9

1. x = 0.739

2. 2 (x = 0 and x = 2.60), since x/5 > 1 for x > 5

Practice exercise 10

1. sin x

2. sin 2x

3. cos x

Practice exercise 11

1. (1 − cos 2x)/2

2. (2 − √3)/4

3. tan x

Practice exercise 12

1. common denominator; sin x(1 + sin x) + cos2x = 1 + sin x

2. (2 sin x cos x)/(2cos2x) = tan x

Practice exercise 13

1. 5π/12, 7π/12, 17π/12, 19π/12

2. 2π/3, 8π/3

Practice exercise 14

1. 1 − 2s2 = 3s + 2: 2s2 + 3s + 1 = 0: s = −1/2, −1: 7π/6, 3π/2, 11π/6

2. 2 − 2s2 = s + 1: s = ½ or −1: π/6, 5π/6, 3π/2

Practice exercise 15

1. tan 2x = 1: π/8, 5π/8

2. π/3, 2π/3

Practice exercise 16

1. 3 (x = 1.43, 5.27, 7.07); for x > 10 the line is above 1

2. 1.03

Practice exercise 17

1. sin(πt/6) < −½: 7 < t < 11

Practice exercise 18

1. sin x(1 + 2cos x) = 0: 0, 2π/3, π, 4π/3, 2π

2. cross-multiply: sin2x = 1 − cos2x

Practice exercise 19

1. π/3, 2π/3, 4π/3, 5π/3

Practice exercise 20

1. cos x = 2√2/3: sin 2x = 4√2/9, cos 2x = 7/9: tan 2x = 4√2/7

DP MATHEMATICS AA · TOPIC 3: GEOMETRY AND TRIGONOMETRY

3.9

Further Trigonometry

Full notes for HL · Syllabus AHL 3.9, 3.10, 3.11

Contents

1	Reciprocal trigonometric functions
2	Further Pythagorean identities
3	Inverse trigonometric functions
4	Compound angle identities
5	Double angle identity for tangent
6	Proving and using identities
7	Solving equations with the further identities
8	Symmetry properties of trigonometric graphs
9	Graphs of sec, cosec and cot
10	Identities with reciprocal functions
11	Equations with reciprocal functions
12	Inverse trigonometric functions: graphs
13	Evaluating inverse trigonometric expressions
14	Compound angle identities in practice
15	R sin(x + α) form
16	Double angle formula for tan
17	Symmetry properties and transformations
18	Mixed review: worked problems
19	Additional exam-style practice
20	Inverse trig functions in calculus and geometry
21	Proving identities with compound angles
22	Exam-style questions with markscheme
23	Common misconceptions and links
24	Check your understanding
25	Summary and topic map
26	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 Quick check.

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
define sec, cosec and cot and sketch their graphs
use the identities 1 + tan2θ = sec2θ and 1 + cot2θ = cosec2θ
use arcsin, arccos and arctan with their domains and ranges, and sketch their graphs
use the compound angle identities for sin, cos and tan
use the double angle identity for tan
prove identities and solve equations using these results
use symmetry properties such as sin(π − x) = sin x and cos(π/2 − x) = sin x

This booklet covers Additional Higher Level (AHL) content (AHL 3.9, 3.10, 3.11). It is studied by HL students only.

Guiding questions

How do reciprocal and inverse trigonometric functions extend the toolkit? How can sines and cosines of sums be expressed in terms of single angles?

How this topic is examined

This AHL content appears mainly on Paper 1, typically in 'show that' identities, exact-value questions (such as tan(π/12)), and equations solved using sec2θ = 1 + tan2θ. Inverse trigonometric functions also appear in calculus (5.14, 5.16).

State the domain and range when sketching inverse trig functions, and give exact values on Paper 1.

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
Solve	find the values of the unknowns, using algebra or technology as allowed
Show that	obtain the given result; every step must be shown and the result must not be assumed
Hence / Hence or otherwise	use the previous result (or another valid method)
Sketch	a diagram showing the general shape and key features, labelled
Prove	a formal, rigorous argument from accepted facts
Justify / Explain	give valid reasons or evidence

How to use these notes

Each section teaches one part of the syllabus. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Why it works	a derivation or proof of a key result, so you understand where formulas come from
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 on Paper 2
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); examples are graded basic, standard and exam level
Practice exercise	graded questions for each section; full answers are at the end of the booklet
Quick check	short questions to check the key ideas

1 Reciprocal trigonometric functions

Secant, cosecant and cotangent.

Think about it

tan x has asymptotes where cos x = 0. Where will the graph of 1/cos x have asymptotes, and what is its minimum positive value?

Each reciprocal function has vertical asymptotes where the original function is zero. sec x has asymptotes at x = π/2 + kπ and range |sec x| ≥ 1; cosec x has asymptotes at x = kπ and range |cosec x| ≥ 1; cot x has asymptotes at x = kπ and range ℝ. Where the original function has a maximum of 1, the reciprocal has a local minimum of 1, and vice versa.

Figure 1. y = sec x has asymptotes where cos x = 0.

WORKED EXAMPLE 1 Exact values (basic)

Find the exact values of sec(π/3), cosec(5π/6) and cot(3π/4).

Solution

sec(π/3) = 1/(1/2) = 2A1

cosec(5π/6) = 1/(1/2) = 2A1

cot(3π/4) = 1/(−1) = −1A1

Practice exercise 1

1. Find the exact value of cosec(π/4) and sec(5π/4).

2. State the period and asymptotes of y = cot x for 0 ≤ x ≤ 2π.

3. Solve sec x = 2 for 0 ≤ x ≤ 2π.

2 Further Pythagorean identities

Divide by cos2 or sin2.

Why it works

Divide cos2θ + sin2θ = 1 by cos2θ: 1 + tan2θ = sec2θ. Divide instead by sin2θ: cot2θ + 1 = cosec2θ.

WORKED EXAMPLE 2 Using the identity (standard)

Given tan θ = 3/4 and π < θ < 3π/2, find sec θ and sin θ.

Solution

sec2θ = 1 + 9/16 = 25/16M1

3rd quadrant: cos negative, sec θ = −5/4A1

cos θ = −4/5, sin θ = tan θ cos θ = −3/5A1

Practice exercise 2

1. Simplify sec2x − tan2x.

2. Given cosec θ = 3 with θ obtuse, find cot θ.

3. Show that (sec x − tan x)(sec x + tan x) = 1.

3 Inverse trigonometric functions

Restricted domains make them one-to-one.

Sine, cosine and tangent are not one-to-one, so to define inverses we restrict their domains (2.5):

Function	Domain	Range
arcsin x (sin−1 x)	−1 ≤ x ≤ 1	−π/2 ≤ y ≤ π/2
arccos x (cos−1 x)	−1 ≤ x ≤ 1	0 ≤ y ≤ π
arctan x (tan−1 x)	x ∈ ℝ	−π/2 < y < π/2

The graph of each inverse is the reflection of the restricted original in y = x. arctan x has horizontal asymptotes y = ±π/2. Note that arcsin(sin x) = x only for x in the range of arcsin: arcsin(sin(5π/6)) = π/6, not 5π/6.

Figure 2. The inverse trigonometric functions on [−1, 1].

WORKED EXAMPLE 3 Exact values and composition (standard)

Find (a) arccos(−½); (b) arctan(−√3); (c) sin(arccos(3/5)).

Solution

(a) in [0, π]: 2π/3A1

(b) in (−π/2, π/2): −π/3A1

Practice exercise 3

1. Find arcsin(−√2/2).

2. Find tan(arcsin(5/13)).

3. Explain why arccos(cos(4π/3)) ≠ 4π/3 and find its value.

4. Sketch y = arctan x, labelling the asymptotes.

4 Compound angle identities

Sines and cosines of sums.

Why it works

Place points P(cos A, sin A) and Q(cos B, sin B) on the unit circle. The angle POQ is A − B. Computing PQ2 with the distance formula gives 2 − 2(cos A cos B + sin A sin B); computing it with the cosine rule gives 2 − 2 cos(A − B). Equating proves cos(A − B) = cos A cos B + sin A sin B. The other identities follow by replacing B with −B or using sin x = cos(π/2 − x). They can also be derived from Euler's form (1.12): ei(A+B) = eiAeiB.

WORKED EXAMPLE 4 Exact value (standard)

Find the exact value of cos(π/12).

Solution

cos(π/3 − π/4) = cos(π/3)cos(π/4) + sin(π/3)sin(π/4)M1

= (1/2)(√2/2) + (√3/2)(√2/2)A1

= (√2 + √6)/4A1

WORKED EXAMPLE 5 Tangent of a difference (exam level)

Find the exact value of tan(π/12).

Solution

tan(π/3 − π/4) = (√3 − 1)/(1 + √3)M1

Multiply by (1 − √3)/(1 − √3): (√3 − 1)(1 − √3)/(1 − 3)M1

= (4 − 2√3)/2 = 2 − √3A1

WORKED EXAMPLE 6 Given ratios (exam level)

Given sin A = 3/5 (A acute) and cos B = −12/13 (B obtuse), find sin(A + B).

Solution

cos A = 4/5; sin B = 5/13A1

sin(A + B) = (3/5)(−12/13) + (4/5)(5/13)M1

= (−36 + 20)/65 = −16/65A1

Practice exercise 4

1. Find the exact value of sin(75°).

2. Simplify sin 3x cos x − cos 3x sin x.

3. Find tan(A + B) given tan A = 2 and tan B = 3.

4. Show that sin(x + π/2) = cos x.

5 Double angle identity for tangent

Set B = A.

Setting B = A in the compound angle identities gives all the double angle identities: sin 2A = 2 sin A cos A, cos 2A = cos2A − sin2A, and tan 2A as above.

WORKED EXAMPLE 7 Double angle for tan (standard)

Given tan θ = ½, find tan 2θ and tan 4θ.

Solution

tan 2θ = 1/(1 − 1/4) = 4/3A1

tan 4θ = (8/3)/(1 − 16/9) = (8/3)/(−7/9) = −24/7M1 A1

Practice exercise 5

1. Given tan x = 3, find tan 2x.

2. Find the exact value of tan(π/8) using tan(π/4) = 2t/(1 − t2).

6 Proving and using identities

Choose identities that match.

WORKED EXAMPLE 8 Proof with reciprocal functions (standard)

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

Solution

LHS = 2cos2x/(2 sin x cos x)M1

= cos x/sin x = cot xA1 AG

WORKED EXAMPLE 9 Triple angle (exam level)

Show that sin 3x ≡ 3 sin x − 4 sin3x.

Solution

sin(2x + x) = sin 2x cos x + cos 2x sin xM1

= 2 sin x cos2x + (1 − 2sin2x) sin xA1

= 2 sin x(1 − sin2x) + sin x − 2sin3x = 3 sin x − 4sin3xA1 AG

Practice exercise 6

1. Prove that sec2x + cosec2x ≡ sec2x cosec2x.

2. Prove that cos(A + B)cos(A − B) ≡ cos2A − sin2B.

7 Solving equations with the further identities

Reduce to one function.

WORKED EXAMPLE 10 Using the secant identity (exam level)

Solve 2 tan2x = 3 sec x for 0 ≤ x ≤ 2π.

Solution

2(sec2x − 1) = 3 sec x, so 2sec2x − 3 sec x − 2 = 0M1 A1

(2 sec x + 1)(sec x − 2) = 0; sec x = −½ impossibleA1 R1

sec x = 2: cos x = ½, x = π/3, 5π/3A1

WORKED EXAMPLE 11 Using a compound angle (exam level)

Solve sin(x + π/6) = 2 cos x for 0 ≤ x ≤ π.

Solution

sin x (√3/2) + cos x (1/2) = 2 cos xM1

(√3/2) sin x = (3/2) cos x, tan x = √3A1

x = π/3A1

Practice exercise 7

1. Solve cot2x = cosec x + 1 for 0 ≤ x ≤ 2π.

2. Solve tan 2x = tan x for 0 < x < π, x ≠ π/2.

8 Symmetry properties of trigonometric graphs

Reflections and translations.

The graphs' symmetries give useful identities. sin is odd and cos is even: sin(−x) = −sin x, cos(−x) = cos x. Complementary angles: sin(π/2 − x) = cos x and cos(π/2 − x) = sin x. Supplementary angles: sin(π − x) = sin x and cos(π − x) = −cos x. Half-turn: sin(π + x) = −sin x, cos(π + x) = −cos x, tan(π + x) = tan x.

Practice exercise 8

1. Simplify cos(π − x) + cos(−x).

2. Show that tan(π/2 − x) = cot x.

9 Graphs of sec, cosec and cot

Reciprocal features.

sec x = 1/cos x has asymptotes where cos x = 0 and turning points at (0, 1), (π, −1); cosec x = 1/sin x has asymptotes at x = kπ and turning points at (π/2, 1), (3π/2, −1); cot x has asymptotes at x = kπ and zeros at π/2 + kπ. None of sec or cosec takes values between −1 and 1.

Figure 3. cosec x and cot x.

WORKED EXAMPLE 12 Range (standard)

State the range of y = 3 sec x − 1.

Solution

sec x ≤ −1 or sec x ≥ 1M1

y ≤ −4 or y ≥ 2A1

Practice exercise 9

1. State the period and asymptotes of y = cosec 2x for 0 ≤ x ≤ π.

2. Solve cosec x = 2 for 0 ≤ x ≤ 2π.

10 Identities with reciprocal functions

Pythagorean variants.

Use 1 + tan2x = sec2x and 1 + cot2x = cosec2x to simplify and to solve equations. Writing everything in terms of sin and cos is often the safest approach.

WORKED EXAMPLE 13 Proof (exam level)

Prove that (sec x − cos x)/sin x ≡ tan x.

Solution

(1/cos x − cos x)/sin x = (1 − cos2x)/(sin x cos x)M1 A1

= sin2x/(sin x cos x) = tan xAG

Practice exercise 10

1. Prove that cosec2x − cot2x ≡ 1.

2. Simplify sec4x − tan4x.

11 Equations with reciprocal functions

Reduce to one function.

Replace tan2 by sec2 − 1 (or cot2 by cosec2 − 1) to obtain a quadratic in one function. Reject values that are impossible (|sec x| < 1 or |cosec x| < 1).

WORKED EXAMPLE 14 Quadratic in sec (exam level)

Solve tan2x + 3 sec x + 3 = 0 for 0 ≤ x ≤ 2π.

Solution

sec2x + 3 sec x + 2 = 0M1 A1

sec x = −1 or −2A1

cos x = −1: π; cos x = −½: 2π/3, 4π/3: x = 2π/3, π, 4π/3A1

Practice exercise 11

1. Solve 2cot2x = 3 cosec x for 0 ≤ x ≤ 2π.

2. Solve sec2x = 4 for 0 ≤ x ≤ π.

12 Inverse trigonometric functions: graphs

Domains and ranges.

arcsin: domain [−1, 1], range [−π/2, π/2]; arccos: domain [−1, 1], range [0, π]; arctan: domain ℝ, range (−π/2, π/2), with horizontal asymptotes y = ±π/2.

Figure 4. arcsin and arccos.

WORKED EXAMPLE 15 Transformed inverse (exam level)

State the domain and range of y = 2 arcsin(x − 1).

Solution

−1 ≤ x − 1 ≤ 1: 0 ≤ x ≤ 2A1

−π ≤ y ≤ πA1

Practice exercise 12

1. State the range of y = arctan(2x) + π/2.

2. Find the domain of y = arccos(3x).

13 Evaluating inverse trigonometric expressions

Use a triangle.

To find sin(arccos(3/5)), let θ = arccos(3/5), so cos θ = 3/5 with θ in [0, π]; draw a right triangle 3, 4, 5 and read sin θ = 4/5.

WORKED EXAMPLE 16 Composite (standard)

Find tan(arcsin(−12/13)).

Solution

θ in [−π/2, 0], sin θ = −12/13, cos θ = 5/13M1 A1

tan θ = −12/5A1

WORKED EXAMPLE 17 Sum of inverse functions (exam level)

Show that arctan(1/2) + arctan(1/3) = π/4.

Solution

tan(A + B) = (1/2 + 1/3)/(1 − 1/6) = 1M1 A1

A + B in (0, π): A + B = π/4R1 AG

Practice exercise 13

1. Find cos(arctan 2).

2. Find arcsin(sin(2π/3)).

14 Compound angle identities in practice

Exact values and simplification.

Use sin(A ± B), cos(A ± B) and tan(A ± B) to find exact values of angles such as 15°, 75°, 105°, and to simplify expressions like sin x cos 2x + cos x sin 2x.

WORKED EXAMPLE 18 Exact value (standard)

Find the exact value of cos 105°.

Solution

cos(60° + 45°) = (1/2)(√2/2) − (√3/2)(√2/2)M1

= (√2 − √6)/4A1

WORKED EXAMPLE 19 Simplifying (standard)

Simplify cos 3x cos x + sin 3x sin x.

Solution

cos(3x − x)M1

= cos 2xA1

Practice exercise 14

1. Find tan 75° exactly.

2. Simplify sin(x + π/3) + sin(x − π/3).

15 R sin(x + α) form

Combining sine and cosine.

a sin x + b cos x can be written R sin(x + α) with R = √(a2 + b2) and tan α = b/a (α chosen in the right quadrant). This gives the maximum (R), minimum (−R), and solves equations of the form a sin x + b cos x = c.

WORKED EXAMPLE 20 Combined form (exam level)

Write 3 sin x + 4 cos x as R sin(x + α) and solve 3 sin x + 4 cos x = 2 for 0 ≤ x ≤ 2π.

Solution

R = 5, tan α = 4/3: α = 0.927A1 A1

sin(x + 0.927) = 0.4: x + 0.927 = 2.730 or 6.695M1

x = 1.80 or 5.77A1

Practice exercise 15

1. Find the maximum and minimum of 5 sin x − 12 cos x.

2. Write sin x + √3 cos x as R sin(x + α).

16 Double angle formula for tan

Applications.

tan 2A = 2 tan A/(1 − tan2A). It gives exact values such as tan(π/8) and solves equations with tan 2x and tan x.

WORKED EXAMPLE 21 Equation (exam level)

Solve tan 2x = 3 tan x for 0 < x < π, x ≠ π/2.

Solution

2t/(1 − t2) = 3tM1

t = 0 or 2 = 3 − 3t2: t2 = 1/3A1

t = 0 gives no solution in (0, π); x = π/6, 5π/6A1

Practice exercise 16

1. Given tan x = 1/2, find tan 2x and tan 4x.

17 Symmetry properties and transformations

Relating trig functions.

Identities such as sin(π/2 − x) = cos x, cos(π − x) = −cos x and tan(π + x) = tan x correspond to reflections and translations of the graphs. They are useful for simplifying expressions and for finding additional solutions of equations.

WORKED EXAMPLE 22 Using symmetry (standard)

Given sin(π/7) = k, express sin(6π/7) and cos(5π/14) in terms of k.

Solution

sin(π − π/7) = kA1

cos(π/2 − π/7) = sin(π/7) = kA1

Practice exercise 17

1. Simplify sin(π + x) + cos(π/2 + x).

18 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 23 Mixed problem (exam level)

(a) Show that sec x − tan x ≡ 1/(sec x + tan x). (b) Hence solve sec x + tan x = 2 for 0 ≤ x < π/2.

Solution

(a) (sec x − tan x)(sec x + tan x) = sec2x − tan2x = 1M1 AG

(b) sec x − tan x = 1/2; add: 2 sec x = 5/2M1 A1

cos x = 4/5: x = 0.644A1

Practice exercise 18

1. Solve sin(2x + π/6) = cos x for 0 ≤ x ≤ π/2.

2. Find the exact value of sin(arctan(1/2) + arctan(1/3)).

19 Additional exam-style practice

Paper 1 and Paper 2.

WORKED EXAMPLE 24 Paper 1 question (exam level)

Given sin A = 4/5 (A acute) and cos B = −5/13 (B obtuse), find tan(A − B) exactly.

Solution

tan A = 4/3, tan B = −12/5A1 A1

(4/3 + 12/5)/(1 − 16/5) = (56/15)/(−11/5)M1

= −56/33A1

Practice exercise 19

1. Solve 3 sec2x = 4 for 0 ≤ x ≤ 2π.

20 Inverse trig functions in calculus and geometry

Angles from ratios.

Inverse trigonometric functions express angles in terms of lengths: the angle subtended by an object of height h at distance x is arctan(h/x). Their derivatives (5.14) and the integrals giving them (5.16) are important in later topics.

WORKED EXAMPLE 25 Viewing angle (exam level)

A screen 3 m tall has its base 1 m above eye level. Write the viewing angle θ at distance x and find θ when x = 4.

Solution

θ = arctan(4/x) − arctan(1/x)A1

x = 4: 0.785 − 0.245 = 0.540 rad (30.96°)A1

Practice exercise 20

1. Find x that maximizes the viewing angle (use technology).

21 Proving identities with compound angles

Expand and simplify.

Expand each compound angle, then simplify using Pythagorean identities. Products such as sin(A + B) sin(A − B) often simplify to differences of squares.

WORKED EXAMPLE 26 Product (exam level)

Prove that sin(A + B) sin(A − B) ≡ sin2A − sin2B.

Solution

(sin A cos B)2 − (cos A sin B)2M1 A1

= sin2A(1 − sin2B) − (1 − sin2A)sin2B = sin2A − sin2BA1 AG

Practice exercise 21

1. Prove that tan(π/4 + x) − tan(π/4 − x) ≡ 2 tan 2x.

22 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

Question 1 (Paper 1 style) (about 15 minutes)

(a) Show that tan(π/4 + x) = (1 + tan x)/(1 − tan x). [2]

(b) Hence find the exact value of tan(7π/12). [3]

Markscheme

Part	Answer	Marks
(a)	(tan(π/4) + tan x)/(1 − tan(π/4)tan x)	M1
	tan(π/4) = 1 gives the result	AG A1
(b)	x = π/3: (1 + √3)/(1 − √3)	M1
	rationalize: (1 + √3)2/(1 − 3)	M1
	−2 − √3	A1
(c)	1 + t = 3t(1 − t), so 3t2 − 2t + 1 = 0	M1 A1
	discriminant 4 − 12 < 0	R1
	no solutions	A1

Examiner's comments. In (c), recognizing that the quadratic has no real roots, and stating so clearly, earns the reasoning mark.

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

Let f(x) = arccos(2x − 1).

(a) Find the domain and range of f. [3]

(b) Find the exact value of f(1/4). [2]

Markscheme

Part	Answer	Marks
(a)	−1 ≤ 2x − 1 ≤ 1	M1
	0 ≤ x ≤ 1	A1
	0 ≤ f(x) ≤ π	A1
(b)	arccos(−½)	M1
	2π/3	A1
(c)	2x − 1 = ½	M1
	x = 3/4	A1

Examiner's comments. The range of arccos must be stated as [0, π].

23 Common misconceptions and links

Students often think…	Actually…
sec x = 1/sin x	sec x = 1/cos x
arcsin(sin x) = x for all x	only for −π/2 ≤ x ≤ π/2
sin(A + B) = sin A + sin B	use the compound angle formula
sin−1x = 1/sin x	sin−1 means arcsin; 1/sin x is cosec x

How 3.9 connects to the rest of the course

Topic	Connection
3.6 Identities	Pythagorean and double angle
2.5 Inverse functions	restricted domains
1.12 Complex numbers	deriving identities
5.14 Further differentiation	derivatives of inverse trig functions

24 Check your understanding

Section A: multiple choice

1. sec(π/3) =

A. 2 B. 1/2 C. √3 D. 2/√3

2. 1 + tan2x =

A. cosec2x B. cot2x C. sec2x D. 1

3. The range of arcsin is

A. [0, π] B. [−π/2, π/2] C. [−1, 1] D. ℝ

4. cos(A + B) =

A. cos A cos B + sin A sin B B. cos A + cos B C. sin A cos B + cos A sin B D. cos A cos B − sin A sin B

5. arctan x has asymptotes

A. x = ±π/2 B. y = ±π/2 C. x = ±1 D. y = 0

Section B: structured

6. Find the exact value of sin(π/12). [3]

7. Solve 3 sec2x = 5 tan x + 1 for 0 ≤ x ≤ π, to 3 s.f. [4]

Answers

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

6. sin(π/3 − π/4) = (√6 − √2)/4.

7. 3tan2x − 5 tan x + 2 = 0: tan x = 1 or 2/3: x = 0.785, 0.588.

25 Summary and topic map

Idea	Remember
Reciprocals	sec = 1/cos, cosec = 1/sin, cot = 1/tan
Pythagorean	1 + tan2 = sec2; 1 + cot2 = cosec2
Inverse functions	arcsin [−π/2, π/2]; arccos [0, π]; arctan (−π/2, π/2)
Compound angles	sin(A ± B), cos(A ± B), tan(A ± B)
tan 2θ	2 tan θ/(1 − tan2θ)
Symmetry	odd sine, even cosine, complementary angles

Before the exam, make sure you…

•learn the ranges of inverse functions
•use compound angles for exact values
•reduce equations to one function
•reject impossible values (|sec x| < 1)

Further Trigonometry 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 5. Topic map for Further Trigonometry.

26 Answers to practice exercises

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

Practice exercise 1

1. √2; −√2

2. period π; x = 0, π, 2π

3. π/3, 5π/3

Practice exercise 2

1. 1

2. cot2θ = 8, obtuse so cot θ = −2√2

3. difference of squares: sec2x − tan2x = 1

Practice exercise 3

1. −π/4

2. 5/12

3. the range of arccos is [0, π]: value 2π/3

4. asymptotes y = ±π/2, through (0, 0)

Practice exercise 4

1. (√6 + √2)/4

2. sin 2x

3. (2 + 3)/(1 − 6) = −1

4. sin x cos(π/2) + cos x sin(π/2) = cos x

Practice exercise 5

1. −3/4

2. t2 + 2t − 1 = 0: t = √2 − 1 (positive)

Practice exercise 6

1. common denominator: (sin2 + cos2)/(cos2 sin2)

2. expand: cos2A cos2B − sin2A sin2B, then replace cos2B and sin2A

Practice exercise 7

1. cosec2x − cosec x − 2 = 0: cosec x = 2 or −1: π/6, 5π/6, 3π/2

2. t(1 + t2) = 0, so t = 0, giving x = 0 or π, which are outside (0, π): no solutions

Practice exercise 8

1. 0

2. sin(π/2 − x)/cos(π/2 − x) = cos x/sin x

Practice exercise 9

1. π; x = 0, π/2, π

2. π/6, 5π/6

Practice exercise 10

1. rearrange 1 + cot2 = cosec2

2. (sec2 − tan2)(sec2 + tan2) = sec2x + tan2x

Practice exercise 11

1. 2cosec2x − 3cosec x − 2 = 0: cosec x = 2: π/6, 5π/6

2. π/3, 2π/3

Practice exercise 12

1. 0 < y < π

2. −1/3 ≤ x ≤ 1/3

Practice exercise 13

1. 1/√5

2. π/3

Practice exercise 14

1. 2 + √3

2. sin x

Practice exercise 15

1. 13; −13

2. 2 sin(x + π/3)

Practice exercise 16

1. 4/3; −24/7

Practice exercise 17

1. −2 sin x

Practice exercise 18

1. cos x = sin(π/2 − x): x = π/9 or π/3

2. sin(π/4) = √2/2

Practice exercise 19

1. cos x = ±√3/2: π/6, 5π/6, 7π/6, 11π/6

Practice exercise 20

1. x = 2 m

Practice exercise 21

1. combine the two fractions: 4t/(1 − t2)

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