e-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersityChapter 2: Logarithmic and exponential functions
When y = −5 There are no solutions to this equation since 3x
3x = −5 is always positive.
Hence, the solution is x = 0.369 to 3 significant figures.
EXERCISE 2E
1 Solve, giving your answers correct to 3 significant figures.
Review a 5x = 18 b 2x = 35 c 32x = 8 d 2x+1 = 25
e 32x−5 = 20 f 3x = 2x+1 g 5x+3 = 74−3x h 41−3x = 3x−2
i 2x = 2(5x ) j 4x = 7(3x ) k 2x+1 = 3(5x ) l 5(2x−3 ) = 4(3x+4 )
2 a Show that 2x+1 + 6(2x−1) = 12 can be written as 2(2x ) + 3(2x ) = 12.
b Hence solve the equation 2x+1 + 6(2x−1) = 12, giving your answer correct to
3 significant figures.
3 Solve, giving your answers correct to 3 significant figures.
a 2x+2 + 2x = 22 b 3x+1 = 3x−1 + 32 c 2x+1 + 5(2x−1) = 24
d 5x + 53 = 5x+2 e 4x−1 = 4x − 43 f 3x+1 − 2(3x−2 ) = 5
4 Use the substitution y = 2x to solve the equation 22x + 32 = 12(2x ). 39
Review 5 Solve, giving your answers correct to 3 significant figures.
a 52x − 5(5x ) + 6 = 0 b 22x + 5 = 6 × 2x
c 32x = 6 × 3x + 7 d 42x + 27 = 12(4x )
6 Use the substitution u = 2x to solve the equation 22x − 5(2x+1) + 24 = 0.
7 Solve, giving your answers correct to 3 significant figures.
a 22x − 2x+1 − 35 = 0 b 32x − 3x+1 = 10
c 2(5x+1) − 52x = 16 d 42x+1 = 17(4x ) − 15
8 Solve, giving your answers correct to 3 significant figures.
a 4x + 2x − 12 = 0 b 3(16x ) − 10(4x ) + 3 = 0
c 4x + 15 = 4(2x+1) d 2(9x ) = 3x+1 + 27
9 For each of the following equations find the value of x correct to 3significant figures.
y
Review a 3x = 7y b 7x = (2.7)y c 42x = 35y
10 Solve the equation x2.5 = 20x1.25, giving your answers in exact form.
11 Solve, giving your answers correct to 3 significant figures.
a 4x − 8 = 2 b 2(3x ) − 4 = 8
c 3 2x − 5 = 2x d 3x + 4 = 3x − 9
e 5(2x ) + 3 = 5(2x ) − 10 f 2x+2 + 1 = 2x + 12
( )g 32 x = 6 3 x + 16 ( )h 4 x = 5 2 x + 14
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PS 12 Given that 24x+1 × 32−x = 8x × 35−2x, find the value of: WEB LINK
a 6x bx Try the To log or not
to log? resource on
PS 13 Solve the equation 8(8x−1 − 1) = 7(4x − 2x+1). the Underground
Mathematics website.
Review2.6 Solving exponential inequalities
To solve inequalities such as 2x . 3 we take the logarithm of both sides of the inequality.
We can take logarithms to any base but care must be taken with the inequality symbol
when doing so.
● If the base of the logarithm is greater than 1 then the inequality symbol remains the same.
● If the base of the logarithm is between 0 and 1 then the inequality symbol must be
reversed.
This is because y = loga x is an increasing function if a . 1 and it is a decreasing
function when 0 , a , 1.
Hence, when taking logarithms to base 10 of both sides of an inequality the inequality
symbol is unchanged.
It is also important to remember that if we divide both sides of an inequality by a negative
number then the inequality symbol must be reversed.
WORKED EXAMPLE 2.13
40
Solve the inequality 0.6x , 0.7, giving your answer in terms of base 10 logarithms.
Review Answer
0.6x , 0.7 Take logs to base 10 of both sides, inequality
symbol is unchanged.
log 0.6x , log 0.7
Use the power rule for logs.
x log 0.6 , log 0.7
Divide both sides by log10 0.6.
x . log 0.7
log 0.6 log 0.6 is negative so the inequality symbol is
reversed.
x . 0.698 (correct to 3 significant figures)
Review
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WORKED EXAMPLE 2.14
Solve the inequality 4 × 32x−1 . 5, giving your answer in terms of base 10 logarithms.
Answer
4 × 32x−1 . 5 Divide both sides by 4.
Take logs to base 10 of both sides and use the
32x−1 . 5 power rule.
4 Expand brackets.
Rearrange.
(2x − 1) log 3 . log 5
4 Simplify.
Review
( 2 log 3 ) x − log 3 . log 5
4
log 5 + log 3
4
x .
2 log 3
log 15
4
x. log 9
EXERCISE 2F
1 Solve the inequalities, giving your answer in terms of base 10 logarithms. 41
2 x
3
a 2x , 5 b 5x ù 7 c . 3 d 0.8x , 0.3
Review 2 Solve the inequalities, giving your answer in terms of base 10 logarithms.
3−x
a 85−x , 10 b 32x+5 . 20 c 2 × 52x+1 ø 3 7 × 5
d 6 . 4
P 3 Solve 5x2 . 2x, giving your answer in terms of logarithms. x ù log 15 .
PS 4 Prove that the solution to the inequality 32x−1 × 21−3x ù 5 is 2
5 In this question you are not allowed to use a calculator.
log 9
8
You are given that log10 4 = 0.60206 correct to 5 decimal places and that
100.206 , 2.
a Find the number of digits in the number 4100.
b Find the first digit in the number 4100.
Review
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2.7 Natural logarithms FAST FORWARD
There is another type of logarithm to a special base called e. The number e is a very
The number e is an irrational number and e ≈ 2.718. important number in
The function y = ex is called the natural exponential function. Mathematics as it has
Logarithms to the base of e are called natural logarithms. very special properties.
ln x is used to represent loge x. You will learn about
these special properties
KEY POINT 2.5 in Chapters 4 and 5.
Review If y = ex then x = ln y
y y = ex
y=x
1 y = ln x
O1 x
The curve y = ln x is the reflection of y = ex in the line y = x.
42 y = ln x and y = ex are inverse functions.
All the rules of logarithms that we have learnt so far also apply for natural logarithms.
We say that y = ex ⇔ x = loge y.
Review
DID YOU KNOW?
The Scottish mathematician John Napier (1550−1617) is credited with
the discovery of logarithms. His original studies involved logarithms to
the base of 1 .
e
Review DID YOU KNOW?
The number e is also known as Euler’s number. It is named after Leonhard Euler (1707−1783), who
devised the following formula for calculating the value of e:
e = 1+ 1 + 1 + 1× 1 3 + 1× 2 1 × 4 + 1× 2 × 1 4 × 5 + …
1 1× 2 2× ×3 3×
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EXERCISE 2G
1 Use a calculator to evaluate correct to 3 significant figures:
a e3 b e2.7 c e0.8 d e−2
2 Use a calculator to evaluate correct to 3 significant figures:
a ln 3 b ln1.4 c ln 0.9 d ln 0.15
3 Without using a calculator, find the value of:
a eln 2 1 ln 9 c 5eln 6 − ln 1
b e 2 de 2
Review 4 Solve. b ln ex = 15 c e3 ln x = 64 d e−ln x = 3
a eln x = 5
5 Solve, giving your answers correct to 3 significant figures.
a ex = 18 b e2x = 25 c ex+1 = 8 d e2x−3 = 16
6 Solve, giving your answers in terms of natural logarithms.
a ex = 13 b e3x = 7 c e2x−1 = 6 d 1 x+3 = 4
e2
7 Solve, giving your answers in terms of natural logarithms.
a ex . 10 b e5x−2 ø 35 c 5 × e2x+3 , 1
8 Solve, giving your answers correct to 3 significant figures. 43
a ln x = 5 b ln x = −4 c ln(x − 2) = 6 d ln(2x + 1) = −2
Review 9 Solve, giving your answers correct to 3 significant figures. b 2 ln(x + 2) − ln x = ln(2x − 1)
a ln(3 − x2 ) = 2 ln x d ln(2x + 1) = 2 ln x + ln 5
f ln(2 + x2 ) = 1 + 2 ln x
c 2 ln( x + 1) = ln( 2x + 3 )
e ln(x + 2) − ln x = 1
10 Express y in terms of x for each of these equations. b ln( y + 2) − ln y = 1 + 2 ln x
a 2 ln( y + 1) − ln y = ln(x + y)
11 Solve, giving your answers in exact form. b e2x − 5ex + 6 = 0
a e2x + 2ex − 15 = 0 d ex − 21e−x = 4
c 6e2x − 13ex − 5 = 0
Review 12 Given that the function f is defined as f : x ֏ 5ex + 2 for x ∈ R, find an expression for f −1(x).
PS 13 Solve 2 ln(3 − e2x ) = 1, giving your answer correct to 3 significant figures.
PS 14 Solve the simultaneous equations, giving your answers in exact form.
a 2 ln x + ln y = 1 + ln 5 b e3x+4y = 2e2x−y
ln10x − ln y = 2 + ln 2 e2x+ y = 8ex+6y
PS 15 Solve ln(2x + 1) ø ln(x + 4).
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Review 2.8 Transforming a relationship to linear form
Review When we collect experimental data for two variables we often wish to find a
mathematical relationship connecting the variables. When the data plots on a graph lie
on a straight line the relationship can be easily found using the equation y = mx + c,
where m is the gradient and c is the y-intercept. It is more common, however, for the
data plots to lie on a curve.
Logarithms can be used to convert some curves into straight lines.
This is the case for relationships of the form y = kxn and y = k(ax ) where k, a and n are
constants.
Logarithms to any base can be used but it is more usual to use those that are most commonly
available on calculators, which are natural logarithms or logarithms to the base 10.
WORKED EXAMPLE 2.15
Convert y = aebx, where a and b are constants, into the form Y = mX + c.
Answer
y = aebx
Taking natural logarithms of both sides gives:
ln y = ln(aebx )
ln y = ln a + ln ebx
44 ln y = ln a + bx
ln y = bx + ln a
Now compare ln y = bx + ln a with Y = mX + c:
ln y = b x + ln a
Y = m X + c
The non-linear equation y = aebx becomes the linear equation:
Y = mX + c, where Y = ln y, X = x, m = b and c = ln a
It is important to note:
● The variables X and Y in Y = mX + c must contain only the original variables x and y.
(They must not contain the unknown constants a and b.)
● The constants m and c must contain only the original unknown constants a and b. (They
must not contain the variables x and y.)
Review WORKED EXAMPLE 2.16 ln y
( )The variables x and y satisfy the equation y = a e−bx2 , where a and b are (0.55, 0.84)
(1.72, 0.26)
constants. The graph of ln y against x2 is a straight line passing through the
points (0.55, 0.84) and (1.72, 0.26) as shown in the diagram.
Find the value of a and the value of b correct to 2 decimal places.
O x2
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Answer
y = a × e−bx2
Taking natural logarithms of both sides gives:
ln y = ln(a × e−bx2)
ln y = ln a + ln e−bx2
ln y = ln a − bx2
ln y = −bx2 + ln a
Review Now compare ln y = −b x2 + ln a with Y = mX + c:
ln y = −b x2 + ln a
Y = m X + c
Gradient = m = 0.26 − 0.84 = −0.4957 … DID YOU KNOW?
1.72 − 0.55
When one variable on
∴ b = 0.50 to 2 decimal places a graph is a log but the
Using Y = mX + c, X = 0.55, Y = 0.84 and m = −0.4957 … other variable is not a
log, it is called a semi-
0.84 = −0.4957 … × 0.55 + c log graph. When both
c = 1.1126 … variables on a graph
are logs, it is called a
∴ ln a = 1.1126 … log–log graph. 45
a = e1.1126…
Hence a = 3.04 and b = 0.50 to 2 decimal places.
Review WORKED EXAMPLE 2.17
x 5 10 20 40 80
y 2593 1596 983 605 372
The table shows experimental values of the variables x and y.
a By plotting a suitable straight-line graph, show that x and y are related by the equation y = k × xn, where
k and n are constants.
b Use your graph to estimate the value of k and the value of n.
Review Answer Take natural logarithms of both sides.
Use the multiplication law.
a y = k × xn Use the power law.
ln y = ln(k × xn )
ln y = ln k + ln xn
ln y = n ln x + ln k
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Review Now compare ln y = n ln x + ln k with Y = mX + c:
ln y = n ln x + ln k
Y = m X + c
Hence the graph of ln y against ln x needs to be drawn where:
● gradient = n
● intercept on vertical axis = ln k
Table of values is
ln x 1.61 2.30 3.00 3.69 4.38
ln y 7.86 7.38 6.89 6.41 5.92
ln y
10
8
6
4
2
46 ln x
O 12345
Review The points form an approximate straight line, so x and y are related by the
equation y = k × xn.
b n = gradient ≈ 5.92 − 7.86 ≈ −0.70 to 2 significant figures.
4.38 − 1.61
ln k = intercept on vertical axis
ln k ≈ 9
k ≈ e9
k ≈ 8100 to 2 significant figures.
DID YOU KNOW?
Review The distributions of a diverse range of natural and man-made
phenomena approximately follow the power law y = axb. These
include the size of craters on the moon, avalanches, species
extinction, body mass, income and populations of cities.
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EXERCISE 2H
1 Given that a and b are constants, use logarithms to change each of these non-linear equations into the form
Y = mX + c. State what the variables X and Y and the constants m and c represent. (Note: there might be
more than one method to do this.)
a y = eax+b b y = 10ax−b c y = a(x−b ) d y = a(bx )
e a = ex2 +by f xa yb = 8 g xa y = b h y = a(e−bx )
Review 2 The variables x and y satisfy the equation y = a × xn, where a and n are ln y
constants. The graph of ln y against ln x is a straight line passing through
the points (0.31, 4.02) and (1.83, 3.22) as shown in the diagram. Find the value (0.31, 4.02)
of a and the value of n correct to 2 significant figures. (1.83, 3.22)
ln x
3 The variables x and y satisfy the equation y = k × en(x−2), where k and n O
are constants. The graph of ln y against x is a straight line passing through ln y
the points (1, 1.84) and (7, 4.33) as shown in the diagram. Find the value of k
and the value of n correct to 2 significant figures. (7, 4.33)
(1, 1.84)
Ox 47
Review 4 Variables x and y are related so that, when log10 y is plotted on the vertical axis and x is plotted on the
horizontal axis, a straight-line graph passing through the points (2, 5) and (6, 11) is obtained.
Review a Express log10 y in terms of x.
b Express y in terms of x, giving your answer in the form y = a × 10bx.
5 Variables x and y are related so that, when ln y is plotted on the vertical axis and ln x is plotted on the
horizontal axis, a straight-line graph passing through the points (2, 4) and (5, 13) is obtained.
a Express ln y in terms of x.
b Express y in terms of x.
6 The variables x and y satisfy the equation 52y = 32x+1. By taking natural logarithms, show that the graph
of ln y against ln x is a straight line, and find the exact value of the gradient of this line and state the
coordinates of the point at which the line cuts the y-axis.
7 The mass, m grams, of a radioactive substance is given by the formula m = m0e−kt, where t is the time in days
after the mass was first recorded and m0 and k are constants.
The table below shows experimental values of t and m.
t 10 20 30 40 50
m 40.9 33.5 27.4 22.5 18.4
a Draw the graph of ln m against t.
b Use your graph to estimate the value of m0 and k.
c The half-life of a radioactive substance is the time it takes to decay to half of its original mass. Find the
half-life of this radioactive substance.
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Review Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3 e-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersitCyoPprye-ssC-aRmebrviidegweCUonipvyersity
Review 8 The temperature, T °C , of a hot drink, t minutes after it is made, can be modelled by the equation
T = 25 + ke−nt, where k and n are constants. The table below shows experimental values of t and T .
Review t 2 4 6 8 10
T 63.3 57.7 52.8 48.7 45.2
a Convert the equation to a form suitable for drawing a straight-line graph.
b Draw the straight-line graph and use it to estimate the value of k and n.
c Estimate:
i the initial temperature of the drink
ii the time taken for the temperature to reach 28 °C
iii the room temperature.
48
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Checklist of learning and understanding
The rules of logarithms
● If y = ax then x = loga y.
● loga a = 1, loga 1 = 0, loga ax = x, aloga x = x
● Product rule: loga (xy) = loga x + loga y
● Division rule: loga x = loga x − loga y
y
● Power rule: loga (x)m = m loga x
Review ● Special case of power rule: loga 1 = −loga x
x
Natural logarithms y y = ex
● Logarithms to the base of e are called natural logarithms.
● e ≈ 2.718 y=x
● ln x is used to represent loge x. 1 y = ln x
O1 x
● If y = ex then x = ln y.
● All the rules of logarithms apply for natural logarithms.
Transforming a relationship to linear form 49
● Logarithms can be used to convert relationships of the form y = kxn and y = k(ax ), where
k , a and n are constants, into straight lines of the form Y = mX + c, where X and Y are
functions of x and of y.
Review
Review
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END-OF-CHAPTER REVIEW EXERCISE 2
1 Solve the inequality 2x . 7, giving your answer in terms of logarithms. [2]
2 Given that ln p = 2 ln q − ln(3 + q) and that q . 0, express p in terms of q not involving logarithms. [3]
3 Solve the inequality 3 × 23x+2 , 8, giving your answer in terms of logarithms. [4]
4 Use logarithms to solve the equation
5x+3 = 7x−1,
giving the answer correct to 3 significant figures. [4]
Review Cambridge International AS & A Level Mathematics 9709 Paper 21 Q1 November 2015
5 Solve the equation 6(4x ) − 11(2x ) + 4 = 0, giving your answers for x in terms of logarithms where appropriate.
[5]
6 Solve the equation ln(5x + 4) = 2 ln x + ln 6. [5]
7 ln y
(6, 10.2)
(0, 2.0)
O ln x
50
The variables x and y satisfy the equation y = Kxm, where K and m are constants. The graph of ln y against
ln x is a straight line passing through the points (0, 2.0) and (6, 10.2), as shown in the diagram. Find the values
Review of K and m, correct to 2 decimal places. [5]
Cambridge International AS & A Level Mathematics 9709 Paper 21 Q3 June 2011
8 i Given that y = 2x, show that the equation
2x + 3(2−x ) = 4
can be written in the form [3]
y2 − 4y + 3 = 0.
ii Hence solve the equation
2x + 3(2−x ) = 4,
giving the values of x correct to 3 significant figures where appropriate. [3]
Cambridge International AS & A Level Mathematics 9709 Paper 21 Q5 June 2010
9 Given that (1.2)x = 6y, use logarithms to find the value of x correct to 3 significant figures. [3]
y
Review
10 The polynomial f(x) is defined by
f(x) = 12x3 + 25x2 − 4x − 12.
i Show that f(−2) = 0 and factorise f(x) completely. [4]
[3]
ii Given that 12 × 27y + 25 × 9y − 4 × 3y − 12 = 0, state the value of 3y and hence find y correct to 3
significant figures.
Cambridge International A Level Mathematics 9709 Paper 31 Q4 June 2011
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11 Solve the equation 4 − 2x = 10, giving your answer correct to 3 significant figures. [3]
Cambridge International A Level Mathematics 9709 Paper 31 Q1 June 2012
12 Use logarithms to solve the equation ex = 3x−2, giving your answer correct to 3 decimal places. [3]
Cambridge International A Level Mathematics 9709 Paper 31 Q1 November 2014
13 Using the substitution u = 3x, solve the equation 3x + 32x = 33x giving your answer correct to 3 significant
figures. [5]
Review Cambridge International A Level Mathematics 9709 Paper 31 Q2 November 2015
14 The variables x and y satisfy the equation 5y = 32x−4.
a By taking natural logarithms, show that the graph of y against x is a straight line and find the exact value
of the gradient of this line. [3]
b This line intersects the x-axis at P and the y-axis at Q. Find the exact coordinates [3]
of the midpoint of PQ.
15 ln y
(3.1, 2.1)
(2.3, 1.7) x 51
O
Review The variables x and y satisfy the equation y = K (bx ), where K and b are constants. The graph of ln y against
x is a straight line passing through the points (2.3, 1.7) and (3.1, 2.1), as shown in the diagram. Find the values
of K and b, correct to 2 decimal places. [6]
Review
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Review52
Review Chapter 3
Trigonometry
In this chapter you will learn how to:
■ understand the relationship of the secant, cosecant and cotangent functions to cosine, sine
and tangent, and use properties and graphs of all six trigonometric functions for angles of any
magnitude
■ use trigonometrical identities for the simplification and exact evaluation of expressions and in
the course of solving equations, and select an identity or identities appropriate to the context,
showing familiarity in particular with the use of:
sec2 θ = 1 + tan2 θ and cosec2 θ = 1 + cot2 θ
• the expansion of sin(A ± B), cos(A ± B) and tan(A ± B)
• the formulae for sin 2A, cos 2A and tan 2A
•• the expression of a sinθ + b cosθ in the forms R sin(θ ± α ) and R cos(θ ± α ).
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PREREQUISITE KNOWLEDGE
Where it comes from What you should be able to do Check your skills
Pure Mathematics 1 Sketch graphs of the sine, cosine and 1 Sketch the graphs of these functions
Coursebook, Chapter 5 tangent functions. for 0° ø x ø 360°.
a y = 2 sin x + 1
Review Pure Mathematics 1 Find exact values of the sine, cosine and b y = tan(x − 90°)
Coursebook, Chapter 5 c y = cos(2x + 90°)
tangent of 30°, 45° and 60° and related
angles. 2 Find the exact values of:
a sin120°
Pure Mathematics 1 Solve trigonometric equations using b cos 225°
Coursebook, Chapter 5 c tan150°
the identities tanθ ≡ sinθ and
sin2 θ + cos2 θ ≡ 1. cosθ 3 Solve for 0° ø x ø 360°.
a 5 sin x = 3 cos x
b 2 cos2 x − sin x = 1
Why do we study trigonometry? REWIND 53
Review In addition to the three main trigonometrical functions (sine, cosine and tangent) there This chapter builds on
are three more commonly used trigonometrical functions (cosecant, secant and cotangent) the work we covered in
that we will learn about. These are sometimes referred to as the reciprocal trigonometrical the Pure Mathematics 1
functions. Coursebook, Chapter 5.
We will also learn about the compound angle formulae, which form a basis for numerous WEB LINK
important mathematical techniques. These techniques include solving equations, deriving
further trigonometric identities, the addition of different sine and cosine functions and Explore the
deriving rules for differentiating and integrating trigonometric functions, which we will Trigonometry:
cover later in this book. Compound angles
station on the
As you know, scientists and engineers represent oscillations and waves using trigonometric Underground
functions. These functions also have further uses in navigation, engineering and physics. Mathematics website.
3.1 The cosecant, secant and cotangent ratios
There are a total of six trigonometric ratios. You have already learnt about the ratios sine,
cosine and tangent. In this section you will learn about the other three ratios, which are
cosecant (cosec), secant (sec) and cotangent (cot).
KEY POINT 3.1
Review The three reciprocal trigonometric ratios are defined as:
cosec θ = 1 sec θ = 1 cot θ = 1 = cos θ
sin θ cos θ tan θ sin θ
Consider the right-angled triangle:
sinθ = y cosθ = x tanθ = y r y
r r x
θ
cosecθ = r secθ = r cotθ = x x
y x y
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The following identities can be found from this triangle:
x2 + y2 = r2 Divide both sides by x2.
Use y = tanθ and r = secθ.
1+ y 2 = r 2
x x xx
1 + tan2 θ ≡ sec2 θ
x2 + y2 = r2 Divide both sides by y2 .
Use x = cotθ and r = cosecθ.
x 2 +1 = r 2
y y yy
Review cot2 θ + 1 ≡ cosec2 θ
We will need these important identities to solve trigonometric equations later in this section.
KEY POINT 3.2
1 + tan2 θ ≡ sec2 θ
cot2 θ + 1 ≡ cosec2 θ
The graphs of y = sinθ and y = cosecθ
Review54 y = sin θ 270 θ y
1y 90 180 360 y = cosec θ
O
–1 1
O
90 180 270 360 θ
–1
The function y = cosecθ is the reciprocal of the function y = sinθ .
The function y = sinθ is zero when θ = 0°, 180°, 360°, … , this means that y = cosecθ
is not defined at each of these points because 1 is not defined. Hence vertical asymptotes
0
need to be drawn at each of these points.
Review
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The graphs of y = cosθ and y = secθ
We can find the graph of y = secθ in a similar way from the graph of y = cosθ.
y
y y = sec θ
1 y = cos θ 1
OO 90 180 270 360 θ
90 180 270 360 θ
Review –1 –1
The graphs of y = tanθ and y = cotθ y
We can find the graph of y = cotθ from the graph of y = tanθ.
y
y = tan θ y = cot θ
O O 55
90 180 270 360 θ 90 180 270 360 θ
Review WORKED EXAMPLE 3.1
Find the exact value of cosec 240°.
Answer
cosec 240° = 1 y
sin 240° SA
= 1 240°
–sin 60°
=1 60° O x
−3
Review 2 TC
= − 2
3
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WORKED EXAMPLE 3.2
Solve sec2 x − tan x − 3 = 0 for 0° ø x ø 360°.
Answer
sec2 x − tan x − 3 = 0 Use 1 + tan2 x ≡ sec2 x.
tan2 x − tan x − 2 = 0 Factorise.
(tan x − 2)(tan x + 1) = 0 y y = tan x
tan x = 2 or tan x = −1 3
Review tan x = 2 ⇒ x = 63.4° (calculator) 2 y=2
or x = 63.4° + 180° = 243.4° 1
tan x = −1 ⇒ x = −45° (not in required range) –45
or x = −45° + 180° = 135° –90 O 63.4 90 180 270 360 x
or x = 135° + 180° = 315° –1 y = –1
The values of x are 63.4°, 135°, 243.4°, 315°. –2
–3
EXERCISE 3A
56 1 Find the exact values of:
a sec 60° b cosec 45° c cot120° d sec 300°
g sec 150° h cot(−30°).
Review e cosec135° f cot 330°
c sec π
2 Find the exact values of: 4
a cosec π b cot π g cot 4π d cot 2π
6 3 3 3
e sec 5π f cosec 7π c cosec x = −3 h sec − π .
6 4 6
3 Solve each equation for 0° ø x ø 360°.
a sec x = 3 b cot x = 0.8 d 3sec x − 4 = 0
4 Solve each equation for 0 ø x ø 2π.
a cosec x = 2 b sec x = −1 c cot x = 2 d 2 cot x + 5 = 0
5 Solve each equation for 0° ø x ø 180°.
Review a cosec 2x = 1.2 b sec 2x = 4 c cot 2x = 1 d 2 sec 2x = 3
6 Solve each equation for the given domains.
a cosec(x − 30°) = 2 for 0° ø x ø 360° b sec(2x + 60°) = −1.5 for 0° ø x ø 180°
d 2 cosec(2x − 1) = 3 for − π ø x ø π
c cot x + π = 2 for 0 ø x ø 2π
4
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7 Solve each equation for −180° ø x ø 180°.
a cosec2 x = 4 b sec2 x = 9 c 9 cot2 1 x = 4
2
d sec x = cos x e cosec x = sec x
f 2 tan x = 3 cosec x
8 Solve each equation for 0° ø θ ø 360°. b 3 cosec2 θ = 13 − cotθ
a 2 tan2 θ − 1 = secθ d cosecθ + cotθ = 2 sinθ
c 2 cot2 θ − cosecθ = 13 f 3 sec2 θ = cosecθ
e tan2 θ + 3secθ = 0
Review 9 Solve each equation for 0° ø θ ø 180°. b 2 sec2 2θ = 3tan 2θ + 1
a secθ = 3 cosθ − tanθ d 2 cot2 2θ + 7 cosec 2θ = 2
c sec4 θ + 2 = 6 tan2 θ
10 Solve each equation for 0 ø θ ø 2π. b 3 cot2 θ + 5 cosecθ + 1 = 0
a tan2 θ + 3secθ + 3 = 0
11 a Sketch each of the following functions for the interval 0 ø x ø 2π. 2 cosec x
i y = 1 + sec x ii y = cot 2x iii y = − π
2
iv y = 1 − sec x v y = 1 + cosec 1 x vi y = sec 2x + π
2 4
b Write down the equation of each of the asymptotes for your graph for part a vi. 57
Review P 12 Prove each of these identities. b cosec x − sin x ≡ cos x cot x
a sin x + cos x cot x ≡ cosec x d (1 + sec x)(cosec x − cot x) ≡ tan x
c sec x cosec x − cot x ≡ tan x
P 13 Prove each of these identities.
a 1 ≡ sin x cos x b sec2 x + sec x tan x ≡ 1 − 1 x
tan x + cot x sin
c 1 − cos2 x ≡ 1 − sin2 x d 1 + tan2 x ≡ sec x cosec x
sec2 x − 1 tan x
e sin x x ≡ cosec x f 1 + sin x ≡ ( tan x + sec x )2
1 − cos2 1 − sin x
g 1 + 1 ≡ 2 cosec2 x h cos x + cos x ≡ 2 sec x
1 + cos x 1 − cos x 1 + sin x 1 − sin x
Review PS 14 Solve each equation for 0° ø θ ø 180°. b 2 cot3 θ + 3 cosec2 θ − 8 cotθ = 0
a 6 sec3 θ − 5 sec2 θ − 8 secθ + 3 = 0
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3.2 Compound angle formulae
EXPLORE 3.1
1 Sunita says that: sin(A + B) = sin A + sinB
Use a calculator with A = 20° and B = 50° to prove that Sunita is wrong.
Review 2 The diagram shows ∆PQR. R
RX is perpendicular to PQ. AB
∠A and ∠B are acute angles. q hp
QR = p, PR = q and RX = h.
a Write down an expression for h in terms of: P X Q
i q and A ii p and B
b Using the formula area of triangle = 1 ab sinC, we can write:
2
Area of ∆PQR = 1 pq sin(A + B).
2
Find an expression in terms of p, q, A and B for the area of:
i ∆PRX ii ∆QRX
c Use your expressions for the areas of triangles PQR, PRX and QRX to prove that:
58 sin(A + B) ≡ sin A cos B + cos A sinB
Review 3 In the identity sin(A + B) ≡ sin A cos B + cos A sin B , replace B by −B to show that:
sin(A − B) ≡ sin A cos B − cos A sinB
4 In the identity sin(A − B) ≡ sin A cos B − cos A sin B , replace A by (90° − A) to show that:
cos(A + B) ≡ cos A cos B − sin A sinB
5 In the identity cos(A + B) ≡ cos A cos B − sin A sin B , replace B by −B to show that:
cos(A − B) ≡ cos A cos B + sin A sinB
6 By writing tan(A + B) as sin(A + B) and tan(A − B) as sin( A − B) , show that:
cos(A + B) cos(A − B)
tan(A + B) ≡ tan A + tanB and tan(A − B) ≡ tan A − tanB
1− tan A tanB 1+ tan A tanB
KEY POINT 3.3
Review Summarising, the six compound angle formulae are:
sin(A + B) ≡ sin A cos B + cos A sin B
sin(A − B) ≡ sin A cos B − cos A sin B
cos(A + B) ≡ cos A cos B − sin A sin B
cos(A − B) ≡ cos A cos B + sin A sin B
tan( A + B) ≡ tan A + tan B
1 − tan A tan B
tan( A − B) ≡ tan A − tan B
1 + tan A tan B
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WORKED EXAMPLE 3.3
Find the exact value of: b cos 42° cos12° + sin 42° sin12°
a sin105°
Answer
a sin105° = sin(60° + 45°)
= sin 60° cos 45° + cos 60° sin 45°
= 3 1 + 1 1
2 2 2 2
Review = 3 +1
22
b cos 42° cos12° + sin 42° sin12° Use cos (A − B) ≡ cos A cos B + sin A sin B.
= cos(42° − 12°)
= cos 30°
=3
2
WORKED EXAMPLE 3.4
Given that sin A = 4 and cos B = 5 , where A is obtuse and B is acute, find the value of: 59
a sin(A + B) 5 13 c tan(A − B)
b cos(A − B)
Review
Answer
S y A y
4 5 x SA
A 13
B 12
–3 O O5 x
TC TC
sin A = 4 cos A = − 3 , tan A = − 4 sin B = 12 , cos B = 5 , tan B = 12
5, 5 3 13 13 5
a sin(A + B) = sin A cos B + cos A sin B
Review = 4 5 + − 3 12
5 13 5 13
= − 16
65
b cos(A − B) = cos A cos B + sin A sin B
= − 3 5 + 4 12
5 13 5 13
= 33
65
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c tan(A − B) = tan A − tan B
1 + tan A tan B
− 4 − 12
3 5
=
4 12
1+ − 3 5
= 56
33
Review WORKED EXAMPLE 3.5
Solve the equation sin(60° − x) = 2 sin x for 0° , x , 360°.
Answer
sin(60° − x) = 2 sin x Expand sin(60° − x).
Use sin 60° = 3 and cos 60° = 1 .
sin 60° cos x − cos 60° sin x = 2 sin x
22
3 cos x − 1 sin x = 2 sin x Multiply both sides by 2.
22 Add sin x to both sides.
Rearrange to find tan x.
3 cos x − sin x = 4 sin x
60
3 cos x = 5 sin x
Review 3 = tan x
5
x = 19.1° or x = 180° + 19.1°
The values of x are 19.1° and 199.1°.
EXERCISE 3B
1 Expand and simplify cos(x + 30°).
2 Without using a calculator, find the exact value of each of the following:
a sin 20° cos 70° + cos 20° sin 70° b sin172° cos 37° − cos172° sin 37°
c cos 25° cos 35° − sin 25° sin 35° d cos 99° cos 69° + sin 99° sin 69°
f tan 82° − tan 52°
Review e tan 25° + tan 20°
1 − tan 25° tan 20° 1 + tan 82° tan 52°
3 Without using a calculator, find the exact value of:
a sin 75° b tan 75° c cos105° d tan(−15°)
e sin π f cos 19π g cot 7π h sin 7π
12 12 12 12
4 Given that cos x = 4 and that 0° , x , 90°, without using a calculator find the exact value of cos(x − 60°).
5
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5 Given that sin A = 4 and sin B = 5 , where A and B are acute, show that sin(A + B) = 63 .
5 13 65
6 Given that sin A = 12 and sin B = 3 , where A is obtuse and B is acute, find the value of:
13 5
a sin(A + B) b cos(A − B) c tan(A + B)
7 Given that cos A = − 4 and sin B = − 8 and that A and B lie in the same quadrant, find the value of:
5 17
a sin(A + B) b cos(A + B) c tan(A − B)
8 Given that tan A = t and that tan(A − B) = 2, find tan B in terms of t.
Review 9 If cos(A − B) = 3 cos(A + B), find the exact value of tan A tan B.
10 a Given that 8 + cosec2 θ = 6 cotθ , find the value of tanθ.
b Hence find the exact value of tan(θ + 45°).
11 a Given that x is acute and that 2 sec2 x + 7 tan x = 17, find the exact value of tan x.
b Hence find the exact value of tan(225° − x).
12 a Show that the equation sin(x + 30°) = 5 cos(x − 60°) can be written in the form cot x = − 3 .
b Hence solve the equation sin(x + 30°) = 5 cos(x − 60°) for −180° , x , 180°.
13 Solve each equation for 0° ø x ø 360°. b cos( x − 60° ) = 3 cos x 61
a cos(x + 30°) = 2 sin x
c sin(30° − x) = 4 sin x d cos(x + 30°) = 2 sin(x + 60°)
Review 14 Solve each equation for 0° ø x ø 180°. b 2 tan(45° − x) = 3tan x
a 2 tan(60° − x) = tan x d sin(x − 45°) = 2 cos(x + 60°)
c sin(x + 60°) = 2 cos(x + 45°) f cos(x + 225°) = 2 sin(x − 60°)
e tan(x + 45°) = 6 tan x
15 Solve the equation tan(x − 45°) + cot x = 2 for 0° ø x ø 180°.
P 16 Use the expansions of cos(5x + x) and cos(5x − x) to prove that cos 6x + cos 4x ≡ 2 cos 5x cos x.
PS 17 Given that sin x + sin y = p and cos x + cos y = q, find an expression for cos(x − y), in terms of p and q.
3.3 Double angle formulae
We can use the compound angle formulae to derive three new identities.
Review Consider sin(A + B) ≡ sin A cos B + cos A sin B
If B = A, sin(A + A) ≡ sin A cos A + cos A sin A
Consider ∴ sin 2A ≡ 2 sin A cos A
If B = A, cos(A + B) ≡ cos A cos B − sin A sin B
cos(A + A) ≡ cos A cos A − sin A sin A
∴ cos 2A ≡ cos2 A − sin2 A
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Consider tan( A + B) ≡ tan A + tan B
If B = A, 1 − tan Atan B
tan( A + A) ≡ tan A + tan A
1 − tan Atan A
∴ tan 2A ≡ 1 2 tan A
− tan2 A
These three formulae are known as the double angle formulae. We can use the identity
sin2 A + cos2 A ≡ 1 to write the identity cos 2A ≡ cos2 A − sin2 A in two alternative
forms:
Using sin2 A ≡ 1 − cos2 A gives: cos 2A ≡ 2 cos2 A − 1
ReviewUsing cos2 A ≡ 1 − sin2 A gives: cos 2A ≡ 1 − 2 sin2 A
Summarising, the double angle formulae are:
KEY POINT 3.4
sin 2A ≡ 2 sin A cos A cos 2A ≡ cos2 A − sin2 A tan 2A ≡ 1 2 tan A
≡ 2 cos2 A − 1 − tan2 A
≡ 1 − 2 sin2 A
62 WORKED EXAMPLE 3.6
Review Given that sin x = − 3 and that 180° , x , 270°, find the exact value of:
5
b cos 2x c tan x
a sin 2x 2
Answer A
y
S
–4 From the diagram:
–3
O x sin x = − 3, cos x = − 4 and tan x = 3
5 C 5 5 4
T
a sin 2x = 2 sin x cos x
Review = 2 − 3 − 4 = 24
5 5 25
b cos 2x = cos2 x − sin2 x
= − 4 2 − − 3 2 = 7
5 5 25
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2 tan x
= 2 Use 2A = x in the double angle formula.
c tan x 1− tan2 x
Rearrange.
2 Factorise.
Solve.
3 = 2 tan x
4 2
1 − tan2 x
2
x x tan2 x + 8 tan x − 3 = 0
3
22
Review 3tan x − 1 tan x + 3 = 0
2 2
tan x = 1 or tan x = −3
23 2
If 180° , x , 270°, then 90° , x , 135°, which means
2
that x is in the second quadrant.
2
∴ tan x = −3
2
WORKED EXAMPLE 3.7 63
Review Solve the equation sin 2x = sin x for 0° ø x ø 360°. Use sin 2x = 2 sin x cos x.
Rearrange.
Answer Factorise.
sin 2x = sin x
y 180 1
2 sin x cos x = sin x 1 y= 2
2 sin x cos x − sin x = 0 270 360 x
0.5 y = cos x
sin x(2 cos x − 1) = 0
Review sin x = 0 or cos x = 1 O
60 90
2
sin x = 0 ⇒ x = 0°, 180°, 360° –0.5
cos x = 1 ⇒ x = 60° –1
2
or x = 360° − 60° = 300°
The values of x are 0°, 60°, 180°, 300°, 360°.
When solving equations involving cos 2x it is important that we choose the most suitable
expansion of cos 2x for solving that particular equation.
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WORKED EXAMPLE 3.8
Solve the equation cos 2x + 3sin x = 2 for 0 ø x ø 2π.
Review Answer Use cos 2x = 1 − 2 sin2 x. y=1
cos 2x + 3sin x = 2 Rearrange.
1
1 − 2 sin2 x + 3sin x = 2 Factorise. y=
2 sin2 x − 3sin x + 1 = 0 y 2
(2 sin x − 1)(sin x − 1) = 0 1 x
3π 2π
sin x = 1 or sin x = 1 0.5 2
2
O π π π y = sin x
sin x = 1 ⇒ x = π –0.5 6 2
26
or x = π − π = 5π –1
66
sin x = 1 ⇒ x = π
2
The values of x are π , π , 5π .
62 6
64
Review EXERCISE 3C
1 Express each of the following as a single trigonometric ratio.
a 2 sin 28° cos 28° b 2 cos2 34° − 1 c 2 tan17°
1 − tan2 17°
2 Given that tan x = 4 , where 0° , x , 90°, find the exact value of:
3
a sin 2x b cos 2x c tan 2x d tan 3x
3 Given that cos 2x = − 527 and that 0° , x , 90°, find the exact value of:
625
a sin 2x b tan 2x c cos x d tan x
4 Given that cos x = − 3 and that 90° , x , 180°, find the exact value of:
5
tan 1 x
a sin 2x b sin 4x c tan 2x d 2
Review 5 Given that 3 cos 2x + 17 sin x = 8, find the exact value of sin x.
6 Solve each equation for 0° ø θ ø 360°.
a 2 sin 2θ = cosθ b 2 cos 2θ + 3 = 4 cosθ c 2 cos 2θ + 1 = sinθ
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7 Solve each equation for 0° ø θ ø 180°. b 3 cos 2θ + cosθ = 2
a 2 sin 2θ tanθ = 1 d 2 tan 2θ = 3 cotθ
c 2 cosec 2θ + 2 tanθ = 3secθ f cot 2θ + cotθ = 3
e tan 2θ = 4 cotθ
8 Express cos2 2x in terms of cos 4x.
9 a Use the expansions of cos(2x + x) to show that cos 3x ≡ 4 cos3 x − 3 cos x.
b Express sin 3x in terms of sin x.
Review 10 Solve tan 2θ + 2 tanθ = 3 cotθ for 0° ø θ ø 180°.
11 a Prove that tanθ + cot θ ≡ 2 .
b sin 2θ
Hence find the exact value of tan π + cot π .
12 12
12 a Prove that 2 cosec 2θ tanθ ≡ sec2 θ.
b Hence solve the equation cosec 2θ tanθ = 2 for −π , θ , π.
13 a Prove that cos 4x + 4 cos 2x ≡ 8 cos4 x − 3.
b Hence solve the equation 2 cos 4x + 8 cos 2x = 3 for −π ø x ø π.
Review 14 a Show that θ = 18° is a root of the equation sin 3θ = cos 2θ. 65
b Express sin 3θ and cos 2θ in terms of sinθ.
c Hence show that sin18° is a root of the equation 4x3 − 2x2 − 3x + 1 = 0. TIP
d Hence find the exact value of sin18°. Solving trigonometric
inequalities is not
PS 15 Find the range of values of θ between 0 and 2π for which cos 2θ . cosθ. in the Cambridge
syllabus. You should,
PS 16 Solve the inequality cos 2θ − 3sinθ − 2 ù 0 for 0° ø θ ø 360°. however, have the skills
necessary to tackle
PS 17 A these more challenging
questions.
b
x a 3x
B C
Prove that a = 4b cos 2x cos x.
Review PS 18 a Use the expansions of cos(2x + x) and cos(2x − x) , to prove that:
cos 3x + cos x ≡ 2 cos 2x cos x
b Solve cos 3x + cos 2x + cos x . 0 for 0° , x , 360°.
PS 19 Solve the inequality cos 4θ + 3 cos 2θ + 1 , 0 for 0° ø θ ø 360°.
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3.4 Further trigonometric identities
This section builds on our previous work by using the compound angle formulae and
the double angle formulae to prove trigonometric identities.
WORKED EXAMPLE 3.9
Prove the identity tan 3x ≡ 3 tan x − tan3 x .
1− 3 tan2 x
Answer
Review LHS ≡ tan(2x + x) Use the compound angle formula for tan.
Use the double angle formula.
≡ tan 2x + tan x Multiply numerator and denominator by
1 − tan 2x tan x (1 − tan2 x).
2 tan x + tan x Expand the brackets and simplify.
1 − tan2 x
≡
1 − tan x 1 2 tan x
− tan2 x
≡ 2 tan x + tan x(1 − tan2 x)
1 − tan2 x − 2 tan2 x
≡ 3tan x − tan3 x
1 − 3 tan2 x
66
≡ RHS
Review WORKED EXAMPLE 3.10
Prove the identity 2 sin(x − y) + y) ≡ cot y − cot x.
cos(x − y) − cos(x
Answer
LHS ≡ 2 sin(x − y) y) Use compound angle formulae.
cos(x − y) − cos(x + Simplify denominator.
Separate fractions.
≡ (cos x cos y 2 sin x cos y − 2 cos x sin y sin x sin y) Simplify fractions.
+ sin x sin y) − (cosx cos y −
≡ 2 sin x cos y − 2 cos x sin y
2 sin x sin y
Review ≡ 2 sin x cos y − 2 cos x sin y
2 sin x sin y 2 sin x sin y
≡ cos y − cos x
sin y sin x
≡ cot y − cot x
≡ RHS
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Review EXPLORE 3.2 1 − cos 2x sin 2x + sin x
sin 2x cos 2x + cos x + 1
Odd one out
1 2
cosec 2x − cot 2x cosec 2x + cot 2x sec2 x sin 2x
sin 2x cot x − 2 cot 2x
1 + cos 2x
Find the trigonometric expression that does not match the other six expressions.
Create as many expressions of your own to match the ‘odd one out’.
(Your expressions must contain at least two different trigonometric ratios.)
Compare your answers with your classmates.
EXERCISE 3D
67
P 1 Prove each of these identities.
Review a tan A + cot A ≡ 2 cosec 2A b 1 − tan2 A ≡ cos 2Asec2A
c tan 2A − tan A ≡ tan Asec 2A d (cos A + 3sin A)2 ≡ 5 − 4 cos 2A + 3sin 2A
e cot 2A + tan A ≡ 1 cosec A sec A f cosec 2A + cot 2A ≡ cot A
2
g sec2A cosec2A ≡ 4 cosec22A h sin(A + B) sin(A − B) ≡ sin2 A − sin2 B
P 2 Prove each of these identities.
a sin A + cos A ≡ 2 cos(A − B) b cos A + sin A ≡ sec 2A + tan 2A
cos B sin B sin 2B cos A − sin A
c 1− tan2 A ≡ cos 2A d cos 2A + sin 2A − 1 ≡ tan A
1+ tan2 A cos 2A − sin 2A + 1
e sin 3A + sin A ≡ cos A f cos 3A − sin 3A ≡ cos A + sin A
2 sin 2A 1 − 2 sin 2A
g cos 2A + 9 cos A + 5 ≡ 2 cos A + 1 h cos3 A − sin3 A ≡ 2 + sin 2A
4 + cos A cos A − sin A 2
Review P 3 Use the fact that 4A = 2 × 2A to show that:
a sin 4A ≡ 8 cos3 A − 4 cos A b cos 4A + 4 cos 2A ≡ 8 cos4 A − 3
sin A
P 4 Prove the identity 8 sin2 x cos2 x ≡ 1 − cos 4x. WEB LINK
P 5 Prove the identity (2 sin A + cos A)2 ≡ 1 (4 sin 2A − 3 cos 2A + 5). Try the Equation or
2 identity?(II) resource
on the Underground
P 6 Use the expansions of cos(3x − x) and cos(3x + x) to prove the identity: Mathematics website.
cos 2x − cos 4x ≡ 2 sin 3x sin x
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3.5 Expressing a sinθ + b cosθ in the form R sin(θθ ± α ) or R cos(θθ ± α )
In this section you will learn how to solve equations of the form a sinθ + b cosθ = c.
EXPLORE 3.3
Use graphing software to confirm that the graph of y = 3sinθ + 4 cosθ for −90° ø θ ø 360° is:
y y = 3 sin θ + 4 cos θ
5
Review −90 O θ
−5 90 180 270 360
1 What are the amplitude and period of this graph?
2 Discuss with your classmates how this graph can be obtained by transforming the graph of:
a y = sinθ b y = cosθ
3 Use your knowledge of transformations of functions to write the function y = 3sinθ + 4 cosθ in the
form:
a y = R sin(θ + α ) b R sin(θ + β )
68 (You will need to use your graph to find the approximate values of α and β.)
ReviewThe formula for replacing a sinθ + b cosθ by R sin(θ + α ) is derived as follows.
Let a sinθ + b cosθ = R sin(θ + α ) where R . 0 and 0° , α , 90°.
R sin(θ + α ) = R(sinθ cosα + cosθ sinα )
∴ a sinθ + b cosθ = R sinθ cosα + R cosθ sinα
Equating coefficients of sinθ: R cosα = a (1)
Equating coefficients of cosθ: R sinα = b (2)
(2) ÷ (1): sin α = b ⇒ tanα = b
cos α a a
Squaring equations (1) and (2) and then adding gives:
a2 + b2 = R2 cos2 α + R2 sin2 α Use cos2 α + sin2 α = 1.
= R2 (cos2 α + sin2 α )
Review = R2
Hence, R = a2 + b2
∴ a sinθ + b cosθ can be written as R sin(θ + α ) where R = a2 + b2 and tanα = b
a
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EXPLORE 3.4
Using a similar approach, show that:
1 a sinθ − b cosθ ≡ R sin(θ − α ) where R = a2 + b2 and tanα = b
a
2 a cosθ + b sinθ ≡ R cos(θ − α ) where R = a2 + b2 and tanα = b
a
3 a cosθ − b sinθ ≡ R cos(θ + α ) where R = a2 + b2 and tanα = b
a
Review Summarising the results from Explore 3.4:
KEY POINT 3.5
a sinθ ± b cosθ ≡ R sin(θ ± α ) and a cosθ ± b sinθ ≡ R cos(θ ∓ α )
where R = a2 + b2 , tan α = b and 0° , α , 90°.
a
WORKED EXAMPLE 3.11
a Express 2 sinθ − 3 cosθ in the form R sin(θ − α ), where R . 0 and 0° , α , 90°. 69
Give the value of α correct to 2 decimal places.
b Hence solve the equation 2 sinθ − 3 cosθ = 2 for 0° , θ , 360°.
Review Answer
a 2 sinθ − 3 cosθ = R sin(θ − α )
∴ 2 sinθ − 3 cosθ = R sinθ cosα − R cosθ sinα
Equating coefficients of sinθ: R cosα = 2 (1)
Equating coefficients of cosθ: R sinα = 3 (2)
(2) ÷ (1): tanα = 3 ⇒ α = 56.31°
2
Squaring equations (1) and (2) and then adding gives:
32 + 22 = R2 ⇒ R = 13
∴ 2 sinθ − 3 cosθ = 13 sin(θ − 56.31°)
Review b 2 sinθ − 3 cosθ = 2
13 sin(θ − 56.31°) = 2
sin(θ − 56.31°) = 2
13
θ − 56.31° = 23.09°, 180° − 23.09°
= 23.09°, 156.91°
θ = 79.4°, 213.2°
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ReviewWe can find the maximum and minimum values of the expression a sinθ + b cosθ by
writing the expression in the form R sin(θ + α ).
From the Pure Mathematics Coursebook 1, Chapter 5, Section 5.4, we know that
−1 ø sin(θ + α ) ø 1, hence −R ø R sin(θ + α ) ø R.
This can be summarised as:
KEY POINT 3.6
The maximum value of a sinθ + b cosθ is a2 + b2 and occurs when sin(θ + α ) = 1.
The minimum value of a sinθ + b cosθ is − a2 + b2 and occurs when sin(θ + α ) = −1.
WORKED EXAMPLE 3.12
Express 2 cosθ − sinθ in the form R cos(θ + α ), where R . 0 and 0° , α , 90°.
Hence, find the maximum and minimum values of the expression 2 cosθ − sinθ and the values of θ in the interval
0° , θ , 360° for which these occur.
Answer (1)
2 cosθ − sinθ = R cos(θ + α )
∴ 2 cosθ − sinθ = R cosθ cosα − R sinθ sinα
70 Equating coefficients of cosθ: R cosα = 2
Equating coefficients of sinθ: R sinα = 1 (2)
Review (2) ÷ (1): tanα = 1 ⇒ α = 26.57°
2
Squaring equations (1) and (2) and then adding gives:
22 + 12 = R2 ⇒ R = 5
∴ 2 cosθ − sinθ = 5 cos (θ + 26.57° )
Maximum value is 22 + (−1)2 and occurs when cos (θ + 26.57°) = 1
θ + 26.57° = 360°
θ = 333.4°
Minimum value is − 22 + (−1)2 and occurs when cos(θ + 26.57°) = −1
θ + 26.57° = 180°
θ = 153.4°
∴ Max value = 5 when θ = 333.4°, min value = − 5 when θ = 153.4°
Review DID YOU KNOW?
The French mathematician Joseph Fourier (1768−1830) found a way to represent any wave-like
function as the sum (possibly infinite) of simple sine waves. The sum of sine functions is called
a Fourier series. These series can be applied to vibration problems. Fourier is also credited with
discovering the greenhouse effect.
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EXERCISE 3E
1 a Express 15 sinθ − 8 cosθ in the form R sin(θ − α ), where R . 0 and 0° , α , 90°.
Give the value of α correct to 2 decimal places.
b Hence solve the equation 15 sinθ − 8 cosθ = 10 for 0° , θ , 360°.
2 a Express 2 cosθ − 3sinθ in the form R cos(θ + α ), where R . 0 and 0° , α , 90°.
Give the exact value of R and the value of α correct to 2 decimal places.
b Hence solve the equation 2 cosθ − 3sinθ = 1.3 for 0° , θ , 360°.
Review 3 a Express 15 sinθ − 8 cosθ in the form R sin(θ − α ), where R . 0 and 0° , α , 90°.
Give the value of α correct to 2 decimal places.
b Hence solve the equation 15 sinθ − 8 cosθ = 3 for 0° , θ , 360°.
c Find the greatest value of 30 sinθ − 16 cosθ as θ varies.
4 a Express 4 sinθ − 6 cosθ in the form R sin(θ − α ), where R . 0 and 0° , α , 90°.
Give the exact value of R and the value of α correct to 2 decimal places.
b Hence solve the equation 4 sinθ − 6 cosθ = 3 for 0° , θ , 180°.
c Find the greatest and least values of (4 sinθ − 6 cosθ )2 − 3 as θ varies.
5 a Express 3sinθ + 4 cosθ in the form R sin(θ + α ), where R . 0 and 0° , α , 90°. 71
Give the value of α correct to 2 decimal places.
Review b Hence solve the equation 3sinθ + 4 cosθ = 2 for 0° , θ , 360°.
c Find the least value of 3sinθ + 4 cosθ + 3 as θ varies.
6 a Express cosθ + 3 sinθ in the form R cos (θ − α ), where R.0 and 0 , α , π .
2
Give the exact values of R and α .
sec2 θ π
Hence prove that 1 ≡ 1 − 3 .
cosθ + 3 sinθ 4
( )b 2
7 a Express 8 sin 2θ + 4 cos 2θ in the form R sin(2θ + α ), where R . 0 and 0° , α , 90°.
Give the exact value of R and the value of α correct to 2 decimal places.
b Hence solve the equation 8 sin 2θ + 4 cos 2θ = 3 for 0° , θ , 360°.
c 10 as θ varies.
Find the least value of (8 sin 2θ + 4 cos 2θ )2
Review 8 a Express cosθ − 2 sinθ in the form R cos(θ + α ) , where R . 0 and 0° , α , 90°.
Give the exact value of R and the value of α correct to 2 decimal places.
b Hence solve the equation cosθ − 2 sinθ = −1 for 0° , θ , 360°.
1 as θ varies.
2 cosθ − 2 sinθ 2
( )c Find the least value of
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9 a Express cosθ − sinθ in the form R cos(θ + α ), where R . 0 and 0 , α , π .
2
Give the exact values of R and α .
b Show that one solution of the equation cosθ − sinθ = 1 6 is θ = 19π and find the other solution in the
interval 0 <θ < 2π. 2 12
c State the set of values of k for which the equation cosθ − sinθ = k has any solutions.
Review 10 a Express sinθ − 3 cosθ in the form R sin(θ − α ), where R > 0 and 0° , α , 90°.
Give the exact value of R and the value of α correct to 2 decimal places.
b Hence solve the equation sinθ − 3 cosθ = −2 for 0° , θ , 360°.
c Find the greatest possible value of 1 + sin 2θ − 3 cos 2θ as θ varies and determine the smallest positive
value of θ for which this greatest value occurs.
11 a Express 5 cosθ + 2 sinθ in the form R cos(θ − α ), where R . 0 and 0° , α , 90°.
Give the value of α correct to 2 decimal places.
b Find the smallest positive angle θ that satisfies the equation 5 cosθ + 2 sinθ = 3.
c Find the smallest positive angle θ that satisfies the equation 5 cos 1θ + 2 sin 1θ = −1.
2 2
12 a Given that 3secθ + 4 cosecθ = 2 cosec 2θ, show that 3sinθ + 4 cosθ = 1.
b Express 3sinθ + 4 cosθ in the form R sin(θ + α ) , where R . 0 and 0° , α , 90°.
72 Give the value of α correct to 2 decimal places.
c Hence solve the equation 3secθ + 4 cosecθ = 2 cosec 2θ for 0° , θ , 360°.
ReviewPS 13 a Express sin(θ + 30°) + cosθ in the form R sin(θ + α ), where R . 0 and 0° , α , 90°.
Give the exact values of R and α .
b Hence solve the equation sin(θ + 30°) + cosθ = 1 for 0° , θ , 360°.
PS 14 a Find the maximum and minimum values of 7 sin2 θ + 9 cos2 θ + 4 sinθ cosθ + 2.
b Hence, or otherwise, solve the equation 7 sin2 θ + 9 cos2 θ + 4 sinθ cosθ = 10 for 0° , θ , 360°.
Review
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Checklist of learning and understanding
Cosecant, secant and cotangent
cosec θ = 1 sec θ = 1 cot θ = 1
sin θ cos θ tan θ
yy y
y = cosec θ
y = sec θ
11 y = cot θ
O θ O θO
−1 90 180 270 360 −1 90 180 270 360 θ
90 180 270 360
Review
Trigonometric identities
● 1 + tan2 x ≡ sec2 x
● 1 + cot2 x ≡ cosec2 x
Compound angle formula
● sin(A + B) ≡ sin A cos B + cos A sin B
● sin(A − B) ≡ sin A cos B − cos A sin B
● cos(A + B) ≡ cos A cos B − sin A sin B 73
● cos(A − B) ≡ cos A cos B + sin A sin B
Review ● tan( A + B) ≡ tan A + tan B
1 − tan A tan B
● tan( A − B) ≡ tan A − tan B
1 + tan A tan B
Double angle formulae
● sin 2A ≡ 2 sin A cos A
● cos 2A ≡ cos2 A − sin2 A
≡ 1 − 2 sin2 A
≡ 2 cos2 A − 1
● tan 2A ≡ 1 2 tan A
− tan2 A
Review Expressing a sin θ + b cos θ in the form R sin(θθ ± α ) or R cos(θθ ± α )
● a sinθ ± b cosθ = R sin(θ ± α )
● a cosθ ± b sinθ = R cos(θ ∓ α )
where R = a2 + b2 and tan α = b
a
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END-OF-CHAPTER REVIEW EXERCISE 3
1 Sketch the graph of y = 3sec(2x − 90°) for 0° , x , 180°. [3]
[5]
2 By expressing the equation cosecθ = 3sinθ + cotθ in terms of cosθ only, solve the equation
for 0° , θ , 180°.
Cambridge International A Level Mathematics 9709 Paper 31 Q3 June 2016
3 Given that cos A = 1 , where 270° , A , 360°, find the exact value of sin 2A. [5]
4 [6]
4 Solve the equation 2 tan2 x + sec x = 1 for 0° ø x ø 360°.
Review 5 Solve the equation 2 cot2 x + 5 cosec x = 10 for 0° , x , 360°. [6]
6 a Prove that sin(x + 60°) + cos(x + 30°) ≡ 3 cos x. [3]
[3]
b Hence solve the equation sin(x + 60°) + cos(x + 30°) = 3 for 0° , x , 360°.
2
7 a Prove that sin(60° − x) + cos(30° − x) ≡ 3 cos x. [3]
[3]
b Hence solve the equation sin(60° − x) + cos(30° − x) = 2 sec x for 0° , x , 360°.
5 [3]
[3]
8 i Show that the equation tan(x + 45°) = 6 tan x can be written in the form 6 tan2 x − 5 tan x + 1 = 0.
ii Hence solve the equation tan(x + 45°) = 6 tan x, for 0° , x , 180°.
Cambridge International AS & A Level Mathematics 9709 Paper 21 Q3 June 2010
9 a Prove that tan(x + 45°) − tan(45° − x) ≡ 2 tan 2x. [4]
[3]
74
b Hence solve the equation tan(x + 45°) − tan(45° − x) = 6 for 0° , x , 180°.
Review 10 i Express 3 cosθ + sinθ in the form R cos(θ − α ), where R . 0 and 0° , α , 90°, giving the exact [3]
value of R and the value of α correct to 2 decimal places.
ii Hence solve the equation 3 cos 2x + sin 2x = 2, giving all solutions in the interval 0° ø x ø 360°. [5]
Cambridge International AS & A Level Mathematics 9709 Paper 21 Q7 November 2013
11 a Prove that cos(60° − x) + cos(300° − x) ≡ cos x. [3]
b Hence
i find the exact value of cos15° + cos 255° [2]
[3]
ii solve the equation cos(60° − x) + cos(300° − x) = 1 cosec x for 0° , x , 180°. [3]
4
sin 2θ − 3 cos 2θ + [4]
12 a Prove the identity 2 sinθ 3 ≡ 4 cosθ + 6 sinθ . [1]
b
Express 4 cosθ + 6 sinθ in the form R cos(θ − α ), where R .0 and 0,α , π . [3]
c 2 [4]
[2]
Give the exact value of R and the value of α correct to 2 decimal places.
Review 2 sin 2θ − 3 cos 2θ + 3 2
Write down the greatest value of sinθ .
13 i Express 4 sinθ − 6 cosθ in the form R sin(θ − α ) , where R . 0 and 0° , α , 90°.
Give the exact value of R and the value of α correct to 2 decimal places.
ii Solve the equation 4 sinθ − 6 cosθ = 3 for 0° ø θ ø 360°.
iii Find the greatest and least possible values of (4 sinθ − 6 cosθ )2 + 8 as θ varies.
Cambridge International AS & A Level Mathematics 9709 Paper 21 Q8 June 2011
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14 i By first expanding sin(2θ + θ ), show that sin 3θ = 3sinθ − 4 sin3 θ . [4]
ii
Show that, after making the substitution x = 2 sinθ , the equation x3 − x + 1 3 = 0 can be written in
iii 36
the form sin 3θ = 3 . [1]
15 a 4
b Hence solve the equation x3 − x + 1 3 = 0, giving your answers correct to 3 significant figures. [4]
6
Cambridge International A Level Mathematics 9709 Paper 31 Q8 November 2014
Prove the identity 1 ≡ sec x. [3]
sin(x + 30°) + cos(x + 60°)
Review Hence solve the equation sin(x + 30°) 2 + 60°) = 7 − tan2 x for 0° , x , 360°. [6]
+ cos(x
16 a Prove the identity cosec4 x − cot4 x ≡ cosec2 x + cot2 x. [3]
b Hence solve the equation cosec4 x − cot4 x = 16 − cot x for 0° , x , 180°. [6]
17 The curves C1 and C2 have equations y = 1 + 4 cos 2x and y = 2 cos2 x − 4 sin 2x.
a Show that the x-coordinates of the points where C1 and C2 intersect satisfy the equation [3]
3 cos 2x + 4 sin 2x = 0.
b Express 3 cos 2x + 4 sin 2x in the form R sin(2x + α ), where R . 0 and 0° , α , 90°.
Give the exact value of R and the value of α correct to 2 decimal places. [3]
c Hence find all the roots of the equation 3 cos 2x + 4 sin 2x = 0 for 0° , x , 180°. [3]
75
Review
Review
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CROSS-TOPIC REVIEW EXERCISE 1
1 Find the set of values of x satisfying the inequality 3 x − 1 , 2x + 1 . [4]
Cambridge International A Level Mathematics 9709 Paper 31 Q1 November 2012
2 Solve the equation 2 3x − 1 = 3x, giving your answers correct to 3 significant figures. [4]
Cambridge International A Level Mathematics 9709 Paper 31 Q2 November 2013
3 i Solve the equation 3x + 4 = 3x − 11 . [3]
Review ii Hence, using logarithms, solve the equation 3 × 2y + 4 = 3 × 2y − 11 , giving the [2]
answer correct to 3 significant figures.
Cambridge International AS & A Level Mathematics 9709 Paper 21 Q1 June 2015
4 i Solve the equation 2 x − 1 = 3 x . [3]
ii Hence solve the equation 2 5x − 1 = 3 5x , giving your answer correct to [2]
3 significant figures.
Cambridge International A Level Mathematics 9709 Paper 31 Q1 June 2016
5 ln y
76 (5, 4.49)
Review (0, 2.14)
Ox [5]
( )The variables x and y satisfy the equation y = A bx , where A and b are constants. The
graph of ln y against x is a straight line passing through the points (0, 2.14) and (5, 4.49),
as shown in the diagram. Find the values of A and b, correct to 1 decimal place.
Cambridge International AS & A Level Mathematics 9709 Paper 21 Q2 June 2012
6 i It is given that x satisfies the equation 32x = 5(3x ) + 14. Find the value of 3x and, using [4]
logarithms, find the value of x correct to 3 significant figures. [1]
ii Hence state the values of x satisfying the equation 32 x = 5(3 x ) + 14.
Cambridge International AS & A Level Mathematics 9709 Paper 21 Q1 November 2016
Review 7 The variables x and y satisfy the equation xn y = C , where n and C are constants.
When x = 1.10, y = 5.20, and when x = 3.20, y = 1.05.
i Find the values of n and C . [5]
ii Explain why the graph of ln y against ln x is a straight line. [1]
Cambridge International A Level Mathematics 9709 Paper 31 Q3 June 2010
8 Given that 3ex + 8e−x = 14, find the possible values of ex and hence solve the equation 3ex + 8e−x = 14
correct to 3 significant figures. [6]
Cambridge International AS & A Level Mathematics 9709 Paper 21 Q3 June 2016
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9 The angles θ and φ lie between 0° and 180°, and are such that tan(θ − φ) = 3 and tanθ + tanφ = 1. [6]
Find the possible values of θ and φ.
Cambridge International A Level Mathematics 9709 Paper 31 Q3 November 2015
10 i Solve the equation 4x − 1 = x − 3 . [3]
ii Hence solve the equation 4y+1 − 1 = 4y − 3 correct to 3 significant figures. [3]
Cambridge International A Level Mathematics 9709 Paper 31 Q4 June 2013
Review 11 The polynomial 4x3 + ax2 + 9x + 9, where a is a constant, is denoted by p(x). It is given [3]
that when p(x) is divided by (2x − 1) the remainder is 10. [4]
i Find the value of a and hence verify that (x − 3) is a factor of p(x).
ii When a has this value, solve the equation p(x) = 0.
Cambridge International AS & A Level Mathematics 9709 Paper 21 Q5 November 2011
12 The polynomial 2x3 − 4x2 + ax + b, where a and b are constants, is denoted by p(x).
It is given that when p(x) is divided by (x + 1) the remainder is 4, and that when p(x)
is divided by (x − 3) the remainder is 12.
i Find the values of a and b. [5]
( )ii When a and b have these values, find the quotient and remainder when p(x) is divided 77
by x2 − 2 .
[3]
Review Cambridge International AS & A Level Mathematics 9709 Paper 21 Q7 November 2012
13 i Express cos x + 3 sin x in the form R cos (x − α ), where R . 0 and 0° , α , 90°, [3]
giving the exact value of R and the value of α correct to 2 decimal places.
ii Hence solve the equation cos 2θ + 3 sin 2θ = 2, for 0° , θ , 90°. [5]
Cambridge International A Level Mathematics 9709 Paper 31 Q6 November 2011
14 It is given that 2 ln(4x − 5) + ln(x + 1) = 3 ln 3.
i Show that 16x3 − 24x2 − 15x − 2 = 0. [3]
ii By first using the factor theorem, factorise 16x3 − 24x2 − 15x − 2 completely. [4]
iii Hence solve the equation 2 ln(4x − 5) + ln(x + 1) = 3 ln 3. [1]
Cambridge International A Level Mathematics 9709 Paper 31 Q6 June 2014
Review 15 The polynomial 8x3 + ax2 + bx − 1, where a and b are constants, is denoted by p(x).
It is given that (x + 1) is a factor of p(x) and that when p(x) is divided by ( 2x + 1) the
remainder is 1.
i Find the values of a and b. [5]
ii When a and b have these values, factorise p(x) completely. [3]
Cambridge International A Level Mathematics 9709 Paper 31 Q6 November 2015
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16 The polynomial 3x3 + 2x2 + ax + b, where a and b are constants, is denoted by p(x).
It is given that (x − 1) is a factor of p(x), and that when p(x) is divided by (x − 2) the
remainder is 10.
i Find the values of a and b. [5]
ii When a and b have these values, solve the equation p(x) = 0. [4]
Cambridge International AS and A Level Mathematics 9709 Paper 21 Q7 November 2010
Review 17 i The polynomial x3 + ax2 + bx + 8, where a and b are constants, is denoted by p(x).
It is given that when p(x) is divided by (x − 3) the remainder is 14, and that when p(x)
is divided by (x + 2) the remainder is 24. Find the values of a and b. [5]
ii When a and b have these values, find the quotient when p(x) is divided by x2 + 2x − 8 [4]
and hence solve the equation p(x) = 0.
Cambridge International AS & A Level Mathematics 9709 Paper 21 Q4 November 2013
18 i Given that (x + 2) and (x + 3) are factors of 5x3 + ax2 + b, find the values of the [4]
constants a and b. [5]
ii When a and b have these values, factorise 5x3 + ax2 + b completely, and hence solve the equation
53y+1 + a × 52y + b = 0, giving any answers correct to 3 significant figures.
Cambridge International AS & A Level Mathematics 9709 Paper 21 Q5 November 2014
78
19 i Find the quotient and remainder when x4 + x3 + 3x2 + 12x + 6 is divided by (x2 − x + 4). [4]
Review ii It is given that, when x4 + x3 + 3x2 + px + q is divided by (x2 − x + 4), the remainder [2]
is zero. Find the values of the constants p and q.
iii When p and q have these values, show that there is exactly one real value of x satisfying [3]
the equation x4 + x3 + 3x2 + px + q = 0 and state what that value is.
Cambridge International AS & A Level Mathematics 9709 Paper 21 Q6 November 2015
20 The angle α lies between 0° and 90° and is such that 2 tan2 α + sec2 α = 5 − 4 tanα. [4]
i Show that 3tan2 α + 4 tanα − 4 = 0 and hence find the exact value of tanα.
ii It is given that the angle β is such that cot(α + β ) = 6. Without using a calculator, [5]
find the exact value of cot β.
Cambridge International AS & A Level Mathematics 9709 Paper 21 Q7 November 2014
Review 21 i Express 5 sin 2θ + 2 cos 2θ in the form R sin(2θ + α ), where R . 0 and 0° , α , 90°,
giving the exact value of R and the value of α correct to 2 decimal places.
Hence [3]
ii solve the equation 5 sin 2θ + 2 cos 2θ = 4, giving all solutions in the [5]
[2]
interval 0° ø θ ø 360°,
iii determine the least value of 1 as θ varies.
(10 sin 2θ + 4 cos 2θ )2
Cambridge International AS & A Level Mathematics 9709 Paper 21 Q7 June 2013
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22 The polynomial p(x) is defined by p(x) = ax3 + 3x2 + bx + 12, where a and b are constants.
It is given that (x + 3) is a factor of p(x). It is also given that the remainder is 18 when p(x)
is divided by (x + 2).
i Find the values of a and b. [5]
ii When a and b have these values,
a show that the equation p(x) = 0 has exactly one real root, [4]
b solve the equation p(sec y) = 0 for −180° , y , 180°. [3]
Review Cambridge International AS & A Level Mathematics 9709 Paper 21 Q7 November 2016
79
Review
Review
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Review80
Review Chapter 4
Differentiation
In this chapter you will learn how to:
differentiate products and quotients
■■ use the derivatives of ex, ln x, sin x, cos x, tan x, together with constant multiples, sums,
differences and composites
■ find and use the first derivative of a function, which is defined parametrically or implicitly.
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PREREQUISITE KNOWLEDGE
Where it comes from What you should be able to do Check your skills
Pure Mathematics 1 Differentiate xn together with
Coursebook, Chapter 7 constant multiples, sums and 1 Differentiate with respect to x.
differences.
Pure Mathematics 1 a y = 5x3 − 3 +2 x
Coursebook, Chapter 7 Differentiate composite functions x2
using the chain rule.
Pure Mathematics 1 b y = x8 − 4x5 + x2
Coursebook, Chapter 7 Find tangents and normals to 2x3
curves.
Review Pure Mathematics 1 2 Differentiate with respect to x.
Coursebook, Chapter 8 Find stationary points on curves a (3x − 5)4
and determine their nature.
b 4
1 − 2x
3 Find the equation of the normal to
the curve y = x3 − 5x2 + 2x − 1 at the
point (1, −3).
4 Find the stationary points on the curve
y = x3 − 3x2 + 2 and determine their
nature.
81
Review Why do we study differentiation? REWIND
Review In this chapter we will learn how to differentiate the product and the quotient of two This chapter builds
simple functions. We will also learn how to find and use the derivatives of exponential, on the work from
logarithmic and trigonometric functions. Pure Mathematics 1
Coursebook, Chapters 7
All of the functions that we have differentiated so far have been of the form y = f(x). In and 8 where we learnt
this chapter we will also learn how to differentiate functions that cannot be written in the how to find and use
form y = f(x). For example, the function y2 + 2xy = 4. the derivative of xn.
Lastly, we will learn how to find and use the derivative of a function where the variables x WEB LINK
and y are given as a function of a third variable.
Explore the Calculus
Although these topics might seem like they are just for Pure Mathematics problems, of trigonometry
differentiation is used for calculations to do with quantum mechanics and field theories in and logarithms
Physics. and the Chain rule
and integration by
4.1 The product rule substitution stations
on the Underground
The function y = (x + 1)4(3x − 2)3 can be considered as the product of two separate Mathematics website.
functions:
y = uv where u = (x + 1)4 and v = (3x − 2)3
To differentiate the product of two functions we can use the product rule.
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KEY POINT 4.1
The product rule is:
d ( uv ) = u dv +v du
dx dx dx
Some people find it easier to remember this rule as:
‘(first function × derivative of second function) + (second function × derivative of first function)’
So for y = (x + 1)4(3x − 2)3,
Review dy = (x + 1)4 d (3x − 2)3 + (3x − 2)3 d (x + 1)4
dx dx dx
first differentiate second + second differentiate first
= (x + 1)4 × 9(3x − 2)2 + (3x − 2)3 × 4(x + 1)3
= (x + 1)3(3x − 2)2 [ 9(x + 1) + 4(3x − 2) ]
= (x + 1)3(3x − 2)2(21x + 1)
The product rule from first principles
Consider the function y = uv where u and v are functions of x.
A small increase δx in x leads to corresponding small increases δy, δu and δv iny, u and v.
y + δy = (u + δu) (v + δv) Expand brackets and replace y with uv.
82 uv + δy = uv + uδv + vδu + δuδv Subtract uv from both sides.
δy = uδv + vδu + δuδv Divide both sides by δx.
Review δy =u δv + v δu + δu δv ------------(1)
δx δx δx δx
As δx → 0, then so do δy, δu and δv and
δy → dy , δu → du and δv → dv .
δx dx δx dx δx dx
Equation (1) becomes:
dy = u dv +v du + 0 dv
dx dx dx dx
∴ dy = u dv +v du
dx dx dx
WORKED EXAMPLE 4.1
Review Find dy when y = (2x − 1) 4x + 5 .
dx
Answer
1
y = (2x − 1)(4x + 5)2
dy = (2x − 1) d [(4x 1 + (4x + 1 d (2x − 1)
dx dx dx
+ 5)2 ] 5)2
first differentiate second + second differentiate first
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first differentiate second + second differentiate first
1
2 (4)
( ) = − 1 + − +
(2x 1) (4x 4x + 5 (2)
5)
2
Chain rule.
= 2(2x − 1) + 2 4x + 5 Write as a single fraction.
4x +5 Simplify the numerator.
= 2(2x − 1) + 2(4x + 5)
4x + 5
= 12x + 8
4x + 5
Review WORKED EXAMPLE 4.2
Find the x-coordinate of the points on the curve y = (2x − 3)2 (x + 5)3 where the gradient is 0.
Answer
y = (2x − 3)2(x + 5)3
dy = (2x − 3)2 d [(x + 5)3 ] + (x + 5)3 d [(2x − 3)2 ]
dx dx dx
first differentiate second + second differentiate first
= (2x − 3)2 [3(x + 5)2 (1)] +(x + 5)3 [2(2x − 3)1(2)] 83
Chain rule Chain rule
= 3(2x − 3)2 (x + 5)2 + 4(2x − 3)(x + 5)3 Factorise.
Simplify.
Review = (2x − 3)(x + 5)2[3(2x − 3) + 4(x + 5)]
= (2x − 3)(x + 5)2 (10x + 11)
dy = 0 when (2x − 3)(x + 5)2 (10x + 11) = 0
dx
2x − 3 = 0 x + 5 = 0 10x + 11 = 0
x = 3 x = −5 x = −1.1
2
EXERCISE 4A
1 Use the product rule to differentiate each of the following with respect to x:
a x(x − 2)5 b 5x(2x + 1)3 c x x+2
f x (x2 + 2)3
Review d (x − 1) x + 5 e x3 2x − 1 i (2x − 5)(3x2 + 1)2
g (x − 3)2 (x + 2)5 h (2x − 1)5(3x + 4)4
2 Find the gradient of the curve y = x2 x + 4 at the point (−3, 9).
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Review 3 Find the equation of the tangent to the curve y = (2 − x)3(x + 1)4 at the point where x = 1.
4 Find the gradient of the tangent to the curve y = (x + 2)(x − 1)3 at the point where the curve meets the
y -axis.
5 Find the x-coordinate of the points on the curve y = (3 − x)3(x + 1)2 where the gradient is zero.
6 Find the x-coordinate of the point on the curve y = (x + 2) 1 − 2x where the gradient is zero.
PS 7 a Sketch the curve y = (x − 1)2 (5 − 2x) + 3.
b The curve y = (x − 1)2 (5 − 2x) + 3 has stationary points at A and B. The straight line through A and B
cuts the axes at P and Q. Find the area of the triangle POQ.
4.2 The quotient rule
We can differentiate the function y= x2 − 5 by writing the function in the form
2x + 1
y = (x2 − 5)(2x + 1)−1 and then by applying the product rule.
Alternatively, we can consider y = x2 − 5 as the division (quotient) of two separate
functions: 2x + 1
y= u where u = x2 − 5 and v = 2x + 1.
v
To differentiate the quotient of two functions we can use the quotient rule.
84
KEY POINT 4.2
Review The quotient rule is:
d u = v du − u dv
dx v dx dx
v2
Some people find it easier to remember this rule as
(denominator × derivative of numerator) − (numerator × derivative of denominator)
(denominator)2
The quotient rule from first principles
Consider the function y = u where u and v are functions of x.
v
A small increase δx in x leads to corresponding small increases δy, δu and δv in y, u
and v.
Review
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v + δv
85
δy = u + δu − y Replace y with u .
v + δv v
δy = u + δu − u Combine fractions.
v + δv v
δy = v(u + δu) − u(v + δv) Expand brackets and simplify.
v(v + δv)
δy = vδu − uδv Divide both sides by δx.
v2 + vδv
Review δy = v δu − u δv
δx δx δx
v2 + vδv ---------------- (1)
As δx → 0 , then so do δy, δu and δv and δy → dy , δu → du and δv → dv .
δx dx δx dx δx dx
Equation (1) becomes
d u = v du − u dv
dx v dx dx
v2
WORKED EXAMPLE 4.3
Find the derivative of y = x2 − 5 .
2x + 1
Review Answer
y = x2 − 5
2x + 1
denominator differentiate numerator − numerator differentiate denominator
dy (2x + 1) × d (x2 − 5) − (x2 − 5) × d (2x + 1)
dx dx dx
=
(2x + 1)2
denominator squared
= (2x + 1)(2x) − (x2 − 5)(2)
(2x + 1)2
= 4x2 + 2x − 2x2 + 10
(2x + 1)2
= 2(x2 + x + 5)
(2x + 1)2
Review
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WORKED EXAMPLE 4.4
Find the derivative of y = ( x + 2)3 .
x −1
Answer
y = (x + 2)3
x −1
denominator differentiate numerator − numerator differentiate denominator
Review dy x−1 × d [(x + 2)3 ] − (x + 2)3 × d x − 1
dx dx dx
= ( )x − 1 2
denominator squared
( )=x −1 [3(x + 2)2 (1)] − (x + 2)3 1 (x − 1)− 1 (1)
2 2
x −1
3(x + 2)2 x − 1 − (x + 2)3
= 2 x − 1
Multiply numerator and denominator by
x −1 2 x − 1.
86 = 6(x + 2)2(x − 1) − (x + 2)3 Factorise the numerator.
2(x − 1) x − 1
Review = ( x + 2 )2 [6( x − 1) − (x + 2)]
3
2(x − 1)2
= (x + 2)2(5x − 8)
3
2(x − 1)2
EXPLORE 4.1
In Worked examples 4.3 and 4.4 we differentiated the functions y = x2 − 5 and
y (x + 2)3 2x + 1 a product,
= x −1 using the quotient rule. Now write each of these functions as
such as y = (x2 − 5)(2x + 1)−1, and then use the product rule to differentiate them.
Review Do you obtain the same answers?
Discuss with your classmates which method you prefer for functions such as these:
the quotient rule or the product rule.
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EXERCISE 4B
1 Use the quotient rule to differentiate each of the following with respect to x:
a 2x + 3 b 3x − 5 c x2 − 3 d 3x + 1
x−4 2−x 2x − 1 2 − 5x
e 1 − 2x2 f 5x4 g 3x2 − 7x h (x + 4)2
(x + 4)2 (x2 − 1)2 x2 + 2x + 5 (x2 + 1)3
2 Find the gradient of the curve y= x−5 at the point 2, − 1 .
x+4 2
Review 3 Find the coordinates of the points on the curve y = ( x − 1)2 where the tangent is parallel to the x-axis.
2x + 5
1 − 2x
4 Find the coordinates of the points on the curve y = x−5 at which the gradient is 1.
5 Find the equation of the tangent of the curve y = x−4 at the point where the curve crosses the y -axis.
2x + 1
6 Differentiate with respect to x:
a x b x −1 c 3 − x2 d 5(x − 1)3
5x − 1 2x + 3 x2 − 1 x+2
7 Find the x-coordinate of the point on the curve y = x +1 where the gradient is 0.
x −1
x2 + 1
8 Find the equation of the normal to the curve y = x+2 at the point (−1, 2).
PS 9 The line 2x − 2y = 5 intersects the curve 2x2 y − x2 − 26y − 35 = 0 at three points. 87
a Find the x-coordinates of the points of intersection.
Review b Find the gradient of the curve at each of the points of intersection.
4.3 Derivatives of exponential functions
The derivative of ex
In Chapter 2, we learnt about the natural exponential function f(x) = ex. This function
has a very special property. If we use graphing software to draw the graph of f(x) = ex
together with the graph of its gradient function we find that the two curves are identical.
KEY POINT 4.3
Hence, we have the rule:
d (ex ) = ex
dx
Review
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An explanation of how this rule can be obtained is as follows.
Consider the function f(x) = ex and two points whose x-coordinates are x and x + δx
where δx is a small increase in x.
y f(x) = ex
f(x + dx)
f(x)
O x x + dx x
Review dy = lim f(x + δx) − f(x) = lim ex+δx − ex = lim ex (eδx − 1)
dx (x + δx) − x δx δx
δx → 0 δx → 0 δx → 0
Now consider eδx − 1 for small values of δx. TIP
δx
δx 0.1 0.01 0.001 0.0001 f(x + δx) − f(x)
(x + δx) − x
eδx − 1 lim
δx
δx → 0
1.051709 1.005 017 1.000 500 1.000 050 means the limit of
eδx − 1 f(x + δx) − f(x) as
δx (x + δx) − x
From the table, we can see that as δx → 0, → 1.
δx → 0 .
dy
∴ dx = ex
88 The derivative of ef(x)
Consider the function y = ef(x).
Review Let y = eu where u = f(x)
dy = eu du = f ′(x)
du dx
Using the chain rule: dy = dy × du
dx du dx
= eu × f ′(x)
= f ′(x) × ef(x)
d [ef(x) ] = f ′(x) × e f( x )
dx
KEY POINT 4.4
Review In particular,
d [eax+b ] = a eax+b
dx
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