© boardworks ltd 2006 1 of 35 © boardworks ltd 2006 1 of 35 a-level maths: core 3 for edexcel c3.3...
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© Boardworks Ltd 20061 of 35 © Boardworks Ltd 20061 of 35
A-Level Maths: Core 3for Edexcel
C3.3 Trigonometry 1
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The inverse trigonometric functions
The reciprocal trigonometric functions
Trigonometric identities
Examination-style questionCo
nte
nts
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The inverse trigonometric functions
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The inverse of the sine function
Suppose we wish to find θ such that
In other words, we want to find the angle whose sine is x.
In this context, sin–1 x means the inverse of sin x.
sin θ = x
θ = sin–1 x or θ = arcsin x
This is not the same as (sin x)–1 which is the reciprocal of
sin x, .1sinx
Is y = sin–1 x a function?
This is written as
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The inverse of the sine function
The inverse of this graph is not a function because it is one-to-many:
We can see from the graph of y = sin x between x = –2π and x = 2π that it is a many-to-one function:
y
x
y = sin x
y = sin–1 x
x
y
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The inverse of the sine function
There is only one value of sin–1 x in this range, called the principal value.
However, remember that if we use a calculator to find sin–1 x (or arcsin x) the calculator will give a value between –90° and 90° (or between – ≤ x ≤ if working in radians).2
2
So, if we restrict the domain of f(x) = sin x to – ≤ x ≤ we have a one-to-one function:
2
2
y
2 2
1
–1
x
y = sin x
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y = sin x
x
y
2
2
1–1
y = sin–1 x2
2
1–1
y = sin–1 x
The graph of y = sin–1 x
The graph of y = sin–1 x is the reflection of y = sin x in the line y = x:
The domain of sin–1 x is the same as the range of sin x :
The range of sin–1 x is the same as the restricted domain of sin x :
–1 ≤ x ≤ 1
2 2
1
–1
x
y
2
2– ≤ sin–1 x ≤
(Remember the scale used on the x- and y-axes must be the same.)
Therefore the inverse of f(x) = sin x, – ≤ x ≤ , is also a one-to-one function:
2
2
f –1(x) = sin–1 x
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The inverse of cosine and tangent
We can restrict the domains of cos x and tan x in the same way as we did for sin x so that
if f(x) = cos x for 0 ≤ x ≤ π
– < x <And if f(x) = tan x for 2
2
The graphs cos–1 x and tan–1 x can be obtained by reflecting the graphs of cos x and tan x in the line y = x.
f –1(x) = cos–1 xthen for –1 ≤ x ≤ 1.
f –1(x) = tan–1 xthen for x .
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y = cosx
x0
y
2
1–1
y = cos–1 x
The graph of y = cos–1 x
The domain of cos–1 x is the same as the range of cos x :
The range of cos–1 x is the same as the restricted domain of cos x :
0 ≤ cos–1 x ≤ π
–1 ≤ x ≤ 1
–1
x0
1
2
y
2
1–1
y = cos–1 x
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y = tanx
x
y y = tan x
y = tan–1 x2
2
x2
2
y = tan–1 x2
2
The graph of y = tan–1 x
The domain of tan–1 x is the same as the range of tan x :
The range of tan–1 x is the same as the restricted domain of tan x :
y
x
– < tan–1 x <2
2
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Find the exact value of sin–1 in radians.3
2
Problems involving inverse trig functions
To solve this, remember the angles whose trigonometric ratios can be written exactly:
10
01
10 12
1212
32
32
13
12
tan
cos
sin
90°60°45°30°0°degrees
0radians 6
4
3
2
3
From this tablesin–1 =3
2 3
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Find the exact value of sin–1 in radians.
Problems involving inverse trig functions
22
This is equivalent to solving the trigonometric equation
22cos θ = – for 0 ≤ θ ≤ π
this is the range of cos–1x
We know that cos =4 1
2
Sketching y = cos θ for 0 ≤ θ ≤ π :
–1
θ0
1
2 3
4
4
22
22
From the graph, cos =34 2
2–
22=
So, cos–1 = 342
2
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Find the exact value of cos (sin–1 ) in radians.
Problems involving inverse trig functions
74
Let sin–1 = θ74 so sin θ = 7
4
Using the following right-angled triangle:
θ
47
The length of the third side is 3 so
3
7 + a2 = 16
cos θ = 34
But sin–1 = θ so74
cos (sin–1 ) = 34
74
a = 3
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The inverse trigonometric functions
The reciprocal trigonometric functions
Trigonometric identities
Examination-style questionCo
nte
nts
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The reciprocal trigonometric functions
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The reciprocal trigonometric functions
The reciprocal trigonometric functions are cosecant, secant and cotangent.
They are related to the three main trigonometric ratios as follows:
This is short for cosecant.
cosec x =1
sin x
This is short for secant.
sec x =1
cos x
This is short for cotangent.
cot x =1
tan x
Notice that the first letter of sin, cos and tan happens to be the same as the third letter of the corresponding reciprocal functions cosec, sec and cot.
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Properties of the graph of sec x
The properties of the graphs of sec x, cosec x and cot x can be summarized in the following table:
f(x) = cot x
f(x) = cosec x
f(x) = sec x
odd or
evenperiod
f(x) = 0 when x =
asymptotes at x =
rangedomainfunction
odd180°
odd360°neverf(x) ≤ –1,
f(x) ≥ 1
even360°neverf(x) ≤ –1,
f(x) ≥ 1
90°+180n°,
n
180n°,
n
x
x ≠ 90°+180n° n
f(x)
x
x ≠ 180n°
n
90°+180n°,
n 180n°,
n
x
x ≠ 180n°
n
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Problems involving reciprocal trig functions
Use a calculator to find, to 2 d.p., the value of:
a) sec 85° b) cosec 200° c) cot –70°
a) sec 85° =1
cos85 = 11.47 (to 2 d.p.)
b) cosec 220° =1
sin220 = –1.56 (to 2 d.p.)
c) cot –70° =1
tan 70 = –0.36 (to 2 d.p.)
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Problems involving reciprocal trig functions
Find the exact value of:
a) cosec b) cot c) sec –6 2
3 3
4
a) sin =6 1
2 so, cosec = 26
so, cot = –23 1
3= – 3
3
so, sec – = –34 2
b) tan = –tan (π – )23 2
3
= – tan3
= – 3
θ is in the 2nd quadrant so tan θ = tan (π – θ)
c) cos – =34 –cos(– + π)3
4
4= – cos
θ is in the 3rd quadrant so cos θ = –cos (θ + π)
= – 12
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Given that x is an acute angle and tan x = find the exact values of cot x, sec x and cosec x.
Problems involving reciprocal trig functions
34
Using the following right-angled triangle:
x
4
3
9 +16 = 5
The length of the hypotenuse is5
So
tan x = 34 cos x = 4
5 sin x = 35
Therefore
cot x = 43 sec x = 5
4 cosec x = 53
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Prove that
Problems involving reciprocal trig functions
tan sin sec cos .x x x x
LHS = tan sinx x
sin= sin
cos
xx
x
2sin=
cos
x
x
21 cos=
cos
x
x
Using sin2x + cos2x = 1
21 cos=
cos cos
x
x x
= sec cos = RHSx x
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Problems involving reciprocal trig functions
sec( + 20 ) = 2x
cos (x + 20°) =1
2
x + 20° = 60° or 300°
x = 40° or 280°
1= 2
cos( + 20 )x
Solve sec (x + 20°) = 2 for 0 ≤ x ≤ 360°.
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The inverse trigonometric functions
The reciprocal trigonometric functions
Trigonometric identities
Examination-style questionCo
nte
nts
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Trigonometric identities
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Trigonometric identities
Earlier in the course you met the following trigonometric identities:
We can write these identities in terms of sec θ, cosec θ and cot θ.
1Using
sintan (cos 0)
cos
1
22 2sin cos 1
1cot =
tan
sincos
1=
cos=
sin
So
coscot (sin 0)
sin
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Trigonometric identities
Dividing through by cos2θ gives2
22 2sin cos 1
2 2
2 2 2
sin cos 1+
cos cos cos
2 2tan +1 sec
Dividing through by sin2θ gives2
2 2
2 2 2
sin cos 1+
sin sin sin
2 21+ cot cosec
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Trigonometric identities
2
tanShow that cot .
1 sec
xx
x
2
tanLHS =
1 sec
x
x
2
tan=
1 tan 1
x
x
2
tan=
tan
x
x
1=
tan x
Using sec2x = tan2x + 1
= cot = RHSx
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Trigonometric identities
cosec x = 5
cosec2 x = 25
Using cosec2 x ≡ 1 + cot2 x,
1 + cot2 x = 25
cot2 x = 24
cot x = ±√24 = ±2√6
x is obtuse and so cot x is negative (since tan x is negative in the second quadrant). Therefore:
cot = 2 6x 1tan =
2 6x
Given that x is an obtuse angle and cosec x = 5, find the exact value of tan x.
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Trigonometric equations
Solve 22sec = 2 + tan for 0 360
Using :2 2sec 1+ tan 22(1+ tan ) = 2 + tan
22 + 2tan 2 tan = 0 22tan tan = 0
tan (2tan 1) = 0
θ = 0°, 180°, 360° θ = 26.6°, 206.6° (to 1 d.p.)
The complete solution set is θ = 0°, 26.6°, 180°, 206.6°, 360°.
ortan = 0 tan = ½
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The inverse trigonometric functions
The reciprocal trigonometric functions
Trigonometric identities
Examination-style questionCo
nte
nts
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Examination-style question
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Examination-style question
a) Prove that sec θ ≡ cos θ + sin θ tan θ.
b) Hence solve the equation 2 cos θ = 3 cosec θ – 2 sin θ tan θ in the interval 0° < θ < 360°. Give all solutions in degrees to 1 decimal place.
a)sin
= cos + sincos
RHS
2 2cos sin= +
cos cos
1=
cos
= sec = LHS