Transcript C3.3 Trigonometry 1

A-Level Maths:
Core 3
for Edexcel
C3.3 Trigonometry 1
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Contents
The inverse trigonometric functions
The inverse trigonometric functions
The reciprocal trigonometric functions
Trigonometric identities
Examination-style question
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The inverse of the sine function
Suppose we wish to find θ such that
sin θ = x
In other words, we want to find the angle whose sine is x.
This is written as
θ = sin–1 x or θ = arcsin x
In this context, sin–1 x means the inverse of sin x.
This is not the same as (sin x)–1 which is the reciprocal of
sin x, 1 .
sin x
Is y = sin–1 x a function?
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The inverse of the sine function
We can see from the graph
of y = sin x between x = –2π
and x = 2π that it is a manyto-one function:
y
y = sin x
x
y
The inverse of this graph is
not a function because it is
one-to-many:
y = sin–1 x
x
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The inverse of the sine function
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 – 2 ≤ x ≤ 2 if working in radians).
There is only one value of sin–1 x in this range, called the
principal value.
So, if we restrict the domain of f(x) = sin x to –  ≤ x ≤  we
2
2
have a one-to-one function:
y
y = sin x
1



2
2
x
–1
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The graph of y = sin–1 x
Therefore the inverse of f(x) = sin x, –  ≤ x ≤  , is also a
2
2
one-to-one function:
f –1(x) = sin–1 x
y
y = sin–1 x
–1

The graph of y = sin x
2
1
is the reflection of
y = sin x
y = sin x in the line y = x:
x
(Remember the scale
1 2
 2 –1
–1
used on the x- and y-axes
 2
must be the same.)
The domain of sin–1 x is the same as the range of sin x :
–1 ≤ x ≤ 1
The range of sin–1 x is the same as the restricted domain of
sin x :
– 2 ≤ sin–1 x ≤ 2
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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
then
f –1(x) = cos–1 x
And if
f(x) = tan x
then
f –1(x) = tan–1 x
for
for
for
for
0≤x≤π
–1 ≤ x ≤ 1.
– 2 < x < 2
x .
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.
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The graph of y = cos–1 x
y=
cos–1 x
y


2
1
–1
0
–1
1

2

x
y = cosx
The domain of cos–1 x is the same as the range of cos x :
–1 ≤ x ≤ 1
The range of cos–1 x is the same as the restricted domain of
cos x :
0 ≤ cos–1 x ≤ π
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The graph of y = tan–1 x
y
y = tan
tanxx

y = tan–1 x
2
 2

x
2
 2
The domain of tan–1 x is the same as the range of tan x :
x
The range of tan–1 x is the same as the restricted domain of
tan x :
–  < tan–1 x < 
2
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Problems involving inverse trig functions
Find the exact value of sin–1
3
2
in radians.
To solve this, remember the angles whose trigonometric ratios
can be written exactly:
radians
0

6

4

3

2
degrees
0°
30°
45°
60°
90°
sin
0
1
2
1
tan
0
3
2
1
3
3
2
1
2
1
cos
1
2
1
2
1
3
From this table
sin–1
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3
2
0

= 
3
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Problems involving inverse trig functions
Find the exact value of sin–1 
2
2
in radians.
This is equivalent to solving the trigonometric equation
cos θ = – 2
for 0 ≤ θ ≤ π
2
this is the range of cos–1x
We know that cos 4 = 1 = 22
2
Sketching y = cos θ for 0 ≤ θ ≤ π :
1
From the graph, cos 34 = – 22
2
2

0
2
2
–1
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

4
2
3
4

θ
So, cos–1 
2
2
= 3
4
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Problems involving inverse trig functions
Find the exact value of cos (sin–1
Let sin–1
7
4
=θ
so
7
4
) in radians.
7
4
sin θ =
Using the following right-angled triangle:
7 + a2 = 16
a=3
4
7
θ
3
The length of the third side is 3 so
cos θ =
But sin–1
7
4
= θ so
cos (sin–1
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3
4
7
)
4
=
3
4
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Contents
The reciprocal trigonometric functions
The inverse trigonometric functions
The reciprocal trigonometric functions
Trigonometric identities
Examination-style question
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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:
1
cosec x =
sin x
This is short
for cosecant.
1
sec x =
cos x
This is short
for secant.
1
cot x =
tan x
This is short
for cotangent.
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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The graph of sec x
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The graph of cosec x
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The graph of cot x
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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:
function domain
range
odd
period or
even
asymptotes f(x) = 0
at x =
when x =
f(x) =
sec x
x
f(x) ≤ –1,
x ≠ 90°+180n°
f(x) ≥ 1
n
90°+180n°,
n
never
360°
even
f(x) =
cosec x
x
x ≠ 180n°
n
f(x) ≤ –1,
f(x) ≥ 1
180n°,
n
never
360°
odd
f(x) =
cot x
x
x ≠ 180n°
n
f(x) 
180n°,
n
90°+180n°,
n
180°
odd
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Transforming the graph of f(x) = sec x
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Transforming the graph of f(x) = cosec x
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Transforming the graph of f(x) = cot x
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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°
1
a) sec 85° =
cos85
c) cot –70°
= 11.47 (to 2 d.p.)
1
b) cosec 220° =
= –1.56 (to 2 d.p.)
sin 220
1
c) cot –70° =
= –0.36 (to 2 d.p.)
tan  70
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Problems involving reciprocal trig functions
Find the exact value of:
a) cosec 6
a) sin  =
1
2
cosec  = 2
so,
6
θ is in the 2nd quadrant
so tan θ = tan (π – θ)
= – tan 
3
=– 3
cot 2 = – 1 = – 3
so,
c) cos – 3 = –cos(– 3 + π)
= – cos 
=– 1
4
3
3
3
θ is in the 3rd quadrant
so cos θ = –cos (θ + π)
4
2
so,
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4
3
6
b) tan 2 = –tan (π – 2 )
3
3
4
c) sec – 3
b) cot 2
sec – 3 = – 2
4
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Problems involving reciprocal trig functions
Given that x is an acute angle and tan x =
values of cot x, sec x and cosec x.
3
4
find the exact
Using the following right-angled triangle:
5
3
The length of the hypotenuse is
x
9 +16 = 5
4
So
tan x =
3
4
cos x =
4
5
sin x =
cot x =
4
3
sec x =
5
4
cosec x =
3
5
Therefore
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5
3
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Problems involving reciprocal trig functions
Prove that tan x sin x  sec x  cos x.
LHS = tan x sin x
sin x
=
sin x
cos x
sin2 x
=
cos x
1  cos2 x
=
cos x
Using sin2x + cos2x = 1
1
cos2 x
=

cos x cos x
= sec x  cos x = RHS
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Problems involving reciprocal trig functions
Solve sec (x + 20°) = 2 for 0 ≤ x ≤ 360°.
sec( x + 20 ) = 2
1
=2
cos( x + 20 )
1
cos (x + 20°) =
2
x + 20° = 60° or 300°
x = 40° or 280°
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Contents
Trigonometric identities
The inverse trigonometric functions
The reciprocal trigonometric functions
Trigonometric identities
Examination-style question
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Trigonometric identities
Earlier in the course you met the following trigonometric
identities:
sin
 tan (cos  0)
1
cos
sin2   cos2   1
2
We can write these identities in terms of sec θ, cosec θ and cot θ.
Using 1
1 = 1
cos
cot  =
=
sin 
tan
sin
cos 
So
cos
 cot 
sin
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(sin  0)
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Trigonometric identities
sin2   cos2   1
2
Dividing 2 through by cos2θ gives
sin2  cos2 
1
+

cos2  cos2  cos2 
tan2  +1  sec 2 
Dividing 2 through by sin2θ gives
sin2  cos2 
1
+

2
2
sin  sin  sin2 
1+ cot 2   cosec 2
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Trigonometric identities
tan x
Show that
  cot x.
2
1  sec x
tan x
LHS =
1  sec 2 x
=
tan x
1  tan2 x  1
Using sec2x = tan2x + 1
tan x
=
 tan2 x
=
1
 tan x
=  cot x = RHS
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Trigonometric identities
Given that x is an obtuse angle and cosec x = 5, find the
exact value of tan x.
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:
1
cot x = 2 6  tan x = 
2 6
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Trigonometric equations
Solve 2sec 2  = 2 + tan for 0    360
Using sec 2   1+ tan2  :
2(1+ tan2  ) = 2 + tan
2 + 2tan2   2  tan = 0
2tan2   tan = 0
tan (2tan  1) = 0
tan = 0
or
θ = 0°, 180°, 360°
tan = ½
θ = 26.6°, 206.6° (to 1 d.p.)
The complete solution set is θ = 0°, 26.6°, 180°, 206.6°, 360°.
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Contents
Examination-style question
The inverse trigonometric functions
The reciprocal trigonometric functions
Trigonometric identities
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.
RHS = cos + sin
a)
sin
cos
cos2  sin2 
=
+
cos cos
1
=
cos
= sec  = LHS
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The reciprocal trigonometric functions
2cos = 3cosec   2sin tan
b)
2cos + 2sin tan = 3cosec 
2(cos + sin tan ) = 3cosec 
Using the result given in part a):
2sec  = 3cosec 
2
3
=
cos sin
sin 3
=
cos 2
tan = 32
 = 56.3, 236.3 (to 1 d.p.)
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