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C3 Trigonometry
In C2 you were introduced to radian measure and had to find areas
of sectors and segments. In addition to this you solved trigonometric
equations using the identities below.
Sin2   Cos 2   1
Sin
Cos
By the end of this unit you should:
Have a knowledge of secant, cosecant and cotangent and of arcsin,
arcos and arctan. Their relationship to sine, cosine and tangent and
their respective graphs including appropriate restrictions of the
domain.
Have a knowledge of 1  Cot 2   Co sec 2  and Tan 2   1  Sec 2  .
Have a knowledge of double angle formulae and “r” formulae.
Tan 
New trigonometric functions
The following three trigonometric functions are the reciprocals of sine,
cosine and tan. The way to remember them is by looking at the third
letter.
Secx 
1
Cosx
Co sec x 
Cotx 
1
Sinx
1
Tanx
These three trigonometric functions are use to derive two more
identities.
New Identities
Starting with Sin2   Cos 2   1 and by dividing by Sin2  gives:
Sin2  Cos 2 
1


2
2
Sin  Sin  Sin2 
Using the new functions outlined above and the fact that Cot 
this becomes:
1  Cot 2   Co sec 2  .
Returning to Sin2   Cos 2   1 and by dividing by Cos 2  gives:
Therefore:
Sin2  Cos 2 
1


2
2
Cos  Cos  Cos 2 
Cos
Sin
Tan 2   1  Sec 2  .
The three identities will be used time and again. Try to remember them
but you should also be able to derive them as outlined above.
Sin2   Cos 2   1
1  Cot 2   Co sec 2 
Tan 2   1  Sec 2 
Example 1
Solve for 0    360 the equation
5 tan2   sec   1 ,
giving your answers to 1 decimal place.
You should have come across questions of this type in C2 using the
identity cos2   sin2   1 . The given equation has a single power of secθ
therefore we must use an identity to get rid of the tan2θ.
tan2   1  sec2 
So the equation becomes:
5 sec2   5  sec   1
5 sec2   sec   6  0
We now have a quadratic in sec so by factorising:
5 sec2   sec   6  0
(5 sec   6)(sec   1)  0
sec  =
-6
5
sec   1
cos=
-5
6
cos=1
  146.4 ,213.6
  0 ,360 
Inverse Trigonometric Functions.
Functions are introduced in C3 and we use the concept of inverses to find
the following functions (remember that the inverse of a function in
graphical terms is its reflection in the line y = x). The domain of the
original trigonometric function has to be restricted to ensure that it is
still one to one. It is also worth remembering that the domain and range
swap over as you go from the function to the inverse. ie in the first case
π
π
 sin x 
the domain of sinx is restricted to 
and this becomes the
2
2
range of the inverse function.
y=arcsinx
Domain
1  x 1
Range

π
π
 arcsin x 
2
2
Example
Find
arcsin0.5 = y
Simply swap around
Siny = 0.5
y = П/6
y=arccosx
Domain
1  x 1
Range
0  arccos x  π
y=arctanx
Domain
x 
Range


2
 arctan x 

2
Addition Formulae
A majority of the formulae in C3 need to be learnt. One’s in red are in
the formula book.
Sin (A  B )  SinACosB  CosASinB
Cos (A  B )  CosACosB  SinASinB
Tan (A  B ) 
TanA TanB
 TanATanB
The examples below use addition formulae.
Example
Given that Sin A =
reflex find:
a)
Sin (A + B)
4
12
and that Cos B =
where A is obtuse and B is
13
5
b)
Cos (A – B)
c)
Cot (A – B)
Before we start the question it is advisable to draw the graphs of Sin x
and Cos x.
Since A is obtuse the cosine of A must be negative and by using the
5
Pythagorean triple Cos A =
. The angle B is slightly more tricky. We
13
are told that B is reflex but we know that Cos B is positive. Therefore B
must be between 270° and 360° and so Sin B is negative.
3
Hence Sin B =
.
5
We are now ready to attempt part (a)
4
12
5
3
Sin A =
Cos A =
Sin B =
Cos B =
13
13
5
5
a)
Using the formula above to find Sin (A + B)
Sin (A + B) = Sin A Cos B + Sin B Cos A
b)
c)
Sin (A + B) =
12 4  3  5
 

13 5
5
13
Sin (A + B) =
33
65
Cos (A – B) = Cos A Cos B + Sin A Sin B
Cos (A – B) =
 5 4 12  3
 

13 5 13 5
Cos (A – B) =
 56
65
Cot (A – B) =
1
1 TanATanB

Tan (A  B ) TanA TanB
The graphs above show the range of values that A and B lie within so we
need to consider the Tan graph at these points.
I have included the diagrams below to help in my calculations.
13
12
5
A
B
5
Therefore
4
3
 12
since for obtuse angles the tan graph is
5
negative and:
Tan A =
Tan B =
4
for the same reason.
3
So finally
Cot (A – B) =
1 TanATanB
TanA TanB
 12  4

5
3
Cot (A – B) =
 12 4

5
3
 63
63
Cot (A – B) = 15 
 16 16
15
1
The above example may appear to be a little mean by being non calculator
but it is an opportunity to really start thinking about the angles and
graphs involved.
Double angle Formulae
The Addition Formulae are used to derive the double angle formulae.
In all cases let A=B, therefore:
Sin A  SinACosA
Cos A  Cos A  Sin A
Tan A 
TanA
  TanA
The second of the identities above is combined with Sin2   Cos 2   1 to
express Sin2θ and Cos2θ in terms of Cos2θ. This is vital when you are
asked to integrate Sin2θ and Cos2θ.
If we start by trying to express the right hand side of the identity
Cos   Cos   Sin  in terms of Cos2θ only.
Cos 2  Cos 2  Sin 2
- Sin2  Cos2  
Cos 2  2Cos 2  1
Rearranging to make Cos2θ the subject:
Cos  

Cos   

A similar approach is used to find the identity for Sin2θ:
Sin  

  Cos  

Please note that this derivation has been tested on a C3 past paper.
Example
Given that sin x =
3
, use an appropriate double angle formula to find the
5
exact value of sec 2x.
From earlier work you should remember that Secx 
Sec 2 x 
1
and hence
Cosx
1
.
Cos2 x
Using the identity above:
Cos2  Cos 2  Sin 2
Therefore:
and the fact that: Cos 2  1 - Sin 2
and
Cos 2  1  2 Sin 2
Sec 2 x 

1
1  2 Sin 2
1
3
1  2 
5
2

5
7
Here’s one to complete yourselves!
Prove that
cot 2x + cosec 2x  cot x,
(x ≠
n
,n
2
).
Sum and Difference Formulae
SinA  SinB  2Sin
AB
AB
Cos
2
2
SinA  SinB  2Cos
AB
AB
Sin
2
2
CosA  CosB  2Cos
AB
AB
Cos
2
2
CosA  CosB  2Sin
AB
AB
Sin
2
2
The following examples deal with a variety of the identities outlined
above. Be warned, some are more complex than others.
Example 2
Given that tan 2x = ½, show that tan x = 2  5
Using the double angle formula:
tan2x 
2tan x
1  tan2 x
1
2tan x

2 1  tan2 x
1  tan2 x  4 tan x
tan2 x  4 tan x  1  0
Using the quadratic formula to solve a quadratic in tan:
tan2 x  4 tan x  1  0
tan x 
4  42  4
2
tan x 
4  20 4  2 5

2
2
tan x  2  5
Example 3
(i)
Given that cos(x + 30)º = 3 cos(x – 30)º, prove that tan x º = 3
.
2
Using double angle formulae:
cosx cos30 – sinx sin30 = 3cosxcos30 + 3sinxsin30
2cosx cos30 = -4 sinx sin30 sin 30 = ½,
3 cosx = -2sinx
cos 30 =
3
2
tan x º = (ii)
(a)
3
2
Prove that
1  cos 2
= tan θ .
sin 2
Using the fact that cos2  1  2sin2 
Rewriting the fraction:
1  cos2 1  2sin2   1

sin2
2sin  cos 

sin 
 tan 
cos 
(b)
Verify that θ = 180º is a solution of the equation
sin 2θ = 2 – 2 cos 2 θ.
Too easy! Simply let  = 180º
(c)
Using the result in part (a), or otherwise, find the
other two solutions, 0 < θ < 360º , of the equation
sin 2θ = 2 – 2 cos 2θ.
sin 2θ = 2 – 2 cos 2θ
sin 2θ = 2(1 – cos 2θ)
1 1  cos2

2
sin2
Therefore from part (a):
tanθ = 0.5
θ = 26.6º, 206.6º
Example 4
Find the values of tan θ such that
2 sin2θ - sinθsecθ = 2sin2θ - 2.
Remembering that sec  
equation becomes:
1
and the double angle formulae, the
cos 
2sin2   tan   4 sin  cos   2 Dividing by cos2 
2tan2   tan  sec2   4 tan   2sec2 
sec2  1  tan2 
2tan2   tan   tan3   4 tan   2  2tan2 
tan3   4 tan2   5 tan   2  0
We now have a trigonometric polynomial:
t3  4 t2  5t 2  0
By using factor theorem (t – 1) is a factor, therefore:
(t – 1)(t2 – 3t + 2) = (t – 1)(t – 1)(t – 2)
Hence
tan θ = 1
tan θ = 2
Example 5
(i)
Given that sin x =
3
, use an appropriate double angle formula to
5
find the exact value of sec 2x.
We can use a 3,4,5 Pythagorean triangle to show that cos x =
sec 2x =
1
cos2x
4
5
1
1

2
cos2x cos x  sin2 x

(ii)
1
16 9

25 25

25
7
Prove that
cot 2x + cosec 2x  cot x,
(x 
n
, n
2
).
Left hand side becomes:
cos2x
1

sin2x sin2x
using cos2x  2cos2 x  1 and common denomenator of sin2x
2cos2 x
2cos2 x
cos x


 cot x
sin2x
2sin xcos x sin x
R Formulae
Expressions of the type aSin  bCos can be written in terms of sine or
cosine only and hence equations of the type aSin   bCos  c can be
solved. The addition formulae outlined above are used in the derivation.
In most you cases you will be told which addition formula to use.
Example 6
f(x) = 14cosθ – 5sinθ
Given that f(x) = Rcos(θ + α), where R ≥ 0 and 0    90 ,
a)
find the value of R and α.
b)
Hence solve the equation
14cosθ – 5sinθ = 8
for 0    360 , giving your answers to 1 dp.
c)
Write down the minimum value of 14cosθ – 5sinθ.
d)
Find, to 2 dp, the smallest value positive value of θ for which
this minimum occurs.
a)
Using the addition formulae:
Rcos(θ + α) = R(cosθcosα – sinθsinα)
Therefore since Rcos(θ + α) = f(x)
R(cosθcosα – sinθsinα) = 14cosθ – 5sinθ
Hence
Rcosα = 14
Dividing the two
tanα = 5/14
Rsinα = 5
α = 19.7º
Using Pythagoras
R = √(142 + 52) = 14.9
Therefore
b)
f(x) = 14.9cos(θ + 19.7)
Hence solve the equation
14cosθ – 5sinθ = 8
Therefore:
14.9cos(θ + 19.7) = 8
cos(θ + 19.7) = 0.5370
θ = 37.9º
The cos graph has been translated 19.7º to the left, the second
solution is at 282.7º (can you see why?)
Example 7
a)
b)
c)
a)
Express 3 sin2x + 7 cos2x in the form R sin(2x + α), where R > 0

and 0 < α < . Give the values of R and α to 3 dp.
2
Express 6 sinxcosx + 14 cos2x in the form a cos2x + b sin2x +c,
where a, b and c are constants to be found.
Hence, using your answer to (a), deduce the maximum value of
6sinxcosx + 14cos2x.
R sin(2x + α) = R(sin2xcosα + sinαcos2x).
Hence
R(sin2xcosα + sinαcos2x) = 3 sin2x + 7 cos2x
Therefore:
3 = Rcosα. Because the sin2x is being multiplied by the
and 7 = Rsinα
α = 1.17c
tanα =
By Pythagoras
R = √(72 + 32) = 7.62
Therfore:
b)
7
3
Dividing the two
3 sin2x + 7 cos2x = 7.62 sin(2x + 1.17)
Express 6 sinxcosx + 14 cos2x in the form a cos2x + b sin2x +c
The question is using part (a) but you have to remember your
identities:
1
6 sinxcosx = 3 sin2x
cos2x =  cos2x + 1 
2
Therefore:
c)
14cos2x = 7cos2x + 7
6 sinxcosx + 14 cos2x = 3 cos2x + 7 sin2x + 7
Hence, using your answer to (a), deduce the maximum value of
6sinxcosx + 14cos2x.
From (a)
3 sin2x + 7 cos2x = 7.62 sin(2x + 1.17)
Therefore:
6 sinxcosx + 14 cos2x = 7.62 sin(2x + 1.17) + 7
Using the right hand side this is a sine curve of amplitude 7.62,
it has also been translated 7 units up. Therefore its maximum
value will be 14.62.
Questions of the type Rcos(x  ) or Rsin(x  ) are definitely going to
be on a C3 paper. Other trig questions require a little bit of proof and
the use of identities.
Example 8
Solve, for 0 < θ < 2π,
sin 2θ + cos 2θ + 1 = √6 cos θ ,
giving your answers in terms of π.
Since there is a single power of cos θ I will aim to write as
much of the equation in cos θ
cos2x  2cos2 x  1
2sin θcos θ + 2cos2 θ = √6 cos θ
Factorising gives:
Cos θ(2sin θ + 2cos θ - √6) =0
Therefore:
cos θ = 0
or
θ=
 3
,
2 2
2sin θ + 2cos θ - √6 = 0
sin θ + cos θ =
Use of Rsin(θ + α)
6
2
R = √2
√2sin(θ +
6

)=
2
4
α=

4
θ=
 5
,
12 12
This is a little challenging but I’m sure that parts of it are
accessible.