EG4011 Unit 1 Trigonometry
What is Trigonometry?
Trigonometry provides the link between angles and lengths of shapes, using the basic
trigonometric functions of Sine, Cosine and Tangent.
How does it help me; where is it used?
Trigonometric functions are building-block functions, along with polynomial,
exponential and logarithmic functions, that are used in engineering to mathematically
model systems: they occur in integrals, differential equations, matrices, vector
analysis and complex analysis, which in turn can appear in structural engineering,
hydraulics and are used extensively in surveying, to name a few.
What are the contents of this unit?
I
We define the functions Sine, Cosine and Tangent, and together with the Sine
rule, Cosine Rule and Pythagoras’ theorem, we predict the angular and
length properties of triangular shapes.
II
We develop the Cartesian coordinate system, and define the properties of
Sine, Cosine and Tangent within the system.
III
We deal with the algebra of compound functions in Sine, Cosine and Tangent.
IV
We define the radian system of angular measurement.
V
We illustrate the graphical representation of the functions Sine, Cosine and
Tangent.
VI
We define odd and even functions applied to Sine, Cosine and Tangent .
VII
We define inverse trigonometric functions.
VIII
We define the Wave function, and show how it can be used to solve, or
simplify trigonometric formulae.
IX
We prepare the way for Integration of trigonometric combinations by
showing the Power Series approximation of Sine, Cosine and Tangent, and
by providing formulae that reduce a multiplicative combination of
trigonometric functions to a much simpler one of the addition of
trigonometric functions.
X
For those interested, there are appendices detailing the formulation of the
Sine an Cosine Rules
Page 1.1
BFL Pearce July 2021/EG4011
Suppose miners are trapped at point C in the mine shaft ABC shown. There is need
for a vertical shaft sunk to point C. How would you calculate the depth DC.
D
A
11.85°
B
246m
21.25°
213m
C
We shall use this problem in our tutorial to demonstrate MatLab and Octave.
______________________________________________________________________________
Basic Trigonometric functions.
Using a right-angled triangle. (Fig 1.1)
BC 4
= the cosine of the angle α
AC 5
AB 3
sin  =
=
the sine of the angle α
AC 5
AB 3
tan  =
= the tangent of the angle α
BC 4
A
cos  =

5
3

B
4
Notice: diagrammatically,
the right-angle is signified
by the square symbol.
C
Fig 1.1. A right-angled Triangle
Where, for angle α, BC is the adjacent side, AB is the opposite side, and AC is the
hypotenuse.
Pythagoras’ Theorem: The Square of the Hypotenuse is equal to the sum of the
squares on the other two sides.
Applied to the figure above:
AC 2 = CB 2 + BA2
 5 2 = 4 2 + 32
.
Note:
Note :
AC
1
sec  =
=
the secant (reciprocal cosine)
BC cos 
function of the angle α
AC
1
cos ec =
=
the cosecant (reciprocal sine)
AB sin 
function of the angle α
BC
1
cot  =
=
the cotangent (reciprocal
AB tan 
tangent) function of the angle α
sin  =
Page 1.2
BC
= cos
AC
AB
cos  =
= sin 
AC
BC
tan  =
= cot 
AB
BFL Pearce July 2021/EG4011
The Cosine Rule
Cosine Rule - three permutations:
This formula can be applied to any
triangle. (Fig 1.2) See Appendix A.
Pythagoras’ Theorem being a special
case.
AB 2 = AC 2 + BC 2 − 2 AC.BC.cos ACB
AC 2 = AB 2 + BC 2 − 2 AB.BC.cos ABC
BC 2 = AC 2 + AB 2 − 2 AC. AB.cos CAB
A
D
B
Notice: any triangle can be formed
into a pair of right-angled
triangles by ‘dropping’ a
perpendicular from one vertex
(point) to its opposite side.
C
Fig 1.2. A pair of right-angled Triangles
developed from a typical Triangle
The Sine rule
This can be applied to any triangle. (Fig 1.3) See Appendix B.
A
The Sine Rule
D
sin  sin  sin 
=
=
AB
AC
BC
E
O
B
C
F
Fig 1.3 Triangle highlighting Sine Rule
Trigonomentical ratios throughout the Cartesian Plane
2nd
SINE
+ve
1st
ALL
+ve
3rd
TAN
+ve
4th
COS
+ve
In 1st quadrant : sin  = +ve , cos  = +ve , tan  = +ve
In 2nd quadrant : sin  = +ve , cos  = −ve , tan  = −ve
In 3rd Quadrant : sin  = −ve , cos  = −ve , tan  = +ve
In 4th quadrant : sin  = −ve , cos  = +ve , tan  = −ve
Page 1.3
BFL Pearce July 2021/EG4011
Trigonometrical functions applied to combinations of angles
Compound Angles:
Note:
cos( A + B) = cos A cos B − sin Asin B
cos( A − B) = cos A cos B + sin Asin B
sin( A + B) = sin Acos B + sin B cos A
sin( A − B) = sin A cos B − sin B cos A
tan A + tan B
tan( A + B) =
1 − tan A tan B
tan A − tan B
tan( A − B) =
1 + tan A tan B
cos(− B) = cos B ,
sin( −B) = − sin B
Note: Arguably, the most useful
relationship in trigonometry is
cos2 A + sin 2 A = 1 for any angle A.
Example
5
5 y
x  = , (Fig
4
4 x
2
2 y
1.4.) and y = x  = , where
3
3 x
PQ 5
ST 2
tan  =
= , tan  =
= ,
OT 3
OQ 4
then calculate the angle α using the
tan A − tan B
formula tan( A − B) =
1 + tan A tan B
Given two lines y =
P
y
y=mx
y=nx
S



O
Q
T
x
Fig 1.4 Angle between two lines from the compound Tangent formula
tan( ) = tan( −  ) =
tan  − tan 
1 + tan  tan 

 15 − 10 
 5 
5 2 
−






4 3  = tan −1  12  = tan −1  12 
  = tan −1 
 1 +  5  2  
 1 +  10  
  22  
  4  3  
  12  
  12  
    
  
 
 5 
= tan −1   = 12.804 or 180 + 12.804 = 192.804
 22 
Every inverse Trig function has two possible solutions: here , we can eliminate one
of those possibilities: on inspection of Fig1.4 the angle is less than 90 , and so is
12.804
Page 1.4
BFL Pearce July 2021/EG4011
Graphical representation (Profile) of the functions sine, cosine and tangent





      










Fig 1.5 Profile of cosine x





       









Fig 1.6 Profile of sine x
Note cosine curve is a phase shift
of - π/2(-90º).





      










Fig 1.7 Profile of tangent x
Page 1.5
BFL Pearce July 2021/EG4011
Odd and Even Functions
Define Odd Function : f ( − x ) = − f ( x )
e.g sin( −35 ) = − sin( 35 ) = −0.5736
sin( − x ) = sin(0 − x ) = sin(0 )cos( x ) − cos(0 )sin( x )
= 0 − sin( x ) = − sin( x )
sin( x  2 ) = sin( x )cos( 2 )  cos( x )sin( 2 )
= sin( x )  0 = sin( x )
Properties of cosine
Cosine is an even function
with period 2
i.e cos( − x ) = cos( x )
cos( x  2 ) = cos x
Properties of sine
Sine is an odd function
with period 2
i.e sin( − x ) = − sin( x )
sin( x  2 ) = sin x
Define Even Function : f ( − x ) = f ( x )
 
 
e.g cos  −  = cos   = cos ( 45 ) = 0.7071
 4
4
cos( − x ) = cos(0 − x ) = cos(0 )cos( x ) + sin(0 )sin( x )
= cos( x ) + 0 = cos( x )
cos( x  2 ) = cos( x )cos( 2 ) sin( x )sin( 2 )
= cos( x ) 0 = cos( x )
Properties of tangent
Note:
tan( − x ) =
sin( − x ) − sin( x )
=
= − tan( x )
cos( − x ) cos( x )
 
 
 = − tan   = tan (60 ) = 1.7321
 3
3
e.g. tan  −
tan( x   ) =
Tangent is an odd
function with period 
i.e tan( − x ) = − tan( x )
tan( x   ) = tan x
tan( x )  tan(  ) tan( x )  0
=
= tan( x )
1 tan( x )tan(  )
1 0
Odd and Even functions have importance in Fourier Series expansions
Inverse Trigonometric Functions : sin−1( x ),cos −1( x ),tan−1( x )
The functions sin(x), cos(x) and tan(x) have the possibility of two solutions, based on
these functions being positive in two quadrants, and negative in 2 quadrants.
We have to eliminate one of the possibilities, and recognise that the calculator will
produce angular results: positive will be in 1st quadrant, negative in the 4th quadrant.
Example
If sin( x ) = 0.5  x = sin−1(0.5 ) .
The calculator will give the answer 30 , but equally, sine is positive in the 2nd
quadrant, where 180 − 30 = 150 , and as such 150 is valid.
Similarly, if cos( x ) = −0.25  x = cos −1( −0.25 ) .
The calculator will give the answer 104.478 in the 2nd quadrant; but equally,
cosine is negative in the 3rd quadrant, where 360 − 104.478 = 255.522 , is valid.
Page 1.6
BFL Pearce July 2021/EG4011
Wave functions
Wave functions are trigonometric functions f(t) of the variable ‘time’, of the form
Asin( t +  ) , Asin( t −  ) , Acos( t +  ) , Acos( t −  ) .
A is the amplitude: a multiplier that allows the natural range [-1, +1] of sin or cos
to be expanded to [-A, +A];
 - the angular frequency, is associated with the period of the function.
If f ( t ) = Asin( t +  ) and t =  + 2 , then the function will be identical to
f ( t ) = Asin(  +  ) .
2

is the wavelength, or period of the function;
is the frequency of the
t=

2
function.
In f ( t ) = Asin( t +  ) ,  is the phase shift from the Asin( t ) function.
Example of Wave functions
Given f ( t ) =
3

sin( 4t + ) ,
2
4
shown below are the profiles of f(t) superimposed on the sin(t) function.





      










Fig 1.8 f(t) superimposed on the sin(t) function
Notice with the f(t) profile: the amplitude increase; the profile cuts the t-axis when
4t +
 t=
when n = 0, t = −

16
;

4
= n
( n −  )
4 , for n = 0,1,2,3........
4
n = 1, t =
3
;
16
n = 2, t =
Page 1.7
7
;
16
n = 3, t =
11
16
BFL Pearce July 2021/EG4011
Further example of the use of Wave Function to combine a sine and a cosine
function
Suppose we have sin( t ) − 2cos( t ) .
This combination can be written as any of the four functions
Asin( t   ) , or
Acos( t   ) , where A and  are unknowns.
Notice we have functions of
If we had
t only, and all functions are of t .
sin( 3t ) − 2cos( 3t ) , then we would need Asin( 3t   ) , or Acos( 3t   ) .
It matters little which of the four functions Asin( t   ) , or Acos( t   ) are used.
Take Acos( t −  ) = A(cos( t )cos(  ) + sin( t )sin(  )) ,
then sin( t ) − 2cos( t ) = Acos( t −  ) = A(cos( t )cos(  ) + sin( t )sin(  ))
Comparing coefficients left and right sides of the equation gives two equations:
Asin(  ) = 1
Acos(  ) = −2
(1.1)
(1.2)
sine is +ve, cosine is –ve implies 2nd quadrant
Dividing (1.1) by (1.2) to eliminate A :
tan(  ) = −
1
 1
  = tan −1  −  = − 26.565 4th quad ,153.435 2nd quad
2
 2
So  = 153.435 or2.678 rads
The two equations give A2 (sin 2 ( t ) + cos 2 ( t )) = ( 1 )2 + ( −2 )2 = 5  A = 5
So sin( t ) − 2 cos( t ) = 5 cos( t − 2.678 )
Further example of the use of Wave Function to solve a sine and a cosine
combination
Suppose we have sin( 3t ) − 2cos( 3t ) = 1 .
Take Asin( 3t −  ) = A(sin( 3t )cos(  ) − sin(  )cos( 3t )) , then
sin( 3t ) − 2cos( 3t ) = Asin( 3t −  ) = A(sin( 3t )cos(  ) − sin(  )cos( 3t ))
Comparing coefficients left and right sides of the equation gives two equations:
Acos(  ) = 1
Asin(  ) = 2
(1.3)
(1.4)
sine is +ve, cosine is +ve implies 1st quadrant
Page 1.8
BFL Pearce July 2021/EG4011
Dividing (1.3) by (1.4) to eliminate A :
tan(  ) = −
1
  = tan −1 ( 2 ) = 63.435 1st quad ,243.435 3rd quad
2
So  = 63.435 or1.107 rads
The two equations give A2 (sin 2 ( t ) + cos 2 ( t )) = ( 1 )2 + ( 2 )2 = 5  A = 5
So sin( 3t ) − 2 cos( 3t ) = 5 sin( 3t − 63.435 ) = 1
Therefore sin( 3t − 63.435 ) =
1
 1 
 ( 3t − 63.435 ) = sin −1 

5
 5
 1 
 has two solutions, which are equally valid, namely
 5
26.565,1st quad , and 153.435 , 2nd quad
But sin −1 
Therefore ( 3t − 63.435 ) = 26.565 + 360k  t =
And
26.565 + 63.435 + 360k 
= 30 + 120k 
3
153.435 + 63.435 + 360k 
= 72.29 + 120k 
3
where k = 0,1,2
( 3t − 63.435 ) = 26.565 + 360k  t =
k = 0  t = 30, and 72.290 ,
k = 1  t = 150, and 192.290
k = 2  t = 270, and 312.290
Power Series approximations for sine, cosine and tangent
These are infinite series representations of the main trigonometric functions.
sin  =  −
cos  = 1 −
3 5
3!
+
5!
2 4
2!
+
n −1
− .......( −1 )
− .......( −1 )n −1
4!
 3 2 5
tan  =  + +
+ ....
3 15
 2n −1
( 2n − 1 )!
 2n −2
( 2n − 2 )!
+ ....
+ ....
 must be in radians: it cannot
be in degrees, as degrees
raised to a power has no
meaning.
With very small values of  ,
these approximate to sin = ,
cos  = 1 , and tan =
These series expansions are developed from the Maclaurins series , as seen in Unit
8 Differentiation
Page 1.9
BFL Pearce July 2021/EG4011
Converting products of sines and cosines into the addition or subtraction of sines,
or cosines : a very useful tool for simplifying Integration
Example
Adding and subtracting pairs of sines or
cosines in terms of half angles
To show
A B
A B
− ) cos( + ))
2 2
2 2
Take half angles of A and B
1
1
sin A = sin[ ( A + B ) + ( A − B )]
2
2
A+ B
A− B
= sin(
)cos(
)
2
2
(1.5)
A− B
A+ B
+ sin(
)cos(
)
2
2
sin A − sin B = 2(sin(
Note we can write
A=
1
1
( A + B) + ( A − B)
2
2
and, similarly
1
1
B = ( A + B) − ( A − B)
2
2
A B
A B
+ ) cos( − ))
2 2
2 2
A B
A B
sin A − sin B = 2(sin( − ) cos( + ))
2 2
2 2
A B
A B
cos A + cos B = 2(cos( + ) cos( − ))
2 2
2 2
A B
A B
cos A − cos B = −2(sin( + ) sin( − ))
2 2
2 2
sin A + sin B = 2(sin(
1
1
sin B = sin[ ( A + B ) − ( A − B )]
2
2
A+ B
A− B
A− B
A+ B
= sin(
)cos(
) − sin(
)cos(
)
2
2
2
2
(1.6)
subtracting (1.6) from (1.5) gives us
sin A − sin B = 2(sin(
The other relationships follow the
pattern of re-writing A and B as
combinations of half angles.
A B
A B
− ) cos( + ))
2 2
2 2
Example
Given sin(5x)cos(3x) write this in terms of sin A + sin B
Using the relationship sin A + sin B = 2(sin(
A B
A B
+ ) cos( − ))
2 2
2 2
As we have sin(x) cos(y) pattern:
1
5x = ( A + B )  A + B = 10x
(1.7)
2
1
3x = ( A − B )  A − B = 6 x
and
(1.8)
2
Adding (1.7) and (1.8) gives 2A = 16 x  A = 4x and substituting this value back into
either (1.7) or (1.8) gives us
B = −2x
then
So
sin( 5x )cos( 3x ) =
1
1
[sin 4x + sin( −2x )] = (sin 4x − sin 2x )
2
2
Page 1.10
BFL Pearce July 2021/EG4011
Conversion from Cartesian plane coordinates to polar coordinates
Note: Cartesian coordinates
are two lengths typically x
and y.
Polar coordinates are a
length and an angle, typically
A point P has coordinates in the rectangular xy
(Cartesian) system represented by the lengths
x = OQ and y = PQ = ON
PQ PQ
=
 PQ = r.sin  and
r and 
OP
r
OQ OQ
cos =
=
 OQ = r. cos
OP
r
Now by Pythagoras for triangle OQP we have the relationship that
Notice sin  =
( PQ ) 2 + (OQ) 2 = (OP) 2 = r 2  r = ( PQ ) 2 + (OQ) 2
Note we can use the relationship cos2  + sin 2  = 1 by squaring both equations we
have
r 2 (cos2  + sin 2  ) = ( PQ) 2 + (OQ) 2 = r 2
Given r and lengths PQ and OQ we can calculate sin  =
cos =
PQ
PQ
  = sin −1 (
) , or
r
r
OQ
OQ
  = cos−1 (
) . r and α are the polar coordinates of point P.
r
r
Example
Given PQ=0.866 and OQ=0.5 calculate polar coordinates r and α.
(a)
y − coord : r sin  = PQ = 0.866
and
(b)
x − coord : r cos  = OQ = 0.5
Squaring both these eqns. and adding them gives us
r 2 = 0.8662 + 0.52 = 0.75 + 0.25 = 1  r = +1 (we discount the –ve root)
Using eqn (a) sin  =
0.866
  = sin −1 0.866 = 60
1
However sin is +ve in both 1st and 2nd quadrants so equally likely is that  =120
0.5
= 0.5 = +ve . Now cos is +ve in both 1st and 3rd
1
quadrants ie  = 60 , or equally likely,  = 240
Examine eqn (b) this  cos =
The common answer is that  = 60 .
Page 1.11
BFL Pearce July 2021/EG4011
Appendix 1A
Proof of Cosine Rule
A
D
B
C
Fig 1.9 using a perpendicular to divide a triangle into two right-angled triangles
As a proof, note that dividing triangle ABC into two right-angled triangles (Fig 1.9.)
allows us to use Pythagoras’ theorem for triangles ABD and BDC
Now
AB 2 = AD 2 + BD2 and BC 2 = BD2 + DC 2
Taking one equation from the other gives a combined equation
AB 2 − BC 2 = AD 2 + BD2 − BD2 − DC 2 = AD 2 − DC 2
re-arranging by adding BC 2 to both sides gives
AB 2 = AD 2 − DC 2 + BC 2
But
AD 2 = ( AC − DC) 2 = AC 2 − 2. AC.DC + DC 2
Substituting for AD 2 gives
AB 2 = AC 2 − 2 AC.DC + DC 2 − DC 2 + BC 2 = AC 2 + BC 2 − 2 AC.DC
Now
DC = BC.cos
And so
AB 2 = AC 2 + BC 2 − 2 AC.BC.cos
which is the 1 result: the other two variants can be obtained by moving around the
triangle changing the angle α.
Page 1.12
BFL Pearce July 2021/EG4011
Appendix 1B
Proof of Sine Rule
A
D
E
O
B
C
F
Fig 1.10 Triangle highlighting Sine Rule
By definition sin  =
AF
AF
, sin  =
, (Fig 1.10.) and we can re-arrange these, using
AC
AB
AF as the common unknown
AF = AC.sin  = AB sin 
dropping the unknown AF (we never needed to calculate it), we have
sin  sin 
=
AB
AC
BD
BD
Further, by definition sin  =
, sin  =
, and we can re-arrange these, using BD
BC
AB
as the common unknown
BD = BC.sin  = AB sin 
dropping the unknown BD (we never needed to calculate it), we have
sin  sin 
=
AB
BC
Linking the two sets of relationships yields
sin  sin  sin 
=
=
, as required
AB
BC
AC
Page 1.13
BFL Pearce July 2021/EG4011
Recommended Reading
Breach, M (2011) Fundamental Maths for Engineering and Science, Basingstoke:
Palgrave MacMillan ISBN 978-0-230-25208-0
Chapters 15,16 and 22
Croft A., and Davison R. (2008) Mathematics for Engineers.3rd edn. Harlow: Pearson
ISBN 978-0-13-205156-9
Chapters 9, 10
Singh, K. (2011) Engineering Mathematics Through Applications. 2nd edn.
Basingstoke: Palgrave ISBN 978-0-230-27479-2
Chapter 4 – Trigonometry and Waveforms
Page 1.14
BFL Pearce July 2021/EG4011