Engineering Maths 1 Trigonometry

1. Unit 1 Trigonometry
Pythagoras Theorem, Trigonometric ratios (sin, cos, tan, cot, sec, cosec),
Fractional and Surd forms of trigonometric ratios, trigonometric
waveforms, (frequency, amplitude, phase angle (leading and lagging)),
Trigonometric identities and equations, compound angles

2. Learning Outcomes
When this chapter is completed the student will be able to
• Use Pythagoras’ theorem on trigonometric functions.
• Express trigonometric ratios in surd form.
• Plot trigonometric waveforms.
• Calculate the phase angle(leading/lagging) of trigonometric waveforms.
• Use trigonometric identities to solve equations
• Convert trigonometric expression into the form Rsin(wt±𝛼)

3. Introduction
• This chapter is going to cover Pythagoras theorem and its relationship to
trigonometric functions of cosine, sine, tangent, cosec, cotan and sec.
• This will be followed by surd form of trigonometric function, the drawing
of waveforms for trigonometric functions, determining of phase angle
and finally solving equations using trigonometric identities and compound
angle formula.

4. 1.1 Pythagoras Theorem, Trigonometric ratios (sin,
cos, tan, cot, sec, cosec)
Known concepts
• Pythagoras Theorem for a right angled triangle
Equation:𝑐2
= 𝑏2
+ 𝑎2
c b
θ a
• Trigonometric ratios of the acute angle: (i) sinθ =opp/hypo=b/c
(ii) cosθ=adj/hypo=a/c (iii) tanθ=opp/adj=b/a=sinθ/cosθ
(iv)cotθ=
𝑐𝑜𝑠θ
𝑠𝑖𝑛θ
=
1
𝑡𝑎𝑛θ
(v) secθ=
1
𝑐𝑜𝑠𝜃
(vi) cosecθ=
1
𝑠𝑖𝑛θ
The ratios (iv), (v) and (vi) are called reciprocal ratios

5. Examples
• 1
• 2
• 3

6. 1.2 Fractional and Surd forms of trigonometric ratios,
• Using the triangles ABC and PQR known
concepts
1.Sin300 =
1
2
, 𝑐𝑜𝑠300 =
3
2
, 𝑡𝑎𝑛300 =
1
3
2. Sin600 =
3
2
, 𝑐𝑜𝑠600 =
1
2
, 𝑡𝑎𝑛600 =
3
1
3.Sin450 =
1
2
, 𝑐𝑜𝑠450 =
1
2
, 𝑡𝑎𝑛450 = 1
• From the ratios general relations:
sinθ =cos(900 − 𝜃) and cos𝜃 = sin 900 − 𝜃
• Example 4(a)
(b) 3sin30-2cos60
(c) 5tan60-3sin60

7. Example 5
(a) An electricity pylon stands on a horizontal ground. At a point 80m from
the base of the pylon, the angle of elevation of the top of the pylon is
230. Calculate the height of the pylon to the nearest meter (34m)
(b) A surveyor measures the angle of elevation of the top of a
perpendicular building as 190. He moves 120m nearer the building and
finds the angle of elevation as 470. 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑡ℎ𝑒 ℎ𝑒𝑖𝑔ℎ𝑡 of the
building. (60.85m)
(c) -assignment
The angle of depression of a ship viewed at a particular instant from the top
of 75m vertical cliff is 300. 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑑𝑖𝑠𝑡𝑎nce of the ship from the base of
the cliff at this instant. The ship is sailing away from the cliff at constant
speed and 1 minute later its angle of depression from the top of cliff is 200
.
Determine the speed of the ship in km/h

8. 1.3 Trigonometry for non right angled triangles
• Known concepts
1. The sine Rule:
𝑐
𝑠𝑖𝑛𝑒𝐶
=
𝑏
𝑠𝑖𝑛𝐵
=
𝑎
𝑠𝑖𝑛𝐴
2. The cosine Rule:
𝑐2
= 𝑎2
+ 𝑏2
− 2𝑎𝑏𝑐𝑜𝑠𝐶 or 𝑎2
= 𝑏2
+ 𝑐2
− 2𝑏𝑐𝑐𝑜𝑠𝐴
Or 𝑏2 = 𝑐2 + 𝑎2 − 2𝑎𝑐 𝑐𝑜𝑠𝐵
3. Area of a triangle: (i) Area=
1
2
𝑥 𝑏𝑎𝑠𝑒𝑥 ℎ𝑒𝑖𝑔ℎ𝑡
(ii) Area=
1
2
𝑎𝑏𝑠𝑖𝑛𝐶 𝑜𝑟
1
2
𝑎𝑐𝑠𝑖𝑛𝐵 𝑜𝑟
1
2
𝑏𝑐𝑠𝑖𝑛𝐴
(iii) Area = 𝑠 𝑠 − 𝑎 𝑠 − 𝑏 𝑠 − 𝑐 𝑤ℎ𝑒𝑟𝑒 𝑠 =
(𝑎+𝑏+𝑐)
2

9. 1.4 Practical Problem Situations
1. Solve the triangle XYZ and find their area (a) x=10.0cm , y=8.0cm, z=7.0cm.
[ X=83.33,Y=52.62 Z= 44.05, area 27.8𝑐𝑚2
]
(b) x=21mm, y=34mm, z=42mm, Z=29.770, 53.520, 96.720, 𝑎𝑟𝑎 = 355𝑚𝑚2
2. A room 8.0m wide has a span roof of which slopes at 330
on one side and
400𝑜𝑛 𝑡ℎ𝑒 other. Find the length of the roof slopes correct to the nearest
centimeter. [4.56m and 5.38m]
3. A man leaves a point walking at 6.5 km/h in a direction E200
𝑁. A cyclist
leaves the same point at the same time in a direction E400𝑆 traveling at a
constant speed. Find the average speed of the cyclist if the walker and cyclist
are 80km apart after 5 hours.[18.23km/hr]
4. Two voltage phasors 𝑉1 𝑎𝑛𝑑 𝑉2are shown . Find
Their resultant and the angle it makes with 𝑉1
[131.4V 32.550]

10. Practical Problem Situations cont’d
4 Ass 1. The diagram shows an inclined jib crane 10m long.
PQ is 4m long. Determine the inclination of the jib
to the vertical and the tie QR.
5. A building site is in the form of a quadrilateral as shown
And its area is 1510𝑚2
. Determine the perimeter of the site
[163.4m)
6.Ass 1 A vertical aerial stands on the horizontal ground.
A surveyor positioned due east of the aerial measures the
elevation of the top as 480
. He moves due south 30m and measures the
elevation as 440. Determine the height of the aerial.[58.65m]

11. 1.5 Trigonometric waveforms, (frequency,
amplitude, phase angle (leading and lagging)
• Graphs of y =Rsin A, Rsin2A or Rcos A or RSin(A±α)
• The graphs are drawn using the 1 unit arm rotating anticlockwise and
tabulate values from measurements or using values from calculator

12. Cosine
• Y=cosx

13. Graphs cont’d
• Graphs
• Amplitude= maximum value of peak of graph
• The graphs repeat their values after a certain value= periodic functions
• The angular duration/length taken to repeat values =period/cycle(3600 = 2𝜋
for sin or cos)
• In general if y =sinpA or cos pA the period = 3600/p or 2π/p(radians)

14. Examples 6
• Sketch the graphs and state their amplitude and period
(a) y= 3sinA (b) y = 4 cos2A from A =0 to 2π radians (c) y= 4sin2x from x =0 to
3600
• Lagging and Leading angles
A periodic function expressed as y=Rsin(A±𝛼) 𝑜𝑟 𝑅𝑐𝑜𝑠(𝐴 ± 𝛼) where α is a phase
displacement compared to y =RsinA or R cosA
The phase displacement is called a lagging(-α) if value is attained late on the graph
The phase angle is called leading(+α) if value is attained earlier on the graph(left to
right)
The graph y=Rsin(A+α) leads the graph of RsinA
For graph y1=Rsin(pA+α) compare to y2 =RsinpA For maximum y1
Sin(pA+α)=1 gives pA+α=90 i.e A=(90-α)/p
For max y2: sinpA=1 give pA= 90 i.e A=90/p phase angle difference=α/p
Example 7: Sketch the graph (i) y= 5sin(A-300) (ii) y=7sin (2A-
𝜋
3
) from 0 to 360
State amplitude , period and phase angle of the wave forms

15. Sinusoidal form y=Rsin(ωt±𝛼)
• The waveform plotted with the horizontal time units. The phasor is assumed
to rotate at an angular speed ω rad/s anti-clockwise and α, the phase angle is
also in radians.
• ωt gives the angular displacement in radians
• The phasor completes its period also called
Cycle =3600 = 2𝜋 𝑟𝑎𝑑 in time T sec called
Periodic time.
• The angular velocity ω=
2𝜋
𝑇
or T=
2𝜋
𝜔
• Since 1 cycle is made in T secs therefore number of cycles in 1 sec called
frequency (f) =
1
𝑇
=
𝜔
2𝜋
cycles/s[Hetz(Hz)]
or ω =2πf

16. Examples 8
a. An alternating current is given by i= 30sin(100πt+0.27) amperes.. Find the
amplitude, periodic time, frequency and phase angle in degrees and minutes.
b-ass 1. An oscillating mechanism has a maximum displacement of 2.5m and a
frequency of 60Hz. At time t=0 the displacement is 90cm. Express the displacement in
the general form Asin(ωt±𝛼).
c-ass 1. The instantaneous value of voltage in an ac circuit at any time t seconds is
given by v= 340sin(50πt-0.541) volts. Determine (i) amplitude, periodic time,
frequency, and phase angle in degrees.(ii) Value of voltage when t=0 and 10 ms. (iii)
Time when the voltage first reaches 200V (iv)Time when the voltage is a maximum
Sketch one cycle of the waveform
d. For the problems find amplitude, periodic time, frequency and phase angle stating
whether it is leading or lagging sinωt of the alternating quantities given.
(i) i=40sin(50πt+0.29)mA (ii) y=75sin(40πt-0.54) cm (iii) v=300sin(200πt -0.412)V

17. Trig Identities
• Known Concepts:
1. Base identities: (i) tanθ=
𝑠𝑖𝑛𝜃
𝑐𝑜𝑠𝜃
(ii)cotθ=
𝑐𝑜𝑠θ
𝑠𝑖𝑛θ
=
1
𝑡𝑎𝑛θ
(iii) secθ=
1
𝑐𝑜𝑠𝜃
iv) Cosec θ=
1
𝑠𝑖𝑛θ
2. From Pythagoras’ theorem (i) 𝑠𝑖𝑛2𝜃 + 𝑐𝑜𝑠2𝜃 =1
(ii) 1 +𝑡𝑎𝑛2𝜃 = 𝑠𝑒𝑐2𝜃 (iii) 𝑐𝑜𝑡2𝜃 + 1 = 𝑐𝑜𝑠𝑒𝑐2𝜃
Examples 9 : Prove the following trigonometric identities
(i) Sinx cotx =cos x (ii)
1
1−𝑐𝑜𝑠2𝜃
= 𝑐𝑜𝑠𝑒𝑐𝜃 (iii)
cos 𝑥−𝑐𝑜𝑠3𝑥
sin 𝑥
= 𝑠𝑖𝑛𝑥 𝑐𝑜𝑠𝑥
(iv)
𝑠𝑖𝑛2𝑥(𝑠𝑒𝑐𝑥+𝑐𝑜𝑠𝑒𝑐𝑥)
𝑐𝑜𝑠𝑥 𝑡𝑎𝑛𝑥
= 1 + 𝑡𝑎𝑛𝑥

18. 1.6 Trigonometric Equations
• The task is to determine angles with the same real number value in a
given range mostly 00
≤ 𝑥 ≤ 3600
.
• The figure shows angles with same cos sin or tan of angles and their sign

19. Examples 9
1. Solve the equations for all values between 00
𝑎𝑛𝑑 3600
.
(a) (i) 8𝑠𝑖𝑛2𝜃 + 2sinθ = 0 𝑖𝑖 5𝑐𝑜𝑠2𝑡 + 3𝑠𝑖𝑛𝑡 − 3 = 0
(b) (i) 18𝑠𝑒𝑐2𝐴 − 3𝑡𝑎𝑛𝐴 = 21 𝑖𝑖 3c𝑜𝑠𝑒𝑐2𝜃 − 5 − 4𝑐𝑜𝑡𝜃 = 0
(c) (i) 12𝑠𝑖𝑛2𝜃 − 6 = 𝑐𝑜𝑠𝜃 𝑖𝑖 − 𝑎𝑠𝑠 1 4𝑐𝑜𝑡2𝐴 − 6𝑐𝑜𝑠𝑒𝑐𝐴 + 6 = 0

20. 1.7 The compound Angle Formulae
• Known Concepts
1. Sin(A+B) =sinACosB+ cosAsin B
2. Sin(A-B) =SinACosB - CosAsinB
3. Cos(A+B)= CosACosB –SinASinB
4. Cos(A-B) = CosACosB+SinASinB
5. Tan (A+B)=
𝑡𝑎𝑛𝐴+𝑡𝑎𝑛𝐵
1−𝑡𝑎𝑛𝐴𝑡𝑎𝑛𝐵
6. Tan(A-B)=
𝑡𝑎𝑛𝐴−𝑡𝑎𝑛𝐵
1+𝑡𝑎𝑛𝐴𝑡𝑎𝑛𝐵

21. Examples 10
1. Prove that cos(y-π)+sin(y+
𝜋
2
)=0
2. Show that (a) tan(x+
𝜋
4
) tan 𝑥 −
𝜋
4
= −1
(b)-ass 1 sin x +
𝜋
3
+ sin 𝑥 +
2𝜋
3
= 3𝑐𝑜𝑠𝑥
(c) sin
3𝜋
2
− ∅ = 𝑐𝑜𝑠∅
(3) Express the following in the form Rsin(ωt+α)
(i) 4.6sinωt-7.3cosωt (ii) -2.7sinωt-4.1cosωt
(4) Express 3sinθ+5cosθ in the form Rsin(θ+α) and hence solve the equation
3sinθ+5cosθ =4 for values of θ from00
𝑡𝑜 3600
.
(5)Solve the equation 3.5cosA-5.8sinA =6.5 for 00 ≤ 𝜃 ≤ 3600
NB: Value of α is obtained by checking the location of the quadrant in which
the phasor falls

22. 1.8 Further Problems
1.The third harmonic of a wave motion is given by 4.3cos3θ-6.9sin3θ.
Express this in the form Rsin(3θ±𝛼) [8.13sin(3θ+2.584)
2. The displacement x metres of amass from a fixed point about which is
oscillating is given by x = 2.4 sinωt+3.2cosωt, where t is time in seconds.
Express x in the form Rsin(ωt+α) [x=4.0sin(ωt+0.927)]
3. Two voltages 𝑣1 = 5cosω𝑡 𝑎𝑛𝑑 𝑣2 = −8𝑠𝑖𝑛𝜔𝑡 are inputs to an
analogue circuit. Determine the expression for the output if this is given by
𝑣1 + 𝑣2 [V=9.434sin(ωt+2.583)
4. Solve the following equations 𝑓𝑜𝑟 00< 𝐴 < 3600
(i) 3cosA+2sinA=2.8 (ii) 12cosA-4sinA=11
[(i) 72.730 𝑜𝑟 354.630 𝑖𝑖 11.150 𝑜𝑟 311.980
• Reference: