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Unit 1Trigonometry.pptx
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
β’ Use Pythagorasβ theorem on trigonometric functions.
β’ Express trigonometric ratios in surd form.
β’ Plot trigonometric waveforms.
β’ Calculate the phase angle(leading/lagging) of trigonometric waveforms.
β’ Use trigonometric identities to solve equations
β’ Convert trigonometric expression into the form Rsin(wtΒ±πΌ)
3. Introduction
β’ This chapter is going to cover Pythagoras theorem and its relationship to
trigonometric functions of cosine, sine, tangent, cosec, cotan and sec.
β’ 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
β’ Pythagoras Theorem for a right angled triangle
Equation:π2
= π2
+ π2
c b
ΞΈ a
β’ Trigonometric ratios of the acute angle: (i) sinΞΈ =opp/hypo=b/c
(ii) cosΞΈ=adj/hypo=a/c (iii) tanΞΈ=opp/adj=b/a=sinΞΈ/cosΞΈ
(iv)cotΞΈ=
πππ ΞΈ
π ππΞΈ
=
1
π‘ππΞΈ
(v) secΞΈ=
1
πππ π
(vi) cosecΞΈ=
1
π ππΞΈ
The ratios (iv), (v) and (vi) are called reciprocal ratios
6. 1.2 Fractional and Surd forms of trigonometric ratios,
β’ Using the triangles ABC and PQR known
concepts
1.Sin300 =
1
2
, πππ 300 =
3
2
, π‘ππ300 =
1
3
2. Sin600 =
3
2
, πππ 600 =
1
2
, π‘ππ600 =
3
1
3.Sin450 =
1
2
, πππ 450 =
1
2
, π‘ππ450 = 1
β’ From the ratios general relations:
sinΞΈ =cos(900 β π) and cosπ = sin 900 β π
β’ 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. π·ππ‘ππππππ π‘βπ βπππβπ‘ 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. πΉπππ π‘βπ πππ π‘πnce 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
β’ Known concepts
1. The sine Rule:
π
π ππππΆ
=
π
π πππ΅
=
π
π πππ΄
2. The cosine Rule:
π2
= π2
+ π2
β 2πππππ πΆ or π2
= π2
+ π2
β 2πππππ π΄
Or π2 = π2 + π2 β 2ππ πππ π΅
3. Area of a triangle: (i) Area=
1
2
π₯ πππ ππ₯ βπππβπ‘
(ii) Area=
1
2
πππ πππΆ ππ
1
2
πππ πππ΅ ππ
1
2
πππ πππ΄
(iii) Area = π π β π π β π π β π π€βπππ π =
(π+π+π)
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 27.8ππ2
]
(b) x=21mm, y=34mm, z=42mm, Z=29.770, 53.520, 96.720, πππ = 355ππ2
2. A room 8.0m wide has a span roof of which slopes at 330
on one side and
400ππ π‘βπ 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
π. A cyclist
leaves the same point at the same time in a direction E400π 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 π1 πππ π2are shown . Find
Their resultant and the angle it makes with π1
[131.4V 32.550]
10. Practical Problem Situations contβd
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 1510π2
. 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)
β’ Graphs of y =Rsin A, Rsin2A or Rcos A or RSin(AΒ±Ξ±)
β’ The graphs are drawn using the 1 unit arm rotating anticlockwise and
tabulate values from measurements or using values from calculator
13. Graphs contβd
β’ Graphs
β’ Amplitude= maximum value of peak of graph
β’ The graphs repeat their values after a certain value= periodic functions
β’ The angular duration/length taken to repeat values =period/cycle(3600 = 2π
for sin or cos)
β’ In general if y =sinpA or cos pA the period = 3600/p or 2Ο/p(radians)
14. Examples 6
β’ Sketch the graphs and state their amplitude and period
(a) y= 3sinA (b) y = 4 cos2A from A =0 to 2Ο radians (c) y= 4sin2x from x =0 to
3600
β’ Lagging and Leading angles
A periodic function expressed as y=Rsin(AΒ±πΌ) ππ π πππ (π΄ Β± πΌ) where Ξ± is a phase
displacement compared to y =RsinA or R cosA
The phase displacement is called a lagging(-Ξ±) if value is attained late on the graph
The phase angle is called leading(+Ξ±) if value is attained earlier on the graph(left to
right)
The graph y=Rsin(A+Ξ±) leads the graph of RsinA
For graph y1=Rsin(pA+Ξ±) compare to y2 =RsinpA For maximum y1
Sin(pA+Ξ±)=1 gives pA+Ξ±=90 i.e A=(90-Ξ±)/p
For max y2: sinpA=1 give pA= 90 i.e A=90/p phase angle difference=Ξ±/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(ΟtΒ±πΌ)
β’ The waveform plotted with the horizontal time units. The phasor is assumed
to rotate at an angular speed Ο rad/s anti-clockwise and Ξ±, the phase angle is
also in radians.
β’ Οt gives the angular displacement in radians
β’ The phasor completes its period also called
Cycle =3600 = 2π πππ in time T sec called
Periodic time.
β’ The angular velocity Ο=
2π
π
or T=
2π
π
β’ Since 1 cycle is made in T secs therefore number of cycles in 1 sec called
frequency (f) =
1
π
=
π
2π
cycles/s[Hetz(Hz)]
or Ο =2Οf
16. Examples 8
a. An alternating current is given by i= 30sin(100Οt+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(ΟtΒ±πΌ).
c-ass 1. The instantaneous value of voltage in an ac circuit at any time t seconds is
given by v= 340sin(50Οt-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 sinΟt of the alternating quantities given.
(i) i=40sin(50Οt+0.29)mA (ii) y=75sin(40Οt-0.54) cm (iii) v=300sin(200Οt -0.412)V
18. 1.6 Trigonometric Equations
β’ The task is to determine angles with the same real number value in a
given range mostly 00
β€ π₯ β€ 3600
.
β’ The figure shows angles with same cos sin or tan of angles and their sign
20. 1.7 The compound Angle Formulae
β’ Known Concepts
1. Sin(A+B) =sinACosB+ cosAsin B
2. Sin(A-B) =SinACosB - CosAsinB
3. Cos(A+B)= CosACosB βSinASinB
4. Cos(A-B) = CosACosB+SinASinB
5. Tan (A+B)=
π‘πππ΄+π‘πππ΅
1βπ‘πππ΄π‘πππ΅
6. Tan(A-B)=
π‘πππ΄βπ‘πππ΅
1+π‘πππ΄π‘πππ΅
21. Examples 10
1. Prove that cos(y-Ο)+sin(y+
π
2
)=0
2. Show that (a) tan(x+
π
4
) tan π₯ β
π
4
= β1
(b)-ass 1 sin x +
π
3
+ sin π₯ +
2π
3
= 3πππ π₯
(c) sin
3π
2
β β = πππ β
(3) Express the following in the form Rsin(Οt+Ξ±)
(i) 4.6sinΟt-7.3cosΟt (ii) -2.7sinΟt-4.1cosΟt
(4) Express 3sinΞΈ+5cosΞΈ in the form Rsin(ΞΈ+Ξ±) and hence solve the equation
3sinΞΈ+5cosΞΈ =4 for values of ΞΈ from00
π‘π 3600
.
(5)Solve the equation 3.5cosA-5.8sinA =6.5 for 00 β€ π β€ 3600
NB: Value of Ξ± 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 4.3cos3ΞΈ-6.9sin3ΞΈ.
Express this in the form Rsin(3ΞΈΒ±πΌ) [8.13sin(3ΞΈ+2.584)
2. The displacement x metres of amass from a fixed point about which is
oscillating is given by x = 2.4 sinΟt+3.2cosΟt, where t is time in seconds.
Express x in the form Rsin(Οt+Ξ±) [x=4.0sin(Οt+0.927)]
3. Two voltages π£1 = 5cosΟπ‘ πππ π£2 = β8π ππππ‘ are inputs to an
analogue circuit. Determine the expression for the output if this is given by
π£1 + π£2 [V=9.434sin(Οt+2.583)
4. Solve the following equations πππ 00< π΄ < 3600
(i) 3cosA+2sinA=2.8 (ii) 12cosA-4sinA=11
[(i) 72.730 ππ 354.630 ππ 11.150 ππ 311.980
β’ Reference: