sine is:
(a) 0.6792 (b)−0.1483
y⫽sin x y
T′ S′ S
T R
⫺1.0
⫺0.5 90°
60°
60° 120°
120° 150°
180°
210°
210°
240°
270°
270°
300° 330°
330° Angle x° 30°
1.0 0.5
0 360°
Fig. 20.11
2. Solve the following equations for values ofxbetween 0◦and 360◦:
(a) x=cos−10.8739 (b)x=cos−1(−0.5572) 3. Find the angles between 0◦to 360◦whose tangent is:
(a) 0.9728 (b)−2.3418
20.3 The production of a sine and cosine wave
In Fig. 20.11, letORbe a vector 1 unit long and free to rotate anticlockwise aboutO. In one revolution a circle is produced and is shown with 15◦sectors. Each radius arm has a vertical and a horizontal component. For example, at 30◦, the vertical component isTSand the horizontal component isOS.
From trigonometric ratios, sin 30◦= TS
TO = TS
1 , i.e. TS=sin 30◦ and cos 30◦=OS
TO=OS
1 , i.e. OS=cos 30◦
The vertical component TS may be projected across toTS, which is the corresponding value of 30◦on the graph ofyagainst anglex◦. If all such vertical components asTS are projected on to the graph, then a sine wave is produced as shown in Fig. 20.11.
If all horizontal components such asOS are projected on to a graph ofyagainst anglex◦, then acosine waveis produced.
It is easier to visualize these projections by redrawing the circle with the radius armORinitially in a vertical position as shown in Fig. 20.12.
From Figs. 20.11 and 20.12 it is seen that a cosine curve is of the same form as the sine curve but is displaced by 90◦(or π/2 radians).
Trigonometric waveforms 155
y⫽cos x
O′ S′ y
T S
R
⫺0.5
⫺1.0 1.0 0.5
90° 60°
45°
15° 0°
0° 60°
120°
120° 180°
150°
180° 210° 225°
255° 285° 315°
240° 330°
360° 300°
Angle x° 30°
0
Fig. 20.12
20.4 Sine and cosine curves
Graphs of sine and cosine waveforms
(i) A graph of y=sinA is shown by the broken line in Fig. 20.13 and is obtained by drawing up a table of values as in Section 20.1. A similar table may be produced for y=sin 2A.
A◦ 0 30 45 60 90 120
2A 0 60 90 120 180 240
sin 2A 0 0.866 1.0 0.866 0 −0.866
A◦ 135 150 180 210 225 240
2A 270 300 360 420 450 480
sin 2A −1.0 −0.866 0 0.866 1.0 0.866
A◦ 270 300 315 330 360
2A 540 600 630 660 720
sin 2A 0 −0.866 −1.0 −0.866 0
A graph ofy=sin 2Ais shown in Fig. 20.13.
y⫽sin A
y⫽sin 2A
360° A° 1.0
⫺1.0
270° 180°
90° 0
y
Fig. 20.13
(ii) A graph ofy=sin12A is shown in Fig. 20.14 using the following table of values.
A◦ 0 30 60 90 120 150 180
1
2A 0 15 30 45 60 75 90
sin12A 0 0.259 0.500 0.707 0.866 0.966 1.00
A◦ 210 240 270 300 330 360
1
2A 105 120 135 150 165 180
sin12A 0.966 0.866 0.707 0.500 0.259 0 y⫽sin A
360° A° 1.0
⫺1.0
270° 180°
90° 0
y y⫽sin21A
Fig. 20.14
(iii) A graph of y=cosA is shown by the broken line in Fig. 20.15 and is obtained by drawing up a table of val- ues. A similar table may be produced fory=cos 2Awith the result as shown.
(iv) A graph ofy=cos12Ais shown in Fig. 20.16 which may be produced by drawing up a table of values, similar to above.
Periodic time and period
(i) Each of the graphs shown in Figs. 20.13 to 20.16 will repeat themselves as angle A increases and are thus calledperiodic functions.
156 Basic Engineering Mathematics
y⫽cos A y⫽cos 2A
360° A° 1.0
⫺1.0
270° 180°
90° 0
y
Fig. 20.15
y⫽cos A
360° A° 1.0
⫺1.0
270° 180°
90° 0
y
y⫽cos A
2 1
Fig. 20.16
(ii) y=sinA and y=cosA repeat themselves every 360◦ (or 2πradians); thus 360◦ is called the period of these waveforms. y=sin 2Aand y=cos 2Arepeat themselves every 180◦(orπradians); thus 180◦is the period of these waveforms.
(iii) In general, ify=sinpAory=cospA(wherepis a constant) then the period of the waveform is 360◦/p(or 2π/prad).
Hence ify=sin 3Athen the period is 360/3, i.e. 120◦, and ify=cos 4Athen the period is 360/4, i.e. 90◦
Amplitude
Amplitude is the name given to the maximum or peak value of a sine wave. Each of the graphs shown in Figs. 20.13 to 20.16 has an amplitude of+1 (i.e. they oscillate between+1 and−1).
However, ify=4 sinA, each of the values in the table is multi- plied by 4 and the maximum value, and thus amplitude, is 4.
Similarly, ify=5 cos 2A, the amplitude is 5 and the period is 360◦/2, i.e. 180◦
Problem 4. Sketch y=sin 3A between A=0◦ and A=360◦
Amplitude=1 and period=360◦/3=120◦. A sketch ofy=sin 3Ais shown in Fig. 20.17.
y⫽sin 3A
360° A° 1.0
⫺1.0
270° 180°
90° 0
y
Fig. 20.17
Problem 5. Sketch y=3 sin 2A from A=0 to A=2π radians
Amplitude=3 and period=2π/2=πrads (or 180◦) A sketch ofy=3 sin 2Ais shown in Fig. 20.18.
y⫽3 sin 2A
A°
⫺3 3
360° 270°
180° 90°
0 y
Fig. 20.18
Problem 6. Sketchy=4 cos 2xfromx=0◦tox=360◦
Amplitude=4 and period=360◦/2=180◦. A sketch ofy=4 cos 2xis shown in Fig. 20.19.
Problem 7. Sketchy=2 sin35Aover one cycle.
Amplitude=2; period=360◦
3 5
=360◦×5 3 =600◦. A sketch ofy=2 sin35Ais shown in Fig. 20.20.
Trigonometric waveforms 157
y⫽4 cos 2x
360° x°
⫺4 4
270° 180°
90° 0
y
Fig. 20.19
360° A°
⫺2 2
540° 600° 180°
0 y
y⫽2 sin35A
Fig. 20.20
Lagging and leading angles
(i) A sine or cosine curve may not always start at 0◦. To show this a periodic function is represented byy=sin(A±α) or y=cos(A±α) whereαis a phase displacement compared withy=sinAory=cosA.
(ii) By drawing up a table of values, a graph ofy=sin(A−60◦) may be plotted as shown in Fig. 20.21. If y=sinA is assumed to start at 0◦theny=sin(A−60◦) starts 60◦later (i.e. has a zero value 60◦later). Thusy=sin(A−60◦) is said tolagy=sinAby 60◦
360°
270° A°
⫺1.0 1.0
180° 90°
0 y
y⫽sin (A⫺60°) y⫽sin A
60°
60° Fig. 20.21
(iii) By drawing up a table of values, a graph ofy=cos(A+45◦) may be plotted as shown in Fig. 20.22. If y=cosA is assumed to start at 0◦ then y=cos(A+45◦) starts 45◦ earlier (i.e. has a maximum value 45◦ earlier). Thus y=cos(A+45◦) is said toleady=cosAby 45◦
360°
270° A°
⫺1.0
180° 90°
0 y
y⫽cos (A⫹45°) y⫽cos A
45°
45° Fig. 20.22
(iv) Generally, a graph of y=sin(A−α) lags y=sinA by angleα, and a graph ofy=sin(A+α) leadsy=sinAby angleα
(v) A cosine curve is the same shape as a sine curve but starts 90◦earlier, i.e. leads by 90◦. Hence
cosA=sin(A+90◦)
Problem 8. Sketch y=5 sin(A+30◦) from A=0◦ to A=360◦
Amplitude=5 and period=360◦/1=360◦.
5 sin(A+30◦) leads 5 sinAby 30◦(i.e. starts 30◦earlier).
A sketch ofy=5 sin(A+30◦) is shown in Fig. 20.23.
360°
270° A°
⫺5 5
180° 90°
0 y
y⫽5 sin (A⫹30°) y⫽5 sin A
30°
30°
Fig. 20.23
158 Basic Engineering Mathematics
Problem 9. Sketchy=7 sin(2A−π/3) over one cycle.
Amplitude=7 and period=2π/2=πradians.
In general, y=sin(pt−α) lags y=sinpt by α/p, hence 7 sin(2A−π/3) lags 7 sin 2Aby (π/3)/2, i.e.π/6 rad or 30◦. A sketch ofy=7 sin(2A−π/3) is shown in Fig. 20.24.
360°
270° A°
⫺7 7
180° 90°
0 y
y⫽7sin (2A⫺π/3) π/6
π/6
π/2 π 3π/2 2π y⫽7sin 2A
Fig. 20.24
Problem 10. Sketch y=2 cos(ωt−3π/10) over one cycle.
Amplitude=2 and period=2π/ωrad.
2 cos(ωt−3π/10) lags 2 cosωtby 3π/10ωseconds.
A sketch ofy=2 cos(ωt−3π/10) is shown in Fig. 20.25.
t
⫺2 2
0 y
π/v
π/2v 3π/2v 2π/v 3π/10v rads
y⫽2cos vt y⫽2cos(vt⫺3π/10)
Fig. 20.25
Now try the following exercise