Download Math 105: Trigenometry

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2. Trigonometry and f(x)=0
ECE 111
Math 105: Trigonometry and f(x)=0
Introduction
From Wikipedia,
Trigonometry (from Greek trigonon, "triangle" and metron, "measure"[1]) is a branch of mathematics that studies
relationships involving lengths and angles of triangles. The field emerged in the Hellenistic world during the 3rd century
BC from applications of geometry to astronomical studies. www.Wikipedia.com
Trigonometry is fundamental to electrical and computer engineering.
Power is transmitted as a 60Hz sine wave
Analysis of filters, such as the bass boost on your stereo, relies upon expressing an audio signal in
terms of a sum of sinusoids,
AC motors, such as a quad-copter motor, are driven by sinusoidal signals where the frequency of
the sine wave determines the speed of the motor
Analysis of systems described by differential equations (read: everything) depends upon being able
to use complex numbers - which have their origin in sin() and cos() functions.
Likewise, trigonometry may seem like an archaic topic which deals only with architecture and triangles.
Actually, it's much more than that.
sin(), cos(), tan()
Trigonometry is the study of the unit circle. If you draw a unit circle and take a point an angle  from the
x-axis, then
The x-coordinate of that point is cos 
The y-coordinate of that point is sin 
If you extend the line from the origin to the point on the unit circle to x=1, the length of the line to
the x-axis is tan 
y
1
cos(q)
tan(q)
0.5
sin(q)
q
0
-1.5
-1
-0.5
-20
0.5
1
1.5
x
2
-0.5
-1
sin(), cos(), and tan() are all related to the unit circle
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It you let the angle increase with time as
  t
then you get a sine wave. In Matlab:
>>
>>
>>
>>
>>
>>
t = [0:0.01:6]';
w = 1;
x = cos(w*t);
y = sin(w*t);
plot(t,x,t,y)
xlabel('Time (seconds)');
1 rad/sec sine wave: cos(t) (blue) and sin(t) (red)
Note that
cos() and sin() go between -1 and +1. This isn't surprising since these are just the x and y
coordinates as you go around the unit circle.
The period of cos() and sin() is 2 (6.28 seconds).
The default units for cos() and sin() is radians. If you want to use degrees, the conversion is
360 degrees = 2 radians
cycle
1 second  1Hz  2 rad
sec
Pretty much, anything English isn't natural. You'll find in engineering that the math works out a lot nicer
if you use natural units - such as radians.

If you increase the frequency, you get a sine wave that is quicker. A 1Hz sine wave  2 rad
sec  looks like
the following:
>>
>>
>>
>>
>>
w = 2*pi;
x = cos(w*t);
y = sin(w*t);
plot(t,x,'b',t,y,'r')
xlabel('Time (seconds)');
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1Hz Sine Wave: cos(6.28t) (blue) and sin(6.28t) (red)
What also shouldn't be surprising is that if you plot cos() vs sin() you get a circle
>> x = cos(w*t);
>> y = sin(w*t);
>> plot(x,y)
cos(t) vs. sin(t) produces the unit circle
It also shouldn't surprising that
cos 2 t  sin 2 t  1
That just says that the circle you're using has a radius of one. That's sort of the definition of cos() and
sin().
Polar Coordinates
Given any point, you can experss it in Cartesian coordinates with its x and y values:
P = (x, y)
You can also express this in polar form as
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ECE 111
P = r
y
1
(x, y)
0.5
r
q
0
-1.5
-1
-0.5
-20
0.5
1
1.5
x
2
-0.5
-1
A point P can be expressed in Cartesian coordinates (x,y) or polar coordinates (r )
The conversion from one to the other is
x  r cos 
y  r sin 
or
r  x2  y2
  arctan y, x
There are two arctan() functions in Matlab
atan(y/x) returns the angle from -pi/2 to +pi/2 (-90 degrees to +90 degrees)
atan2(y, x) returns the angle
The problem with atan is that if both x and y are negative, the signs cancel. To get the actual angle, you
need to use atan2()
With polar coordinates, you can plot some pretty neat functions.
4-Leaf Clover:
r  cos 2
Matlab Code:
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>>
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q = [0:0.001:1]' * 2 * pi;
r = cos(2*q);
x = r .* cos(q);
y = r .* sin(q);
plot(x,y);
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4-Leaf Clover:
ECE 111
r  cos 2
A spiral
r
In Matlab:
>>
>>
>>
>>
>>
q = [0:0.001:1]' * 2 * pi * 10;
r = q;
x = r .* cos(q);
y = r .* sin(q);
plot(x,y);
Spiral: r  
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Lissajieu Figures:
x  cosa
y  sin3a  
0  a  2
This is a lot more fun if you let  vary with time (see video for this). The figure with   0 looks like the
following:
Lissajieu Figure: x = cos(a), y = sin(3a)
To make it rotate, use the following code:
>> a = [0:0.001:1]' * 2 * pi;
>> for i=1:1000
phi = i/100;
x = cos(a);
y = sin(3*a + phi);
plot(x,y);
pause(0.01);
end
Complex Numbers
In electrical engineering, we deal with sine waves extensively. The problem with sine waves is you really
need two coordinates: cos() and sin(). To get around this, there's a thing called complex numbers. Any
given point is expressed as its real part (x) and its complex part (y)
P = a + jb
To keep the two separate, a term 'j' is added to the second part (the y-coordinate). 'j' is defined as
j  1
and (a, b) are defined as
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a is the real part of P (think of it as the x-coordinate)
b is the imaginary part of P (think of it as the y-coordinate).
Complex numbers make moving around the (x,y) plane a lot simpler: you can express a point with a
single (complex) variable.
The Matlab functions associated with complex numbers are:
j:
1
P = 2 + j*3;
Input the point 2+j3
real(P)
The real part of P:
imag(P)
The imaginary part of P:
abs(P)
The magnitude of P:
angle(P)
The angle of P in radians
the x-coordinate
the y-coordinate
the radius in polar coordinates
You can also express any point in the complex plane in rectangular and polar form
P  x  jy  r
x  r cos 
y  r sin 
r  x2  y2
  arctany/x
Fun with Complex Numbers: Shoot Game
Using complex numbers can make some computations a lot simpler. For example, write a MATLAB
program to launch a cannon ball. Assume:
The position of the cannon ball at any given time be P = x + jy. The real part of P is the X
coordinate, the imaginary part of P is the y-coordinate.
This initial position of the cannon ball is at the origin:
P(t=0) = 0+j0
The initial velocity at t = 0 is
v(t=0) = r
Gravity pulls down with a force of 9.8 m/s2
The ground is flat: when the imaginary part of P goes negative, you hit the ground.
Numerically, you can solve this easily in Matlab. At any given time, the acceleration is 9.8 m/s2
downwards:
a(t) = 0 - j*9.8
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Velocity is the integration of acceleration. Assuming a sampling rate of dt seconds
v(t + dt) = v(t) + a(t)*dt
Position is the integration of velocity
P(t + dt) = P(t) + v(t)*dt
Iterate over and over and you get the position of the cannon ball vs. time.
Matlab Code:
Angle = 45 * pi/180;
Speed = 30;
Target = 90 + j*0;
P = 0 + j*0;
v = Speed*( cos(Angle) + j*sin(Angle) );
a = 0 - j*9.8;
dt = 0.01;
while(imag(P) >= 0)
v = v + a*dt;
P = P + v*dt;
clf
plot([-1,100],[-1,50],'w.');
hold on;
plot([-1,100],[0,0],'b-');
plot(real(Target), imag(Target), 'bx');
plot(real(P),imag(P),'ro');
pause(0.01);
end
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Resulting Display from Shooting a Cannon Ball with Initial Velocity = 30m/s at 45 degres elevation
Now, assume there is wind drag. Wind is a cubic function opposing the direction of the velocity:
f wind    v 3  v
Let  be 0.0001. In Matlab:
Angle = 45 * pi/180;
Speed = 30;
Target = 90 + j*0;
P = 0 + j*0;
v = Speed*( cos(Angle) + j*sin(Angle) );
dt = 0.01;
while(imag(P) >= 0)
wind = -( abs(v)^2 ) * v * 1e-4;
a = wind - j*9.8;
v = v + a*dt;
P = P + v*dt;
clf
plot([-1,100],[-1,50],'w.');
hold on;
plot([-1,100],[0,0],'b-');
plot(real(Target), imag(Target), 'bx');
plot(real(P),imag(P),'ro');
pause(0.01);
end
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Path of a cannon ball without wind drag (pink) and with wind drag (red)
Solving f(x) = 0
This leads to another common problem you encounter in electrical and computer engineering: solving for
a function equal to zero. Take the shoot game from above: determine the angle and speed which results
in you hitting a target 90 meters out. There are several ways to do this:
Interval Halving: Take one shot that's long and one that's short. Your next guess is the midpoint.
Example:
Shot #1: Speed = 30, Angle = 45 degrees.
Error = -11.5287 ( it lands 11.5287 meters short of the target )
Shot #2: Speed = 40, Angle = 45 degrees.
Error = +29.3160
( it hits 29.3160 meters past the target )
Your next guess would be
Shot #3: Speed = 35, Angle = 45 degrees
Error = +9.2588
( it hits 9.2588 meters past the target).
Since the error is long, replace shot #2 (the long shot) with shot #3 and repeat. Your next guess would be
Shot #4: Speed = 32.5, Angle = 45 degrees
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30
Guess #2 (long)
25
20
15
10
Guess #3
(midpoint)
5
0
-5
-10
Guess #1 (short)
-15
30
31
32
33
34
35
36
37
38
39
40
Interval Halving: Start with two guesses: one high, one low. The next guess is the midpoint.
Keep iterating until the error is less than some threshold (0.1 meter here)
California Method: Take one shot that's long and one that's short. Your next guess is interpolated
between the two as
X 
X 3  X 1   error
 E1
For example, with the above data, guess #3 would be
4030
 11.5287
Speed  30   29.316011.5287

Speed  32.8226
This results in
Shot #3: Speed = 32.8226, Angle = 45 degrees
Error = +0.2876
( it hits 0.2876 meters past the target )
Since this shot was long, replace the second shot with this one and repeat. Stop when you get close
enough.
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ECE 111
30
Guess #2 (long)
25
20
15
10
5
0
Guess #3
(interpolated zero crossing)
-5
-10
Guess #1 (short)
-15
30
31
32
33
34
35
36
37
38
39
40
California Method: Start with two guesses - one high the other low. Your next guess is interpolated between these two points.
Newton's Method: Take one shot. Take another slightly faster. Based upon this, determine where the
zero-crossing would be if it were a linear system.

X 2  X 1   X
e  e 1
Example:
Shot #1: Speed = 30, Angle = 45 degrees.
Error = -11.5287 ( it lands 11.5287 meters short of the target )
Shot #2: Speed = 30.1, Angle = 45 degrees.
Error = -11.1437
( it hits 11.143729.3160 meters short of the target )
The next guess would then be
30.130
 11.5287
Speed  30   11.143711.5287

Shot #3: Speed = 30.9952, Angle = 45 degrees.
Error = +1.0139
( it hits 1.0139 meters past of the target )
Repeat with this being shot #1.
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ECE 111
5
Guess #2
0
-5
-10
Guess #1
-15
29
30
31
32
33
Newton's Method: Start with a guess. Perturb this guess slightly to find the sensitivity.
Your next guess is the zero crossing based upon these two.
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