TRIGONOMETRY
π radians ≅ 3.14159265 rad = 180 degrees = 180°
full (complete) circle = 2π = 360°
Units:
Special Values:
sin(θ)
0°
0
cos(θ)
1
3 /2
tan(θ)
0
1/ 3
30° (π/6)
½
45° (π/4)
1/ 2
1/ 2
1
60° (π/3)
3 /2
½
3
90° (π/2)
1
0
±∞
Derivatives:
∂[sin(x)]/∂x = cos(x)
∂[cos(x)]/∂x = −sin(x)
∂[tan(x)]/∂x = sec2(x)
Chain rule:
∂[sin(u)]/∂x = cos(u) ∂u/∂x
example: ∂[sin(2x)]/∂x = 2 cos(2x)
Sum Rules:
sin(A ± B) = sin(A) cos(B) ± sin(B) cos(A)
cos(A ± B) = cos(A) cos(B) m sin(A) sin(B)
tan(A ± B) = [tan(A) ± tan(B)]/[1 m tan(A) tan(B)]
sin2(A) + cos2(B) = 1
sin(2A) = 2 sin(A) cos(A)
cos(2A) = 2 cos2(A) – 1 = 1 – 2 sin2(A) = cos2(A) – sin2(A)
Taylor Series:
f(x) = f(0) + f’(0)x + f’’(0)x2/(2!) + …
Examples:
where prime denotes a derivative
sin(x) = x – x3/(3!) + x5/(5!) + …
cos(x) = 1 – x2/(2!) + x4/(4!) + …
tan(x) = x + x3/(3!) + 2x5/15 + …
(1 + x)½ = 1 + x/2 – x2/8 + …;
(1 + x)b = 1 + bx + b(b-1)x2/(2!) + …
exp(x) = 1 + x + x2/(2!) + x3/(3!) + …
These are especially important when x is small (typically x << 1).
In this case, sin(x) ≅ x, cos(x) ≅ 1, etc.
COMPLEX NUMBERS
Definition:
Z = a + jb is a Complex Number if:
1) a and b are real numbers
a ≡ Re(Z) is the real part of Z
b ≡ Im(Z) is the imaginary part of Z
2) j is a solution to j2 + 1 = 0 or j2 = −1
Geometry:
Any complex number can be described by a point in a two-dimensional coordinate
system. Consider the complex number Z = 5.0 + j5.0;
6.5
Im aginary A xis
6.0
Y
5.5
5.0
Z = a + jb = 5.0 + j5.0
b
4.5
4.0
3.5
Real A xis
a
3.5
4.0
4.5
5.0
X
5.5
6.0
6.5
We could also use polar coordinates to locate Z:
10
Im agin ary A xis
8
6
Z = a + jb = 5.0 + j5.0
4
|Z |
2
θ
0
0
R eal A xis
2
4
6
8
10
with Z = R cos(θ) + j R sin(θ) = R [cos(θ) + j sin(θ)], where R = (a2 + b2)½ and
θ = tan−1(b/a).
Derivatives:
Consider the derivative of the function f(θ) = cos(θ) + j sin(θ):
∂f(θ)/∂θ = −sin(θ) + j cos(θ) = j [cos(θ) + j sin(θ)]
Thus, we find the derivative of f(θ) with respect to θ gives jf(θ). Can you think of
another function that satisfies ∂f(θ)/∂θ = jf(θ)?
Consider the exponential function exp(cθ) = ecθ,
∂(ecθ)/∂θ = c ecθ .
So, if we let c = j, then this has the same property upon differentiation as f(q) above.
It can be rigorously proven (by expanding both sides below in a Taylor series) that:
ejθ = cos(θ) + j sin(θ)
(which is Euler’s identity)
Thus, we have three ways of writing a complex number Z:
Z = a + jb
Z = R[cos(θ) + j sin(θ)]
Z = R e jθ
Some Operations with Complex Numbers:
Addition and Subtraction –
Z1 ± Z2 = (a1 + jb1) ± (a2 + jb2) = (a1 ± a2) + j (b1 ± b2)
Multiplication –
Z1·Z2 = (a1 + jb1)·(a2 + jb2) = (a1a2 – b1b2) + j (a1b2 + b1a2)
or
Z1·Z2 = R1 e jθ1 · R2 e jθ2 = R1·R2 e j(θ1 + θ2)
Note how much easier multiplication is using the exponential form.
Now consider Euler’s identity above and replace θ by –θ to give
e –jθ = cos(θ) – j sin(θ)
Adding these two equations together yields
cos(θ) = ½( e jθ + e –jθ )
Subtracting these two equations yields
sin(θ) = ½ j ( e –jθ – e jθ ).
Examples:
For what θ is e jθ = 1? Euler’s identity tells us that for θ = 0, 2π, −2π, 4π, −4π, …;
e jθ = 1. For what values of θ will e jθ equal (a) –1, (b) j, and (c) –j? (Try and figure it
out on your own). The answers are:
(a) π, −π, 3π, −3π, …
(b) π/2, −3π/2, 5π/2, …
(c) 3π/2, −π/2, 7π/2, −5π/2, …
You know how to graph complex numbers such as Z1 = 2 e jπ/3 now (I hope!). But how
about Z2 = −2 e jπ/3? If we write Z1 = a + jb, then Z2 = −a – jb or we could write Z2 as
Z2 = (−1) 2 e jπ/3 = (e π) 2 e jπ/3 = 2 e j(π + π/3)
(This tells us to add π to the old angle π/3 to find the new direction)
Z2 = 2 e j4π/3
For instance:
Im a gin ary A x is 1 0
8
Z 1 = 5 .0 + j5 .0
6
b
4
2
−a
-1 0
-8
-6
-4
-2
0
0
2
-2
-4
−b
Z 2 = − 5.0 − j5.0
-6
-8
-1 0
4
a
6
8
10
R eal A x is
OR
Im 10
8
Z 1 = 5.0 + j5.0
6
4
|Z 1 |
2
-10
-8
-6
-4
0
-2
θ2 = θ1 + π
θ1
0
2
-2
4
a
6
8
10
Re
|Z 2 | -4
-6
Z 2 = − 5.0 − j5.0
-8
-10
Now, how about Z2 = jZ1? Here, jZ1 = j (a + jb) = -b + ja
OR
jZ1 = j2 e jπ/3
jZ1 = (e jπ/2) 2 e jπ/3 = 2 e j(π/2 + π/3)
(This tells us to add π/2 to the old angle of π/3 to find the new direction)
jZ1 = 2 e j5π/6
For instance:
Im
10
8
Z 2 = jZ 1= −5.0 + j5.0
Z 1 = 5.0 + j5.0
6
b1
a2
4
2
a1
b 2 = −b 1
So,
-6
-4
-2
0
0
2
4
6
Re
-10
-8
8
1)
2)
Multiplication by –1 rotates the “vector” by 180º (or π)
Multiplication by j rotates the “vector” by 90º (or π/2)
10
Complex Conjugation:
The complex conjugate of Z = a + jb is defined to be Z* = a – jb
OR Z = R e jθ and Z* = R e –jθ. Thus, to form the complex conjugate of Z, just change
the sign of j ( i.e., j → −j ).
Note that Z·Z* is always real.
When we write Z in exponential form, Z = R e jθ, powers of Z are easy to find:
Zn = Rn e jnθ
For instance, to find the 1/3 power of 1, we recognize that 11/3 = e j(2π)1/3 = e j2π/3. But, 1
is also equal to e 0 and e j4π, so 11/3 also equals 1 and e j4π/3. All together there are three
possible answers, which is another way of saying that 1 has three cube-roots!
(similarly four 4th-roots, five 5th-roots, and so on)
Im
2
1
2 π /3
-2
-1
0
4 π /3
0
1
2
Re
-1
-2