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The Euler and De Moivre identities are the fundamental identities for
deriving trigonometric formulas. From the identity ejθ=cosθ+jejθ=cosθ+jsinθsinθ and the conjugate identity e-jθ=(ejθ)*=cosθ-je-jθ=(ejθ)*=cosθ-jsinθsinθ, we have the Euler identities for cosθcosθ and sinθsinθ:
cos
θ
=
e
j
θ
+
e
-
j
θ
2
sin
θ
=
e
j
θ
-
e
-
j
θ
2
j
cos
θ
=
e
j
θ
+
e
-
j
θ
2
sin
θ
=
e
j
θ
-
e
-
j
θ
2
j
(1)These identities are illustrated in Figure 1.
The identity ejθ=cosθ+jsin θ
ejθ=cosθ+jsinθ also produces the De Moivre identity:
(
cos
θ
+
j
s
i
n
θ
)
n
=
(
e
j
θ
)
n
=
e
j
n
θ
=
cos
n
θ
+
j
sin
n
θ
.
(
cos
θ
+
j
s
i
n
θ
)
n
=
(
e
j
θ
)
n
=
e
j
n
θ
=
cos
n
θ
+
j
sin
n
θ
.
(2)When the left-hand side of this equation is expanded with the binomial expansion, we obtain the identity
∑
k
=
0
n
n
k
(
cos
θ
)
n
-
k
(
j
sin
θ
)
k
=
cos
n
θ
+
j
sin
n
θ
.
∑
k
=
0
n
n
k
(
cos
θ
)
n
-
k
(
j
sin
θ
)
k
=cosnθ+jsinnθ.
(3)Binomial Coefficients and Pascal's Triangle. The binomial coefficients (kn)(kn) in Equation 3 are shorthand for the number
n
k
=
n
!
(
n
-
k
)
!
k
!
'
k
=
0
,
1
,
...
,
n
.
n
k
=
n
!
(
n
-
k
)
!
k
!
'
k
=
0
,
1
,
...
,
n
.
(4)This number gives the coefficient of xn-kykxn-kyk in the expansion of (x+y)n(x+y)n. How do we know that there are (kn)(kn) terms of the form xn-kykxn-kyk? One way to answer this question is to use Pascal's triangle, illustrated in Figure 2. Each node on Pascal's triangle shows the number of routes that terminate at that node. This number is always the sum of the number of routes that terminate at the nodes just above the node in question. If we think of a left-hand path as an occurrence of an
x
x and a right-hand path as an occurrence of a
y
y, then we see that Pascal's triangle keeps track of the number of occurrences of xn-kykxn-kyk.
Prove (kn)=(n-kn)(kn)=(n-kn).
Find an identity for (k-1n-1)+(n-1k)(k-1n-1)+(n-1k).
Find “half-angle” formulas for cos2θcos2θ and sin2θsin2θ.
Show that
- cos3θ=cos2θ-3cos3θ=cos2θ-3cosθcosθsin2θsin2θ;
- sin3θ=34cos2θsin3θ=34cos2θsinθ-sin3θsinθ-sin3θ.
Use ej(θ1+θ2)=ejθ1ejθ2=(cosθ1+jsinθ1)(cosθ2+jsinθ2)ej(θ1+θ2)=ejθ1ejθ2=(cosθ1+jsinθ1)(cosθ2+jsinθ2)
to prove
- cos(θ1+θ2)=cosθ1cosθ2sinθ1sinθ2cos(θ1+θ2)=cosθ1cosθ2sinθ1sinθ2 ;
- sin(θ1+θ2)=sinθ1cosθ2+sinθ2cosθ1sin(θ1+θ2)=sinθ1cosθ2+sinθ2cosθ1.
(What does "Endorsed by" mean?)
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