Introduction
Multiple angles
Powers of sine / cosine
Summary
DE MOIVRE’S THEOREM:
TRIG IDENTITIES
ALGEBRA 8
INU0114/514 (MATHS 1)
Dr Adrian Jannetta MIMA CMath FRAS
De Moivre’s Theorem: trig identities
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Adrian Jannetta
Introduction
Multiple angles
Powers of sine / cosine
Summary
Objectives
This presentation will cover the following:
• Recap of Binomial expansions and De Moivre’s theorem
• Using De Moivre’s theorem to produce trig identities
• Express multiple angle functions (e.g. sin4θ ) in terms of
single angle functions sin θ and cos θ .
• Express powers of functions sin4 θ in terms of multiple angle
functions (e.g. cos 4θ ).
In this context De Moivre’s theorem is a mathematical tool which
shows us relationships between trig functions. It provides an
alternative (often easier) way of simplifying trig functions that we
might need to apply calculus to (e.g. integration).
De Moivre’s Theorem: trig identities
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Adrian Jannetta
Introduction
Multiple angles
Powers of sine / cosine
Summary
Recap: De Moivre’s Theorem
De Moivre’s theorem is a relationship between complex numbers
and trigonometry.
(cos θ + i sin θ )n = cos nθ + i sin nθ
Previously we used it to evaluate powers and nth roots of a number.
By expanding the LHS and comparing real and imaginary
coefficients it is possible to derive trig identities for powers of sine
and cosine in terms of compound angles.
We will study this application in next.
De Moivre’s Theorem: trig identities
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Adrian Jannetta
Introduction
Powers of sine / cosine
Multiple angles
Summary
Recap: Binomial expansions
Expanding brackets
‹

1 4
Use the binomial theorem to expand x +
x
Recall that the binomial coefficients n = 4 are 1, 4, 6, 4 and 1.

1
x+
x
‹4
‹

1 4
∴ x+
x
 ‹0
 ‹1
 ‹2
1
1
1
3
2
= 1(x )
+ 4(x )
+ 6(x )
x
x
x
 ‹3
 ‹4
1
1
+4(x1 )
+ 1(x0 )
x
x
4
1
= x4 + 4x2 + 6 +
+
x2 x4
4
We will need the binomial series for our work with De Moivre’s
theorem and trig identities.
De Moivre’s Theorem: trig identities
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Adrian Jannetta
Introduction
Multiple angles
Powers of sine / cosine
Summary
De Moivre’s theorem and trig identities
Consider De Moivre’s theore for the case n = 2
(cos θ + i sin θ )2 = cos 2θ + i sin 2θ
If we expand the brackets on the LHS:
(cos θ ) + (i sin θ )2 + 2i sin θ cos θ
2
= cos 2θ + i sin 2θ
2
cos θ − sin θ + 2i sin θ cos θ
= cos 2θ + i sin 2θ
If we compare the real parts of both sides:
cos2 θ − sin2 θ = cos 2θ
(1)
Comparing the imaginary parts gives us
2 sin θ cos θ = sin 2θ
(2)
Equations (1) and (2) are trig identities which you might recognise
from past work.
De Moivre’s theorem provides a way of generating trig identities.
De Moivre’s Theorem: trig identities
5 / 13
Adrian Jannetta
Introduction
Powers of sine / cosine
Multiple angles
Summary
Multiple angle to single angle
Express cos 3θ in terms of sinθ and cos θ .
Construct an equation from De Moivre’s theorem containing the angle
3θ
cos 3θ + i sin3θ = (cos θ + i sinθ )3
We are going to be writing a lot of ‘sines’ and ‘cosines’: so put c = cos θ
and s = sin θ .
Expand the RHS using binomial theorem:
cos 3θ + i sin3θ
=
(c + is)3
=
c3 + 3c2 (is) + 3c(is)2 + (is)3
=
c3 + 3c2 si + 3cs2 i2 + s3 i3
Simplify the powers of i wherever possible:
cos 3θ + i sin3θ
De Moivre’s Theorem: trig identities
=
c3 + 3c2 si − 3cs2 − s3 i
=
c3 − 3cs2 + (3c2 s − s3 )i
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Adrian Jannetta
Introduction
Multiple angles
Powers of sine / cosine
Summary
We grouped the real and imaginary parts on the RHS.
Equating the real part on each side then we get the relationship we
need:
cos 3θ ≡ cos3 θ − 3 cos θ sin2 θ
We didn’t need it, but the method also gives us another identity for
free!
Compare the imaginary parts on both sides of the equation to get:
sin 3θ ≡ 3 cos2 θ sin θ − sin3 θ
De Moivre’s Theorem: trig identities
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Adrian Jannetta
Introduction
Powers of sine / cosine
Multiple angles
Summary
Useful formulae
A complex number in polar form is written z = cos θ + i sin θ .
Raise to the power n and apply De Moivre’s theorem:
zn
=
(cos θ + i sin θ )n
n
=
cos nθ + i sin nθ
∴z
(3)
Substituting a power of −n:
z−n
=
(cos θ + i sin θ )−n =
1
1
=
(cos θ + i sin θ )n cos nθ + i sin nθ
Multiply top and bottom by the complex conjugate:
z−n =
Therefore
cosnθ − i sin nθ
cos nθ − i sin nθ
1
×
=
cos nθ + i sin nθ cosnθ − i sin nθ
cos2 nθ + sin2 nθ
1
z−n =
= cos nθ − i sin nθ
zn
Add (3) and (4) to get:
Subtract (4) from (3):
De Moivre’s Theorem: trig identities
(4)
zn +
1
= 2 cos nθ
zn
(5)
zn −
1
= 2i sin nθ
zn
(6)
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Adrian Jannetta
Introduction
Multiple angles
Powers of sine / cosine
Summary
Deriving trig identities
Express sin4 θ in terms of multiple angles of sin θ and cos θ .
Since we have a function of sin θ we’ll use equation 6. Setting n = 1
we can write:
1
2i sin θ = z −
z
4
We have to make the sin θ term so:

‹
1 4
4
(2i sin θ ) = z −
z
Expand and simplify the LHS. Expand the RHS with the binomial
theorem:
 ‹
 ‹3  ‹4
 ‹2
1
1
1
1
(2i)4 sin4 θ = z4 + 4z3 − + 6z2 −
+ 4z −
+ −
z
z
z
z
1
4
16 sin4 θ = z4 − 4z2 + 6 − +
z2 z4
De Moivre’s Theorem: trig identities
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Adrian Jannetta
Introduction
Powers of sine / cosine
Multiple angles
Summary
We group the powers of z like this:
16 sin4 θ
1
4
= z4 + − 4z2 − + 6
4
2
z ‹

z
‹
1
1
=
z4 +
− 4 z2 +
+6
z4
z2
The grouped terms inside the brackets can be expressed using the
cosine function given in equation 5.
Therefore:
16 sin4 θ
sin4 θ
= (2 cos 4θ ) − 4(2 cos 2θ ) + 6
= 2 cos 4θ − 8 cos 2θ + 6
2 cos 4θ − 8 cos 2θ + 6
=
16
Simplify to get
sin4 θ =
De Moivre’s Theorem: trig identities
cos 4θ − 4 cos 2θ + 3
8
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Adrian Jannetta
Introduction
Multiple angles
Powers of sine / cosine
Summary
Deriving trig identities
Express cos3 4θ in terms of multiple angle functions.
Since we have a function of cos θ we’ll use equation 5. Setting
n = 4 we can write:
1
2 cos 4θ = z4 +
z4
We have to make the cos3 4θ term so:
‹

1 3
3
4
(2 cos 4θ ) = z +
z4
Expand and simplify the LHS. Expand the RHS with the binomial
theorem:
 ‹2  ‹3
 ‹
1
1
1
+ 3z4
+
23 cos3 4θ = (z4 )3 + 3(z4 )2
4
4
z
z
z4
3
1
8 cos3 4θ = z12 + 3z4 + +
4
z
z12
De Moivre’s Theorem: trig identities
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Adrian Jannetta
Introduction
Powers of sine / cosine
Multiple angles
Summary
We group the powers of z like this:
8 cos3 4θ = z12 +
‹

1
1
+ 3 z4 +
z12
z4
The grouped terms inside the brackets can be expressed using the
cosine function given in equation 5.
Therefore:
8 cos3 4θ
= (2 cos 12θ ) + 3(2 cos 4θ )
3
= 2 cos 12θ + 6 cos 4θ
3
= cos 12θ + 3 cos 4θ
8 cos 4θ
4 cos 4θ
Simplify to get
cos3 4θ =
De Moivre’s Theorem: trig identities
cos 12θ + 3 cos 4θ
4
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Adrian Jannetta
Introduction
Multiple angles
Powers of sine / cosine
Summary
Summary
To express multiple angle functions in terms of sin θ and cos θ use
de Moivre’s theorem
cos nθ + i sin θ = (cos θ + i sin θ )n
and expand the RHS (using binomial series if necessary) and
compare the real or imaginary parts to obtain an identity.
To express sink θ or cosk θ in terms of multiple angle functions use
the formulae choose either
1
1
or 2i sin nθ = z −
2 cos θ = z +
z
z
and then use binomial expansions (if necessary) to find sink θ or
cosk θ in terms of z. Then use these formulas
1
1
= 2 cos nθ and zn −
= 2i sin nθ
zn +
zn
zn
De Moivre’s
Theorem:
identities angle identities.
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Adrian Jannetta
to find
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