1.2.3 Applications of de Moivre's Theorem

Application 1: Expanding Trigonometric Powers

One of the most important uses of de Moivre's Theorem is expanding expressions like $(\cos \theta + i \sin \theta)^n$ , which gives a way to express powers of trigonometric functions.

From de Moivre's Theorem:

(\cos \theta + i \sin \theta)^n = \cos(n \theta) + i \sin(n \theta)

Using this, we can find expansions for powers of $\cos \theta$ and $\sin \theta$ . By equating the real and imaginary parts, we can derive useful trigonometric identities.

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Example: Expanding $(\cos \theta + i \sin \theta)^2$

Applying de Moivre's Theorem with $( n = 2 )$ :

(\cos \theta + i \sin \theta)^2 = \cos(2\theta) + i \sin(2\theta)

Expanding the left-hand side algebraically:

(\cos \theta + i \sin \theta)^2 = \cos^2 \theta - \sin^2 \theta + 2i \cos \theta \sin \theta

By comparing real and imaginary parts, we obtain the well-known double-angle identities:

$\cos(2\theta) = \cos^2 \theta - \sin^2 \theta$
$\sin(2\theta) = 2 \cos \theta \sin \theta$

Application 2: Solving Trigonometric Equations

de Moivre's Theorem can also be used to solve trigonometric equations of the form:

\cos(n \theta) + i \sin(n \theta) = z

where $z$ is a complex number. This is especially useful for solving higher powers of trigonometric functions.

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Example: Solving $(\cos \theta + i \sin \theta)^3 = 1$

We need to find the values of $\theta$ for which:

(\cos \theta + i \sin \theta)^3 = 1

Using de Moivre's Theorem:

\cos(3 \theta) + i \sin(3 \theta) = 1 + 0i

This implies that $3\theta = 2k\pi$ for some integer $k$ , since the argument of $1 + 0i$ is $0$ .

Solving for $\theta$

3 \theta = 2k\pi \quad \Rightarrow \quad \theta = \frac{2k\pi}{3}

Thus, the solutions are:

\theta = 0, \frac{2\pi}{3}, \frac{4\pi}{3}

Application 3: Finding Roots of Complex Numbers

As discussed in previous sections, de Moivre's Theorem is highly useful for finding the roots of complex numbers.

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Example: Finding the $4th$ Roots of $16$

We know that $16 = 16 (\cos 0 + i \sin 0)$ is a complex number in modulus-argument form.

Using de Moivre's Theorem, the $4th$ roots of $16$ are given by:

z^{\frac{1}{4}} = 16^{\frac{1}{4}} \left( \cos \frac{0 + 2k\pi}{4} + i \sin \frac{0 + 2k\pi}{4} \right)

z^{\frac{1}{4}} = 2 \left( \cos \frac{2k\pi}{4} + i \sin \frac{2k\pi}{4} \right)

For $k = 0, 1, 2, 3$ , the four roots are:

$z_0 = 2 (\cos 0 + i \sin 0) = 2$
$z_1 = 2 \left( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right) = 2i$
$z_2 = 2 \left( \cos \pi + i \sin \pi \right) = -2$
$z_3 = 2 \left( \cos \frac{3\pi}{2} + i \sin \frac{3\pi}{2} \right) = -2i$ Thus, the $4th$ roots of $16$ are 2, 2i, -2, -2i

Application 4: Trigonometric Identities and Simplifications

de Moivre's Theorem is frequently used to derive and prove trigonometric identities. By expanding and equating the real and imaginary parts, we can establish many identities that are useful in both trigonometry and complex number theory.

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Example: Deriving the Triple Angle Formula

We can use de Moivre's Theorem to derive the triple angle formula for sine and cosine.

Using $n=3$ , we have:

(\cos \theta + i \sin \theta)^3 = \cos(3\theta) + i \sin(3\theta)

Expanding the left-hand side:

(\cos \theta + i \sin \theta)^3 = \cos^3 \theta - 3 \cos \theta \sin^2 \theta + i (3 \cos^2 \theta \sin \theta - \sin^3 \theta)

By comparing real and imaginary parts, we obtain:

$\cos(3 \theta) = \cos^3 \theta - 3 \cos \theta \sin^2 \theta$
$\sin(3 \theta) = 3 \cos^2 \theta \sin \theta - \sin^3 \theta$ These are the triple-angle formulas for sine and cosine.

Application 5: Summing Trigonometric Sequences Using de Moivre's Theorem

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Example: Summing a Cosine Sequence Consider the sequence:

C = 1 + \cos \theta + \cos 2\theta + \dots + \cos n\theta

We want to find a way to simplify this sum. To do this, we can use Euler's formula $e^{i\theta} = \cos \theta + i \sin \theta$ , and de Moivre's Theorem.

Now, let's also consider the sequence of sine terms:

S = \sin \theta + \sin 2\theta + \dots + \sin n\theta

We can combine these two sequences into a complex number:

C + iS = 1 + (\cos \theta + i \sin \theta) + (\cos 2\theta + i \sin 2\theta) + \dots + (\cos n\theta + i \sin n\theta)

Using Euler's formula, this becomes:

C + iS = 1 + e^{i\theta} + e^{i2\theta} + \dots + e^{in\theta}

This is a geometric series with:

First term $a=1$
Common ratio $r = e^{i\theta}$
Number of terms $n+1$

The sum of a geometric series is given by:

S_n = \frac{a(1 - r^n)}{1 - r}

Substituting our values:

S_{n+1} = \frac{1(1 - e^{i(n+1)\theta})}{1 - e^{i\theta}}

Step 1: Simplifying the Denominator

To simplify the denominator, recall that $e^{i\theta} = \cos \theta + i \sin \theta$ .

Therefore:

1 - e^{i\theta} = 1 - (\cos \theta + i \sin \theta) = (1 - \cos \theta) - i \sin \theta

Now, multiply the numerator and denominator by the conjugate of the denominator to simplify it further:

\frac{1 - e^{i(n+1)\theta}}{(1 - \cos \theta) - i \sin \theta} \times \frac{(1 - \cos \theta) + i \sin \theta}{(1 - \cos \theta) + i \sin \theta}

This gives:

= \frac{[1 - \cos(n+1)\theta + i \sin(n+1)\theta]}{(1 - \cos \theta)^2 + \sin^2 \theta} \times (1 - \cos \theta + i \sin \theta)

Step 2: Simplifying the Numerator

Next, expand and simplify the numerator. After expanding, we focus on the real part of the expression to find the sum for $C$ , because we are interested in summing cosine terms.

After some algebra, we get:

C = \frac{1 - \cos \theta - \cos \left[(n+1)\theta\right] + \cos n\theta}{(1 - \cos \theta)^2 + \sin^2 \theta}

Since $\cos^2 \theta + \sin^2 \theta = 1$ , this simplifies to:

C = \frac{1 - \cos \theta - \cos (n+1)\theta + \cos n\theta}{2 - 2 \cos \theta}

Further simplifying:

C = \frac{1}{2} + \frac{\cos n\theta - \cos (n+1)\theta}{2(1 - \cos \theta)}

This is the simplified sum for the cosine sequence.

infoNote

Key Takeaways:

de Moivre's Theorem has many applications, including expanding the powers of trigonometric expressions, solving trigonometric equations, and finding roots of complex numbers.
It simplifies complex operations and is especially useful in deriving trigonometric identities.
The theorem provides a strong connection between complex numbers and trigonometry, making it a powerful tool in both areas.
By using de Moivre's Theorem and Euler's formula, we can simplify the sum of trigonometric sequences like $C = 1 + \cos \theta + \cos 2\theta + \dots + \cos n\theta$
The same method can be used for sine sequences or other similar problems involving trigonometric functions.

Applications of de Moivre's Theorem Simplified Revision Notes for A-Level Edexcel Further Maths Core Pure

1.2.3 Applications of de Moivre's Theorem

Application 1: Expanding Trigonometric Powers

Application 2: Solving Trigonometric Equations

Application 3: Finding Roots of Complex Numbers

Application 4: Trigonometric Identities and Simplifications

Application 5: Summing Trigonometric Sequences Using de Moivre's Theorem

Key Takeaways:

500K+ Students Use These Powerful Tools to Master Applications of de Moivre's Theorem For their A-Level Exams.

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