De Moivre's formula#Formulas for cosine and sine individually

{{Use American English|date = March 2019}}

{{Short description|1 = Theorem: (cos x + i sin x)^n = cos nx + i sin nx}}{{Not to be confused|De Moivre–Laplace theorem}}

In mathematics, de Moivre's formula (also known as de Moivre's theorem and de Moivre's identity) states that for any real number {{mvar|x}} and integer {{mvar|n}} it is the case that

\big(\cos x + i \sin x\big)^n = \cos nx + i \sin nx,

where {{mvar|i}} is the imaginary unit ({{math|i2 {{=}} −1}}). The formula is named after Abraham de Moivre,{{cite journal |last1=Moivre |first1=Ab. de |title=Aequationum quarundam potestatis tertiae, quintae, septimae, nonae, & superiorum, ad infinitum usque pergendo, in termimis finitis, ad instar regularum pro cubicis quae vocantur Cardani, resolutio analytica |journal=Philosophical Transactions of the Royal Society of London |date=1707 |volume=25 |issue=309 |pages=2368–2371 |doi=10.1098/rstl.1706.0037 |s2cid=186209627 |trans-title=Of certain equations of the third, fifth, seventh, ninth, & higher power, all the way to infinity, by proceeding, in finite terms, in the form of rules for cubics which are called by Cardano, resolution by analysis. |language=la |doi-access= }}

  • English translation by Richard J. Pulskamp (2009)

On p. 2370 de Moivre stated that if a series has the form ny + \tfrac{1 - nn}{2 \times 3}ny^3 + \tfrac{1 - nn}{2 \times 3} \tfrac{9 - nn}{4 \times 5}ny^5 + \tfrac{1 - nn}{2 \times 3} \tfrac{9 - nn}{4 \times 5} \tfrac{25 - nn}{6 \times 7}ny^7 + \cdots = a , where n is any given odd integer (positive or negative) and where y and a can be functions, then upon solving for y, the result is equation (2) on the same page: y = \tfrac{1}{2}\sqrt[n]{a + \sqrt{aa-1}} + \tfrac{1}{2}\sqrt[n]{a - \sqrt{aa-1}}. If y = cos x and a = cos nx , then the result is \cos x = \tfrac{1}{2} (\cos(nx) + i\sin(nx))^{1/n} + \tfrac{1}{2}(\cos(nx) - i\sin(nx))^{1/n}

  • In 1676, Isaac Newton found the relation between two chords that were in the ratio of n to 1; the relation was expressed by the series above. The series appears in a letter — Epistola prior D. Issaci Newton, Mathescos Professoris in Celeberrima Academia Cantabrigiensi; … — of 13 June 1676 from Isaac Newton to Henry Oldenburg, secretary of the Royal Society; a copy of the letter was sent to Gottfried Wilhelm Leibniz. See p. 106 of: {{cite book |editor1-last=Biot |editor1-first=J.-B. |editor2-last=Lefort |editor2-first=F. |title=Commercium epistolicum J. Collins et aliorum de analysi promota, etc: ou … |date=1856 |publisher=Mallet-Bachelier |location=Paris, France |pages=102–112 |url=https://books.google.com/books?id=BHBtAAAAMAAJ&pg=PA106 |language=la }}
  • In 1698, de Moivre derived the same series. See: {{cite journal |last1=de Moivre |first1=A. |title=A method of extracting roots of an infinite equation |journal=Philosophical Transactions of the Royal Society of London |date=1698 |volume=20 |issue=240 |pages=190–193 |doi=10.1098/rstl.1698.0034 |s2cid=186214144 |doi-access=free }}; see p 192.
  • In 1730, de Moivre explicitly considered the case where the functions are cos θ and cos nθ. See: {{cite book |last1=Moivre |first1=A. de |title=Miscellanea Analytica de Seriebus et Quadraturis |date=1730 |publisher=J. Tonson & J. Watts |location=London, England |page=1 |language=la |url=https://books.google.com/books?id=bMl5NAAACAAJ&pg=PP5 }} From p. 1: "Lemma 1. Si sint l & x cosinus arcuum duorum A & B, quorum uterque eodem radio 1 describatur, quorumque prior sit posterioris multiplex in ea ratione quam habet numerus n ad unitatem, tunc erit x = \tfrac{1}{2}\sqrt[n]{l + \sqrt{ll-1}} + \tfrac{1}{2}\tfrac{1}{\sqrt[n]{l + \sqrt{ll-1}}}." (If l and x are cosines of two arcs A and B both of which are described by the same radius 1 and of which the former is a multiple of the latter in that ratio as the number n has to 1, then it will be [true that] x = \tfrac{1}{2}\sqrt[n]{l + \sqrt{ll-1}} + \tfrac{1}{2}\tfrac{1}{\sqrt[n]{l + \sqrt{ll-1}}}.) So if arc A = n × arc B, then l = cos A = cos nB and x = cos B. Hence \cos B = \tfrac{1}{2} (\cos(nB) + \sqrt{-1}\sin(nB))^{1/n} + \tfrac{1}{2}(\cos(nB) + \sqrt{-1}\sin(nB))^{-1/n}

See also:

  • {{cite book |last1=Cantor |first1=Moritz |title=Vorlesungen über Geschichte der Mathematik |series=Bibliotheca mathematica Teuberiana, Bd. 8-9 |trans-title=Lectures on the History of Mathematics |date=1898 |publisher=B.G. Teubner |location=Leipzig, Germany |volume=3 |page=624 |url=https://babel.hathitrust.org/cgi/pt?id=chi.11625415&view=1up&seq=644 |language=de }}
  • {{cite journal |last1=Braunmühl |first1=A. von |title=Zur Geschichte der Entstehung des sogenannten Moivreschen Satzes |journal=Bibliotheca Mathematica |date=1901 |volume=2 |pages=97–102 |url=https://babel.hathitrust.org/cgi/pt?id=hvd.32044102937091&view=1up&seq=112 |series=3rd series |trans-title=On the history of the origin of the so-called Moivre theorem |language=de }}; see p. 98. although he never stated it in his works.{{cite book |first1=Margaret L. |last1=Lial |first2=John |last2=Hornsby |first3=David I. |last3=Schneider |first4=Daniels |last4=Callie J. |title=College Algebra and Trigonometry|edition=4th |year=2008 |location=Boston |publisher=Pearson/Addison Wesley |isbn=9780321497444 |page=792}} The expression {{math|cos x + i sin x}} is sometimes abbreviated to {{math|cis x}}.

The formula is important because it connects complex numbers and trigonometry. By expanding the left hand side and then comparing the real and imaginary parts under the assumption that {{mvar|x}} is real, it is possible to derive useful expressions for {{math|cos nx}} and {{math|sin nx}} in terms of {{math|cos x}} and {{math|sin x}}.

As written, the formula is not valid for non-integer powers {{mvar|n}}. However, there are generalizations of this formula valid for other exponents. These can be used to give explicit expressions for the {{mvar|n}}th roots of unity, that is, complex numbers {{mvar|z}} such that {{math|zn {{=}} 1}}.

Using the standard extensions of the sine and cosine functions to complex numbers, the formula is valid even when {{mvar|x}} is an arbitrary complex number.

Example

For x = \frac{\pi}{6} and n = 2, de Moivre's formula asserts that

\left(\cos\bigg(\frac{\pi}{6}\bigg) + i \sin\bigg(\frac{\pi}{6}\bigg)\right)^2 = \cos\bigg(2 \cdot \frac{\pi}{6}\bigg) + i \sin \bigg(2 \cdot \frac{\pi}{6}\bigg),

or equivalently that

\left(\frac{\sqrt{3}}{2} + \frac{i}{2}\right)^2 = \frac{1}{2} + \frac{i\sqrt{3}}{2}.

In this example, it is easy to check the validity of the equation by multiplying out the left side.

Relation to Euler's formula

De Moivre's formula is a precursor to Euler's formula

e^{ix} = \cos x + i\sin x,

with {{mvar|x}} expressed in radians rather than degrees, which establishes the fundamental relationship between the trigonometric functions and the complex exponential function.

One can derive de Moivre's formula using Euler's formula and the exponential law for integer powers

:\left( e^{ix} \right)^n = e^{inx},

since Euler's formula implies that the left side is equal to \left(\cos x + i\sin x\right)^n while the right side is equal to \cos nx + i\sin nx.

Proof by induction

The truth of de Moivre's theorem can be established by using mathematical induction for natural numbers, and extended to all integers from there. For an integer {{mvar|n}}, call the following statement {{math|S(n)}}:

:(\cos x + i \sin x)^n = \cos nx + i \sin nx.

For {{math|n > 0}}, we proceed by mathematical induction. {{math|S(1)}} is clearly true. For our hypothesis, we assume {{math|S(k)}} is true for some natural {{mvar|k}}. That is, we assume

:\left(\cos x + i \sin x\right)^k = \cos kx + i \sin kx.

Now, considering {{math|S(k + 1)}}:

:\begin{alignat}{2}

\left(\cos x+i\sin x\right)^{k+1} & = \left(\cos x+i\sin x\right)^{k} \left(\cos x+i\sin x\right)\\

& = \left(\cos kx + i\sin kx \right) \left(\cos x+i\sin x\right) &&\qquad \text{via induction hypothesis}\\

& = \cos kx \cos x - \sin kx \sin x + i \left(\cos kx \sin x + \sin kx \cos x\right)\\

& = \cos ((k+1)x) + i\sin ((k+1)x) &&\qquad \text{via trigonometric identities}

\end{alignat}

See angle sum and difference identities.

We deduce that {{math|S(k)}} implies {{math|S(k + 1)}}. By the principle of mathematical induction it follows that the result is true for all natural numbers. Now, {{math|S(0)}} is clearly true since {{math|cos(0x) + i sin(0x) {{=}} 1 + 0i {{=}} 1}}. Finally, for the negative integer cases, we consider an exponent of {{math|−n}} for natural {{mvar|n}}.

:\begin{align}

\left(\cos x + i\sin x\right)^{-n} & = \big( \left(\cos x + i\sin x\right)^n \big)^{-1} \\

& = \left(\cos nx + i\sin nx\right)^{-1} \\

& = \cos nx - i\sin nx \qquad\qquad(*)\\

& = \cos(-nx) + i\sin (-nx).\\

\end{align}

The equation (*) is a result of the identity

:z^{-1} = \frac{\bar z}{|z|^2},

for {{math|z {{=}} cos nx + i sin nx}}. Hence, {{math|S(n)}} holds for all integers {{mvar|n}}.

Formulae for cosine and sine individually

{{See also|List of trigonometric identities}}

For an equality of complex numbers, one necessarily has equality both of the real parts and of the imaginary parts of both members of the equation. If {{mvar|x}}, and therefore also {{math|cos x}} and {{math|sin x}}, are real numbers, then the identity of these parts can be written using binomial coefficients. This formula was given by 16th century French mathematician François Viète:

:\begin{align}

\sin nx &= \sum_{k=0}^n \binom{n}{k} (\cos x)^k\,(\sin x)^{n-k}\,\sin\frac{(n-k)\pi}{2} \\

\cos nx &= \sum_{k=0}^n \binom{n}{k} (\cos x)^k\,(\sin x)^{n-k}\,\cos\frac{(n-k)\pi}{2}.

\end{align}

In each of these two equations, the final trigonometric function equals one or minus one or zero, thus removing half the entries in each of the sums. These equations are in fact valid even for complex values of {{mvar|x}}, because both sides are entire (that is, holomorphic on the whole complex plane) functions of {{mvar|x}}, and two such functions that coincide on the real axis necessarily coincide everywhere. Here are the concrete instances of these equations for {{math|n {{=}} 2}} and {{math|n {{=}} 3}}:

:\begin{alignat}{2}

\cos 2x &= \left(\cos x\right)^2 +\left(\left(\cos x\right)^2-1\right) &{}={}& 2\left(\cos x\right)^2-1 \\

\sin 2x &= 2\left(\sin x\right)\left(\cos x\right) & & \\

\cos 3x &= \left(\cos x\right)^3 +3\cos x\left(\left(\cos x\right)^2-1\right) &{}={}& 4\left(\cos x\right)^3-3\cos x \\

\sin 3x &= 3\left(\cos x\right)^2\left(\sin x\right)-\left(\sin x\right)^3 &{}={}& 3\sin x-4\left(\sin x\right)^3.

\end{alignat}

The right-hand side of the formula for {{math|cos nx}} is in fact the value {{math|Tn(cos x)}} of the Chebyshev polynomial {{math|Tn}} at {{math|cos x}}.

Failure for non-integer powers, and generalization

De Moivre's formula does not hold for non-integer powers. The derivation of de Moivre's formula above involves a complex number raised to the integer power {{mvar|n}}. If a complex number is raised to a non-integer power, the result is multiple-valued (see failure of power and logarithm identities).

=Roots of complex numbers=

A modest extension of the version of de Moivre's formula given in this article can be used to find nth root of a complex number for a non-zero integer {{mvar|n}}. If {{mvar|z}} is a complex number, written in polar form as

z=r\left(\cos x+i\sin x\right),

then the {{mvar|n}}-th roots of {{mvar|z}} are given by

r^\frac1n \left( \cos \frac{x+2\pi k}{n} + i\sin \frac{x+2\pi k}{n} \right)

where {{mvar|k}} varies over the integer values from 0 to {{math|{{abs|n}} − 1}}.

This formula is also sometimes known as de Moivre's formula.{{springer|title=De Moivre formula|id=p/d030300}}

=Complex numbers raised to an arbitrary power=

Generally, if z=r\left(\cos x+i\sin x\right) (in polar form) and {{mvar|w}} are arbitrary complex numbers, then the set of possible values is

z^w = r^w \left(\cos x + i\sin x\right)^w = \lbrace r^w \cos(xw + 2\pi kw) + i r^w \sin(xw + 2\pi kw) | k \in \mathbb{Z}\rbrace\,.

(Note that if {{mvar|w}} is a rational number that equals {{math|p / q}} in lowest terms then this set will have exactly {{mvar|q}} distinct values rather than infinitely many. In particular, if {{mvar|w}} is an integer then the set will have exactly one value, as previously discussed.) In contrast, de Moivre's formula gives

r^w (\cos xw + i\sin xw)\,,

which is just the single value from this set corresponding to {{math|1=k = 0}}.

Analogues in other settings

=Hyperbolic trigonometry=

Since {{math|cosh x + sinh x {{=}} ex}}, an analog to de Moivre's formula also applies to the hyperbolic trigonometry. For all integers {{mvar|n}},

: (\cosh x + \sinh x)^n = \cosh nx + \sinh nx.

If {{mvar|n}} is a rational number (but not necessarily an integer), then {{math|cosh nx + sinh nx}} will be one of the values of {{math|(cosh x + sinh x)n}}.{{cite journal|last=Mukhopadhyay|first=Utpal|title=Some interesting features of hyperbolic functions|journal=Resonance|date=August 2006|volume=11|issue=8|pages=81–85|doi=10.1007/BF02855783|s2cid=119753430}}

= Extension to complex numbers =

For any integer {{mvar|n}}, the formula holds for any complex number z=x+iy

:( \cos z + i \sin z)^n = \cos {nz} + i \sin {nz}.

where

: \begin{align} \cos z = \cos(x + iy) &= \cos x \cosh y - i \sin x \sinh y\, , \\

\sin z = \sin(x + iy) &= \sin x \cosh y + i \cos x \sinh y\, . \end{align}

= Quaternions =

To find the roots of a quaternion there is an analogous form of de Moivre's formula. A quaternion in the form

:q = d + a\mathbf{\hat i} + b\mathbf{\hat j} + c\mathbf{\hat k}

can be represented in the form

:q = k(\cos \theta + \varepsilon \sin \theta) \qquad \mbox{for } 0 \leq \theta < 2 \pi.

In this representation,

:k = \sqrt{d^2 + a^2 + b^2 + c^2},

and the trigonometric functions are defined as

:\cos \theta = \frac{d}{k} \quad \mbox{and} \quad \sin \theta = \pm \frac{\sqrt{a^2 + b^2 + c^2}}{k}.

In the case that {{math|a2 + b2 + c2 ≠ 0}},

:\varepsilon = \pm \frac{a\mathbf{\hat i} + b\mathbf{\hat j} + c\mathbf{\hat k}}{\sqrt{a^2 + b^2 + c^2}},

that is, the unit vector. This leads to the variation of De Moivre's formula:

:q^n = k^n(\cos n \theta + \varepsilon \sin n \theta).{{cite journal|last=Brand|first=Louis|title=The roots of a quaternion|journal=The American Mathematical Monthly|date=October 1942|volume=49|issue=8|pages=519–520|jstor=2302858|doi=10.2307/2302858}}

==Example==

To find the cube roots of

:Q = 1 + \mathbf{\hat i} + \mathbf{\hat j}+ \mathbf{\hat k},

write the quaternion in the form

:Q = 2\left(\cos \frac{\pi}{3} + \varepsilon \sin \frac{\pi}{3}\right) \qquad \mbox{where } \varepsilon = \frac{\mathbf{\hat i} + \mathbf{\hat j}+ \mathbf{\hat k}}{\sqrt 3}.

Then the cube roots are given by:

:\sqrt[3]{Q} = \sqrt[3]{2}(\cos \theta + \varepsilon \sin \theta) \qquad \mbox{for } \theta = \frac{\pi}{9}, \frac{7\pi}{9}, \frac{13\pi}{9}.

=2 × 2 matrices=

With matrices, \begin{pmatrix}\cos\phi & -\sin\phi \\ \sin\phi & \cos\phi \end{pmatrix}^n=\begin{pmatrix}\cos n\phi & -\sin n\phi \\ \sin n\phi & \cos n\phi \end{pmatrix} when {{mvar|n}} is an integer. This is a direct consequence of the isomorphism between the matrices of type \begin{pmatrix}a & -b \\ b & a \end{pmatrix} and the complex plane.

References

  • {{cite book |first1=Milton |last1=Abramowitz |first2=Irene A. |last2=Stegun |title=Handbook of Mathematical Functions |year=1964 |publisher=Dover Publications |location=New York |isbn=0-486-61272-4 |page=[https://archive.org/details/handbookofmathe000abra/page/74 74] }}.

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