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

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

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

or equivalently that

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

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

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

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

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

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

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

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)}}:

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

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

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

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

:$\backslash begin\{alignat\}\{2\}$

\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}}.

:$\backslash begin\{align\}$

\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

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

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

{{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:

:$\backslash begin\{align\}$

\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}}:

:$\backslash begin\{alignat\}\{2\}$

\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|T_n(cos x)}} of the Chebyshev polynomial {{math|T_n}} at {{math|cos x}}.

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).

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}}

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

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

(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

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

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

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

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

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}}

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

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

where

:

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

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

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

can be represented in the form

:

In this representation,

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

and the trigonometric functions are defined as

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

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

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

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

:$q^n\; =\; k^n(\backslash cos\; n\; \backslash theta\; +\; \backslash varepsilon\; \backslash sin\; n\; \backslash theta).${{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}}

To find the cube roots of

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

write the quaternion in the form

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

Then the cube roots are given by:

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

With matrices, when {{mvar|n}} is an integer. This is a direct consequence of the isomorphism between the matrices of type and the complex plane.

• {{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] }}.

{{Reflist}}

• [http://demonstrations.wolfram.com/DeMoivresTheoremForTrigIdentities/ De Moivre's Theorem for Trig Identities] by Michael Croucher, Wolfram Demonstrations Project.

{{Spoken Wikipedia|date=2021-06-05|En-De Moivres Formula-article.ogg}}

Category:Theorems in complex analysis

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