Complex numbers: The complex plane
De Moivre's formula
In the notation of imaginary \(e\)-powers various trigonometric formulas can be derived in a simple manner.
Formula of De Moivre \[(\cos\varphi + \mathrm{i}\sin\varphi)^n=\cos(n\,\varphi) + \mathrm{i}\,\sin(n\,\varphi)\qquad\text{for any integer }n\]
This is a consequence of the second equality #\left(\e^{z}\right)^n=\e^{n\cdot z}# from the calculation rules for imaginary \(e\)-powers and two applications of Euler's formula: \[\begin{aligned}\left(\cos(\varphi)+\ii\sin(\varphi)\right)^n &=\left(\e^{\ii\,\varphi}\right)^n \\ &\phantom{abcdwxyz}\color{blue}{\text{Euler's formula}} \\ &= \e^{n\,\varphi\,\ii} \\ &\phantom{abcdwxyz}\color{blue}{\text{product rule of imaginary }e{\text{-powers}}} \\ &=\cos(n\,\varphi)+\ii\,\sin(n\,\varphi)\\ &\phantom{abcdwxyz} \color{blue}{\text{Euler's formula}} \end{aligned}\]
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We now give some applications.
Double-angle formula Apply de Moivre's formule with \(n=2\): \[\begin{aligned}\cos(2\varphi)+\mathrm{i}\,\sin(2\,\varphi) &= (\cos\varphi + \mathrm{i}\sin\varphi)^2 \\ &= \cos^2(\varphi) -\sin^2(\varphi) + \mathrm{i}\cdot 2\cos(\varphi)\sin(\varphi)\end{aligned}\] Thus: \[\cos(2\varphi)=\cos^2(\varphi) -\sin^2(\varphi)\qquad\text{and}\qquad \sin(2\varphi)=2\cos(\varphi)\sin(\varphi)\]
Triple-angle formula Apply de Moivre's formule with \(n=3\): \[\begin{aligned}\cos(3\varphi)+\mathrm{i}\,\sin(3\,\varphi) &= (\cos\varphi + \mathrm{i}\sin\varphi)^3 \\ &= \bigl(\cos^3(\varphi) -3\cos(\varphi)\sin^2(\varphi)\bigr) \\ &{\phantom=}{} + \mathrm{i}\,\bigl(3\cos^2(\varphi)\sin(\varphi)-\sin^3(\varphi)\bigr)\end{aligned}\] Thus: \[\begin{aligned}\cos(3\varphi) &=\cos^3(\varphi) -3\cos(\varphi)\sin^2(\varphi) \\ \sin(3\varphi) &=3\cos^2(\varphi)\sin(\varphi)-\sin^3(\varphi)\end{aligned}\] Because \(\cos^2(\varphi)+\sin^2(\varphi)=1\) we can rewrite the last equations as \[\begin{aligned}\cos(3\varphi) &=4\cos^3(\varphi) -3\cos(\varphi) \\ \sin(3\varphi) &=-4\sin^3(\varphi)+3\sin(\varphi)\end{aligned}\]
Sum and difference formulas The trigonometric sum and difference formulas can also be deduced fairly easyly. An example: \[\begin{aligned}\cos(\alpha+\beta)+\mathrm{i}\,\sin(\alpha+\beta) &= e^{\,\mathrm{i}\,(\alpha+\beta)} \\ &= e^{\,\mathrm{i}\,\alpha}\cdot e^{\,\mathrm{i}\,\beta}\\ &= (\cos\alpha + \mathrm{i}\sin\alpha)\cdot (\cos\beta + \mathrm{i}\sin\beta) \\ &= \bigl(\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)\bigr)\\ &{\phantom=}{} + \mathrm{i}\,\bigl(\cos(\alpha)\sin(\beta)+ \sin(\alpha)\cos(\beta)\bigr)\end{aligned}\] Thus: \[\begin{aligned} \cos(\alpha+\beta) &=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)\\ \sin(\alpha+\beta) &=\cos(\alpha)\sin(\beta)+ \sin(\alpha)\cos(\beta)\end{aligned}\]
Theorem For any integer \(n\) holds \[e^{\,n\,\pi\,\mathrm{i}}=(-1)^n\]
de Moivre's formula yields: \[\begin{aligned}e^{\,n\,\pi\,\mathrm{i}} &= \cos(n\,\pi)+\mathrm{i}\,\sin(n\,\pi) \\ &= (-1)^n+0\,\mathrm{i} \\ &= (-1)^n\end{aligned}\]
