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Product to Sum Trigonometric Identities with Euler’s Formula
June 15, 2022

Table of Contents

Introduction

cos(a) cos(b)

Start by substituting Euler’s formula in for both cosine terms,

(4)   \begin{equation*}\cos ( \alpha ) \cos ( \beta ) = \frac{1}{2} \left( e^{j\alpha} + e^{-j\alpha} \right) \cdot \frac{1}{2} \left( e^{j\beta} + e^{-j\beta} \right)\end{equation*}

Multiplying the right hand terms of (4),

(5)   \begin{equation*}\begin{split}\cos ( \alpha ) \cos ( \beta ) & = \frac{1}{4} \left( e^{j\alpha}e^{j\beta} + e^{j\alpha}e^{-j\beta} + e^{-j\alpha}e^{j\beta} + e^{-j\alpha}e^{-j\beta} \right) \\& = \frac{1}{4} \left( e^{j(\alpha + \beta)} + e^{-j(\alpha + \beta)} + e^{j(\alpha - \beta)} + e^{-j(\alpha - \beta)} \right).\end{split}\end{equation*}

Simplifying (5) with Euler’s formula results in

(6)   \begin{equation*}\frac{1}{4} \left( e^{j(\alpha + \beta)} + e^{-j(\alpha + \beta)} \right) = \frac{1}{2} \cos (\alpha + \beta),\end{equation*}

(7)   \begin{equation*}\frac{1}{4} \left( e^{j(\alpha - \beta)} + e^{-j(\alpha - \beta)} \right) = \frac{1}{2} \cos ( \alpha - \beta),\end{equation*}

such that

(8)   \begin{equation*}\cos (\alpha) \cos (\beta) = \frac{1}{2} \left( \cos(\alpha + \beta) + \cos (\alpha - \beta).\end{equation*}

sin(a) sin(b)

Start by substituting Euler’s formula in for both sine terms,

(9)   \begin{equation*}\sin(\alpha) \sin(\beta) = \frac{1}{2j} \left( e^{j\alpha} - e^{-j\alpha} \right) \cdot \frac{1}{2j} \left( e^{j\beta} - e^{-j\beta} \right).\end{equation*}

Multiplying the right hand terms of (9),

(10)   \begin{equation*}\begin{split}\sin(\alpha) \sin(\beta) & = -\frac{1}{4} \left( e^{j\alpha}e^{j\beta} - e^{j\alpha}e^{-j\beta} - e^{-j\alpha}e^{j\beta} + e^{-j\alpha}e^{-j\beta} \right) \\& = -\frac{1}{4} \left( e^{j(\alpha + \beta)} + e^{-j(\alpha + \beta)} - e^{j(\alpha - \beta)} - e^{-j(\alpha - \beta)} \right) \\& = -\frac{1}{4} \left( e^{j(\alpha + \beta)} + e^{-j(\alpha + \beta)} \right) + \frac{1}{4} \left( e^{j(\alpha - \beta)} + e^{-j(\alpha - \beta)} \right) \\\end{split}\end{equation*}

Simplifying (10) with Euler’s formula results in

(11)   \begin{equation*}\frac{1}{4} \left( e^{j(\alpha + \beta)} + e^{-j(\alpha + \beta)} \right) = \frac{1}{2} \cos (\alpha + \beta),\end{equation*}

(12)   \begin{equation*}\frac{1}{4} \left( e^{j(\alpha - \beta)} + e^{-j(\alpha - \beta)} \right) = \frac{1}{2} \cos ( \alpha - \beta),\end{equation*}

such that

(13)   \begin{equation*}\sin(\alpha) \sin(\beta) = \frac{1}{2} \left( \cos(\alpha - \beta) - \cos(\alpha + \beta) \right).\end{equation*}

cos(a) sin(b)

Start by substituting Euler’s formula in for cosine and sine,

(14)   \begin{equation*}\cos (\alpha) \sin(\beta) = \frac{1}{2} \left( e^{j\alpha} + e^{-j\alpha} \right) \cdot \frac{1}{2j} \left( e^{j\beta} - e^{-j\beta} \right).\end{equation*}

Multiplying the right hand terms of (14),

(15)   \begin{equation*}\begin{split}\cos (\alpha) \sin(\beta) & = \frac{1}{4j} \left( e^{j\alpha}e^{j\beta} - e^{j\alpha}e^{-j\beta} + e^{-j\alpha}e^{j\beta} - e^{-j\alpha}e^{-j\beta} \right) \\& = \frac{1}{4j} \left( e^{j(\alpha+\beta)} - e^{-j(\alpha+\beta)} - e^{j(\alpha - \beta)} + e^{-j(\alpha - \beta)} \right) \\\end{split}\end{equation*}

Simplifying (15) with Euler’s formula results in

(16)   \begin{equation*}\frac{1}{4j} \left( e^{j(\alpha+\beta)} - e^{-j(\alpha+\beta)} \right) = \frac{1}{2} \sin( \alpha + \beta),\end{equation*}

(17)   \begin{equation*}\frac{1}{4j} \left( e^{j(\alpha - \beta)} - e^{-j(\alpha - \beta)} \right) = \frac{1}{2} \sin (\alpha - \beta),\end{equation*}

such that

(18)   \begin{equation*}\cos (\alpha) \sin(\beta) = \frac{1}{2} \left( \sin( \alpha + \beta) - \sin (\alpha - \beta) \right).\end{equation*}

Conclusion

The product to sum trigonometric identities were proven using Euler’s formula. 

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