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Trigonometric Identities the Euler Way
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| "Let's derive the identity for the product of cosine and cosine functions. Using (3) and (4), we have\n", | |
| "\n", | |
| "$$\n", | |
| "\\cos\\alpha \\cos\\beta = \\frac{1}{4}\\left(e^{i\\alpha} + e^{-i\\alpha}\\right)\\left(e^{i\\beta} + e^{-i\\beta}\\right) =\n", | |
| "$$\n", | |
| "\n", | |
| "$$\n", | |
| "= \\frac{1}{4}\\left[e^{i(\\alpha + \\beta)} + e^{i(\\alpha - \\beta)} + e^{-i(\\alpha - \\beta)} + e^{-i(\\alpha + \\beta)}\\right] =\n", | |
| "$$\n", | |
| "\n", | |
| "$$\n", | |
| "= \\frac{1}{2}\\left[\\cos(\\alpha + \\beta) + \\cos(\\alpha - \\beta)\\right].\n", | |
| "$$\n", | |
| "\n", | |
| "# Conclusion\n", | |
| "\n", | |
| "The key advantage of using Euler's formula is that it transforms trigonometric problems into algebraic manipulations with exponentials, and I find this beautiful." | |
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Author
JoeJoe1313
commented
Aug 5, 2025
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