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Trigonometric Identities the Euler Way
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| "**Euler's formula** states that for any real number $\\theta$,\n", | |
| "\n", | |
| "$$\\label{eq:1}\n", | |
| "e^{i \\theta} = \\cos\\theta + i \\sin\\theta, \\tag{1}\n", | |
| "$$\n", | |
| "\n", | |
| "where $i$ is the imaginary unit, defined as $i^2 = -1$. The formula has a geometric interpretation in the complex plane that provides insight into trigonometry and complex numbers, **Figure 1**.\n", | |
| "\n", | |
| "In **Figure 1**, the horizontal axis represents the real part, the vertical axis represents the imaginary part, and any complex number $z = x + i y$ is plotted as the point $(x, y)$. Euler's formula represents a point on the unit circle at angle $\\theta$ from the positive real axis, where:\n", | |
| "\n", | |
| "- **Real part**: $\\cos\\theta$ (horizontal coordinate)\n", | |
| "- **Imaginary part**: $\\sin\\theta$ (vertical coordinate)\n", | |
| "- **Magnitude**: $|e^{i\\theta}| = \\sqrt{\\cos^2\\theta + \\sin^2\\theta} = 1$\n", | |
| "\n", | |
| "As $\\theta$ varies from $0$ to $2\\pi$, the point $e^{i\\theta}$ traces out the entire unit circle. The famous **Euler's identity** emerges naturally:\n", | |
| "\n", | |
| "$$\n", | |
| "e^{i \\pi} = \\cos(\\pi) + i \\sin(\\pi) = -1 + 0 i = -1,\n", | |
| "$$\n", | |
| "\n", | |
| "or\n", | |
| "\n", | |
| "$$\n", | |
| "e^{i \\pi} + 1 = 0.\n", | |
| "$$\n", | |
| "\n", | |
| "Now, let's go back to $\\eqref{eq:1}$. We know that $\\cos(x)$ is an even function, meaning $\\cos(-x) = \\cos(x)$, and $\\sin(x)$ is an odd function, meaning $\\sin(-x) = -\\sin(x)$. Using that, we can derive the following identity:\n", | |
| "\n", | |
| "$$\\label{eq:2}\n", | |
| "e^{-i \\theta} = e^{i (-\\theta)} = \\cos(-\\theta) + i \\sin(-\\theta) = \\cos\\theta - i \\sin\\theta. \\tag{2}\n", | |
| "$$" | |
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Author
JoeJoe1313
commented
Aug 5, 2025
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