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Major refactor: Split notebook into 2 parts, and other visual improvements
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‎trigonometry/r-formula-cosine-first.ipynb‎

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{
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"cells": [
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{
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"cell_type": "markdown",
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"id": "72c44810-5086-4a40-8d8a-f825b064f14d",
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"metadata": {},
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"source": [
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"# Proof of the R-Formula\n",
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"\n",
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"The main goal is to find a unifying expression for\n",
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"\n",
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"$$\n",
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"a\\sin(x) \\pm b\\cos(x)\n",
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"$$\n",
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"\n",
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"And\n",
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"\n",
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"$$\n",
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"a\\cos(x) \\pm b\\sin(x)\n",
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"$$\n",
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"\n",
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"where\n",
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"$$\n",
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"a > 0 \\quad \\text{and} \\quad b > 0\n",
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"$$"
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]
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},
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{
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"cell_type": "markdown",
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"id": "83da1681-d587-498d-9b07-38bac736c4b4",
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"metadata": {},
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"source": [
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"# **Cosine First**: $\\mathbf{a\\cos(x) \\pm b\\sin(x)}$\n",
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"\n",
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"\n",
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"## Finding Patterns in Identities\n",
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"\n",
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"Looking through the different trigonometric identities, we can find by that in the double-angle formula\n",
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"\n",
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"$$\n",
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"\\cos(A \\mp B) = \n",
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"\\cos(A)\\cos(B) \\pm \\sin(A)\\sin(B)\n",
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"$$ \n",
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"\n",
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"Its expanded form of \n",
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"\n",
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"$$\n",
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"\\textcolor{lightgray}{\\cos(A \\mp B)} \n",
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"\\textcolor{lightgray}{=} \n",
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"\\textcolor{green}{\\cos(A)}\n",
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"\\cos(B) \n",
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"\\pm \n",
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"\\textcolor{blue}{\\sin(A)}\n",
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"\\sin(B)\n",
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"$$ \n",
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"\n",
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"looks quite similar to:\n",
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"\n",
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"$$\n",
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"\\boxed{\n",
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" \\textcolor{green}{a}\\cos(x) \\pm \\textcolor{blue}{b}\\sin(x)\n",
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"}\n",
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"$$\n",
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"\n",
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"And has the unifying double-angle form of \n",
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"\n",
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"$$\n",
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"\\cos(A \\mp B) = \n",
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"\\textcolor{green}{\\cos(A)} \n",
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"\\textcolor{lightgray}{\\cos(B)} \n",
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"\\textcolor{lightgray}{\\pm}\n",
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"\\textcolor{blue}{\\sin(A)}\n",
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"\\textcolor{lightgray}{\\sin(B)}\n",
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"$$ \n",
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"\n",
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"## Matching Variables\n",
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"\n",
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"To match with $cos(x)$, we can let either let $A = x$ or $B = x$\n",
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"\n",
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"- I choose to let $B = x$\n",
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"\n",
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"$$\n",
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"\\cos(A \\mp x) = \\cos(A)\\cos(x) \\pm \\sin(A)\\sin(x)\n",
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"$$ \n",
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"\n",
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"It may seem that we can match $a$ with $\\cos(A)$ and $b$ with $\\sin(A)$\n",
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"\n",
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"$$\n",
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"\\textcolor{green}{a} \n",
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"\\textcolor{lightgray}{\\cos(x)} \n",
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"\\textcolor{lightgray}{\\pm} \n",
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"\\textcolor{blue}{b} \n",
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"\\textcolor{lightgray}{\\sin(x)} \n",
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"\\textcolor{lightgray}{=} \n",
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"\\textcolor{green}{\\cos(A)} \n",
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"\\textcolor{lightgray}{\\cos(x)} \n",
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"\\textcolor{lightgray}{\\pm} \n",
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"\\textcolor{blue}{\\sin(A)} \n",
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"\\textcolor{lightgray}{\\sin(x)}\n",
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"$$\n",
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"\n",
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"But actually, their ranges do not match.\n",
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"- $sin(A)$ and $cos(A)$ has the range of $[-1, 1]$\n",
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"- While both `a` and `b` has the range of $(0, \\infty)$\n",
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"\n",
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"Therefore, we need a coefficient on the double-angle formula to make the expressions matchable. \n",
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"- We can either have the coefficients individually\n",
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"\n",
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"$$\n",
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"\\textcolor{gray}{p} \n",
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"\\textcolor{lightgray}{\\cos(A)\\cos(x)} \n",
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"\\textcolor{lightgray}{\\pm} \n",
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"\\textcolor{gray}{q} \n",
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"\\textcolor{lightgray}{\\sin(A)\\sin(x)}\n",
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"$$\n",
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"\n",
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"- Or as we are finding unification, we can have some coefficient $R$ on the unified double-angle form, to distribute across the 2 terms\n",
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"\n",
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"$$\n",
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"R\\cos(A \\mp x) = R\\cos(A)\\cos(x) \\pm R\\sin(A)\\sin(x)\n",
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"$$ \n",
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"\n",
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"Now it can have the nice comparison of \n",
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"\n",
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"$$\n",
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"\\textcolor{green}{a}\n",
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"\\textcolor{lightgray}{\\cos(x)}\n",
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"\\textcolor{lightgray}{\\pm}\n",
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"\\textcolor{blue}{b}\n",
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"\\textcolor{lightgray}{\\sin(x)} \n",
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"\\textcolor{lightgray}{=} \n",
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"\\textcolor{green}{R\\cos(A)} \n",
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"\\textcolor{lightgray}{\\cos(x)} \n",
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"\\textcolor{lightgray}{\\pm} \n",
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"\\textcolor{blue}{R\\sin(A)} \n",
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"\\textcolor{lightgray}{\\sin(x)}\n",
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"$$\n",
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"\n",
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"Therefore, we get 2 equations we can work with:\n",
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"\n",
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"$$\n",
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"\\begin{align}\n",
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" a &= R\\cos(A) \\\\\n",
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" b &= R\\sin(A)\n",
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"\\end{align}\n",
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"$$"
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]
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},
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{
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"cell_type": "markdown",
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"id": "85684cec-5f4a-4b73-b9a7-9e129e521f4c",
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"metadata": {},
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"source": [
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"# Solving for **A** and **R**\n",
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"\n",
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"We want to solve for $A$ and $R$ such that we can complete the unifying formula.\n",
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"\n",
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"\n",
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"## Solving for **A**\n",
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"\n",
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"From the 2 equations, we can find $A$ by combining $\\sin$ and $\\cos$ into $\\tan$:\n",
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"\n",
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"$$\n",
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"\\frac{b}{a} = \\frac{\\cancel{R}\\sin(A)}{\\cancel{R}\\cos(A)} = \\tan(A).\n",
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"$$\n",
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"\n",
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"And then by taking the inverse:\n",
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"\n",
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"$$\n",
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"\\boxed{\n",
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" A = \\arctan\\left(\\frac{b}{a}\\right)\n",
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"}\n",
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"$$\n",
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"\n",
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"- For $\\arctan$, it actually has an infinite domain, so no restrictions here\n",
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"- It is not like $\\arcsin$ or $\\arccos$, where there is an inclusive domain\n",
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"- Or $\\text{arcsec}$ and $\\text{arccsc}$, where there is an exclusive domain\n",
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"- Also notice we could have used $\\cot$ as well in the first place, and then use $\\text{arccot}$\n",
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"\n",
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"$$\n",
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"\\textcolor{lightgray}{\n",
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" \\boxed{\n",
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" A = \\text{arccot} \\left(\\frac{a}{b}\\right)\n",
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" }\n",
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"}\n",
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"$$\n",
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"\n",
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"- But this is a matter of preference\n",
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"\n",
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"Now we have $A$, let's find $R$:\n",
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"\n",
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"## Solving for **R**\n",
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"\n",
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"Going back to the equations:\n",
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"\n",
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"$$\n",
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"a = R\\cos(A), \\quad b = R\\sin(A)\n",
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"$$\n",
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"\n",
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"We can notice that $\\sin(A)$ and $\\cos(A)$ both exist, so we can prepare it to use the Pythagorean identity:\n",
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"\n",
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"$$\n",
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"a^2 = R^2\\cos^2(A), \n",
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"\\quad \n",
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"b^2 = R^2\\sin^2(A)\n",
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"$$\n",
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"\n",
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"> **Take note here**: Because we have squared the equation, we have introduced another extraneous solution into an otherwise single-solution variable\n",
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"\n",
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"\n",
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"Add these together:\n",
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"\n",
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"$$\n",
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"a^2 + b^2 = R^2\\cos^2(A) + R^2\\sin^2(A)\n",
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"$$\n",
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"\n",
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"Factor out $R^2$\n",
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"\n",
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"$$\n",
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"a^2 + b^2 = R^2 \\biggr[ \\cos^2(A) + \\sin^2(A) \\biggr]\n",
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"$$\n",
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"\n",
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"Using the Pythagorean identity $\\cos^2(A) + \\sin^2(A) = 1$:\n",
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"\n",
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"$$\n",
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"a^2 + b^2 = R^2\n",
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"$$\n",
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"\n",
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"Take the square root:\n",
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"\n",
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"$$\n",
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"R = \\pm\\sqrt{a^2 + b^2}\n",
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"$$\n",
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"\n",
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"Thus, right now we have:\n",
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"\n",
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"$$\n",
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"a\\sin(x) \\pm b\\cos(x) = \n",
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"R\\cos(A \\mp x)\n",
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"$$\n",
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"\n",
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"where:\n",
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"\n",
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"$$\n",
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"\\boxed{\n",
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" A = \\arctan\\left(\\frac{b}{a}\\right)\n",
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"}\n",
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"\\quad \\quad \n",
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"R = \\pm\\sqrt{a^2 + b^2}\n",
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"$$\n"
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]
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},
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{
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"cell_type": "markdown",
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"id": "72997fa1-e58d-45e7-88e7-d295d9723d48",
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"metadata": {},
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"source": [
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"# Are There Restrictions Before $R^2 = a^2 + b^2$?\n",
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"\n",
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"Before the step $R^2 = a^2 + b^2$, there is no inherent restriction that forces $R$ to be positive. The square root function $\\sqrt{a^2 + b^2}$ is defined as the **non-negative root**, but the $\\pm$ arises because squaring both sides of an equation introduces an inherent ambiguity."
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]
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},
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{
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"cell_type": "markdown",
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"id": "d6cb31a9-f27c-4266-89c5-14470e5d8c7d",
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"metadata": {},
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"source": [
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"# Resolving the Ambiguity in $R$\n",
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"\n",
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"Notice that we still have ambiguity at $R$, where we can only choose one answer or the other. But let’s dig further:\n",
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"\n",
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"$$\n",
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"R = \\pm\\sqrt{a^2 + b^2}.\n",
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"$$\n",
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"\n",
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"Substituting both cases into our unified equation, we get:\n",
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"\n",
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"$$\n",
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"R\\cos(A \\mp x) \\quad \\text{ and } \\quad -R\\cos(A \\mp x).\n",
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"$$\n",
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"\n",
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"Zooming in on the second case, we can distribute the negative sign on $\\cos$:\n",
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"\n",
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"$$\n",
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"-R\\cos(A \\mp x) = R \\cdot -\\cos(A \\mp x)\n",
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"$$\n",
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"\n",
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"Using the trigonometric identity:\n",
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"\n",
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"$$\n",
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"-\\cos(\\theta) = \\cos(\\theta + \\pi)\n",
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"$$\n",
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"\n",
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"We can rewrite this as:\n",
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"\n",
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"$$\n",
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"R \\cdot -\\cos(A \\mp x) = R \\cos(A \\mp x + \\pi).\n",
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"$$\n",
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"\n",
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"$$\n",
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"R \\cdot -\\cos(A \\mp x) = R \\cos((A + \\pi) \\mp x).\n",
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"$$\n",
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"\n",
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"Now notice that $A$ has a period of $\\pi$, meaning:\n",
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"\n",
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"$$\n",
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"A + \\pi \\quad \\text{is functionally equivalent to} \\quad A.\n",
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"$$\n",
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"\n",
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"And thus\n",
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"\n",
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"$$\n",
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"R\\cos(A \\mp x + \\pi) = R\\cos(A \\mp x).\n",
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"$$\n",
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"\n",
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"We have rephrased the form with negative R, into the form with positive R\n",
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"- Hence, we can always take just the positive part of R\n",
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"\n",
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"$$\n",
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"\\boxed{\n",
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" R = \\sqrt{a^2 + b^2}\n",
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"}\n",
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"$$"
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]
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},
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{
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"cell_type": "markdown",
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"id": "7e70ea64-b61f-49e0-878d-655dbc539641",
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"metadata": {},
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"source": [
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"# Final Formula\n",
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"\n",
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"Thus, the R-formula for $a\\cos(x) \\pm b\\sin(x)$ is:\n",
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"\n",
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"$$\n",
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"a\\cos(x) \\pm b\\sin(x) = R\\cos(A \\mp x)\n",
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"$$\n",
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"\n",
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"where:\n",
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"\n",
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"$$\n",
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"A = \\arctan\\left(\\frac{b}{a}\\right), \\quad R = \\sqrt{a^2 + b^2}\n",
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"$$\n"
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]
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},
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{
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"cell_type": "markdown",
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"id": "5e1b6cc7-fdd7-489a-b854-ef1e9e5a524c",
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"metadata": {},
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"source": [
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"---\n",
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"### We can use the same logic to set up $a\\sin(x) \\pm b\\cos(x)$\n",
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"---"
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]
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3 (ipykernel)",
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"language": "python",
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"name": "python3"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.12.9"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 5
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}

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