From 5c7fd3ae2bf754fd209dfe75001c205d49710a60 Mon Sep 17 00:00:00 2001
From: Youwen Wu <youwenw@gmail.com>
Date: Wed, 5 Feb 2025 01:22:45 -0800
Subject: [PATCH] auto-update(nvim): 2025-02-05 01:22:45

---
 .../by-course/math-4b/course-notes/main.typ   | 49 +++++++++++++++++++
 1 file changed, 49 insertions(+)

diff --git a/documents/by-course/math-4b/course-notes/main.typ b/documents/by-course/math-4b/course-notes/main.typ
index 6de8a5d..e2c5949 100644
--- a/documents/by-course/math-4b/course-notes/main.typ
+++ b/documents/by-course/math-4b/course-notes/main.typ
@@ -786,3 +786,52 @@ $
 $
 
 where $A$ is the amplitude, $theta$ is the phase angle with $cos theta = (c_1)/sqrt(c_1^2 + c_2^2)$
+
+== Some trig review (yuck)
+
+#fact[
+  The following trig identity is useful
+
+  $
+    A cos(omega t) + B sin(omega t) = C cos(omega t - gamma)
+  $
+
+  where $C = sqrt(A^2 + B^2)$, $sin(gamma) = B/C$, and $cos(gamma) = A/C$.
+]<sus-identity>
+
+To find gamma, we can simply use inverse trig functions.
+
+#example[
+  Write $x(t) = -2cos(5t) - sin(5t)$ using only one cosine function:
+
+  First note that $A = -2$, $B - -1$, $omega = 5$. Then by @sus-identity,
+
+  $
+    x(t) = -2cos(5t) - sin(5t) \
+    = sqrt(5) cos(5t - gamma)
+  $
+
+  Now note that
+  $
+    tan(gamma) &= sin(gamma) / cos(gamma) \
+    &= 1 / 2
+  $
+
+  Therefore we have the following:
+
+  $
+    gamma = arctan(1/2)
+  $
+
+  But this is not exactly right.
+
+  Recall that $arctan$ produces outputs only in $[-pi/2, pi/2]$. Therefore if
+  we want $cos$ and $sin$ to be negative, we need to add an additional $pi$
+  term. So in fact
+
+  $
+    gamma = arctan(1/2) - pi
+  $
+
+  Which makes $cos(5t + pi - arctan(1/2))$ negative.
+]