✏️ Score 0 / 0
πŸ“

Pre-Calculus
Study Notebook

Vectors Β· Matrices Β· Polar Graphs Β· Trigonometry

πŸ”Ί Trig Basics ➑️ Vectors πŸ”² Matrices πŸŒ€ Polar Graphs πŸ“Š 20 Questions ✨ Instant Feedback
πŸ“ Unit 1 β€” Trigonometry Fundamentals
🧠 Core Concepts β€” Quick Review
SOH-CAH-TOA\(\sin\theta=\dfrac{\text{opp}}{\text{hyp}},\quad\cos\theta=\dfrac{\text{adj}}{\text{hyp}},\quad\tan\theta=\dfrac{\text{opp}}{\text{adj}}\)
Pythagorean ID\(\sin^2\theta+\cos^2\theta=1\)
Unit Circle key\((x,y)=(\cos\theta,\sin\theta)\)
Radian↔Degree\(180Β°=\pi \text{ rad}\)
Reference AngleAlways positive, in first quadrant
"All Students Take Calculus" β†’ Quadrant I: All+, II: Sin+, III: Tan+, IV: Cos+
1
If \(\sin\theta = \dfrac{3}{5}\) and \(\theta\) is in Quadrant II, what is \(\cos\theta\)?
Easy
⚠️ TRAP: Quadrant II β†’ cosine is NEGATIVE!
πŸ“– EXPLANATION
Use the Pythagorean Identity: \(\cos^2\theta = 1 - \sin^2\theta = 1 - \frac{9}{25} = \frac{16}{25}\)
So \(\cos\theta = \pm\frac{4}{5}\). Since \(\theta\) is in Quadrant II, cosine is negative.
∴ \(\cos\theta = -\dfrac{4}{5}\)
ASTC: In Q2 β†’ only Sin is positive β†’ Cos must be NEGATIVE βœ“
2
Convert \(225Β°\) to radians.
Easy
⚠️ TRAP: Multiply by \(\dfrac{\pi}{180}\), NOT \(\dfrac{180}{\pi}\)!
πŸ“– EXPLANATION
\(225Β° \times \dfrac{\pi}{180Β°} = \dfrac{225\pi}{180} = \dfrac{5\pi}{4}\)
Simplify: GCD(225,180) = 45 β†’ \(\dfrac{225}{180} = \dfrac{5}{4}\)
∴ \(225° = \dfrac{5\pi}{4}\) radians (that's 3rd quadrant, 45° past 180°)
3
What is the exact value of \(\tan\!\left(\dfrac{2\pi}{3}\right)\)?
Tricky ⚠️
Hint: Reference angle of \(\dfrac{2\pi}{3}\) is \(\dfrac{\pi}{3} = 60Β°\)
⚠️ TRAP: 2Ο€/3 is in Q2 β€” tan is NEGATIVE there!
πŸ“– EXPLANATION
\(\frac{2\pi}{3} = 120Β°\) β†’ Reference angle = \(60Β°\), located in Quadrant II
\(\tan 60Β° = \sqrt{3}\), but in Q2, tan is negative (ASTC: only Sin+ in Q2)
∴ \(\tan\!\left(\frac{2\pi}{3}\right) = -\sqrt{3}\)
4
Which expression equals \(\sin(2\theta)\)?
Tricky ⚠️
⚠️ TRAP: \(\sin(2\theta) \neq 2\sin\theta\) !
πŸ“– EXPLANATION
Double Angle Formula: \(\sin(2\theta) = 2\sin\theta\cos\theta\)
Note: \(\cos(2\theta) = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta\) (3 forms!)
Option C is \(\cos(2\theta)\), Option D is also \(\cos(2\theta)\) β€” don't confuse!
sin DOUBLE = 2 Γ— sin Γ— cos (two different functions)
5
The graph of \(y = 3\sin(2x - \pi)\) has amplitude, period, and phase shift of:
Tricky ⚠️
Form: \(y = A\sin(Bx - C)\) β†’ Phase shift = \(\dfrac{C}{B}\)
⚠️ TRAP: Period = \(\dfrac{2\pi}{B}\), NOT just \(2\pi\)!
πŸ“– EXPLANATION
\(y = 3\sin(2x - \pi)\): here A=3, B=2, C=Ο€
β€’ Amplitude = |A| = 3
β€’ Period = \(\dfrac{2\pi}{B} = \dfrac{2\pi}{2} = \pi\)
β€’ Phase shift = \(\dfrac{C}{B} = \dfrac{\pi}{2}\) to the right (positive = right)
Option A has wrong period (2Ο€), Option C has wrong shift direction.
➑️ Unit 2 β€” Vectors
🧠 Vector Quick Reference
Magnitude\(\|\vec{v}\|=\sqrt{v_x^2+v_y^2}\)
Unit vector\(\hat{v}=\dfrac{\vec{v}}{\|\vec{v}\|}\)
Dot product\(\vec{a}\cdot\vec{b}=a_xb_x+a_yb_y=\|\vec{a}\|\|\vec{b}\|\cos\theta\)
Perpendicular\(\vec{a}\perp\vec{b} \Leftrightarrow \vec{a}\cdot\vec{b}=0\)
Component form\(\vec{v}=\|\vec{v}\|\langle\cos\theta,\sin\theta\rangle\)
"DOT = Multiply matching, then ADD" β€” if result = 0, they're perpendicular!
6
Find the magnitude of \(\vec{v} = \langle -3, 4 \rangle\)
Easy
⚠️ TRAP: Don't forget to square the negative! \((-3)^2 = 9\), NOT βˆ’9
πŸ“– EXPLANATION
\(\|\vec{v}\| = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\)
This is a classic 3-4-5 right triangle! Memorize Pythagorean triples: (3,4,5), (5,12,13), (8,15,17)
7
Are \(\vec{a}=\langle 2,-3\rangle\) and \(\vec{b}=\langle 6,4\rangle\) perpendicular?
Find \(\vec{a}\cdot\vec{b}\).
Easy
πŸ“– EXPLANATION
\(\vec{a}\cdot\vec{b} = (2)(6) + (-3)(4) = 12 - 12 = 0\)
Since the dot product = 0, the vectors are perpendicular (orthogonal)! βœ“
DOT = 0 β†’ PERPENDICULAR ("zero dot = right angle")
8
A force of 20 N acts at 30Β° above the positive x-axis. What is its x-component?
Medium
⚠️ TRAP: x-component uses cos, y-component uses sin β€” don't switch!
πŸ“– EXPLANATION
x-component = \(20\cos(30Β°) = 20 \cdot \dfrac{\sqrt{3}}{2} = 10\sqrt{3} \approx 17.3\) N
y-component = \(20\sin(30Β°) = 20 \cdot \dfrac{1}{2} = 10\) N
Component: x→cos (horizontal), y→sin (vertical) — "x is cosine, y is sine"
9
Find the angle between \(\vec{u}=\langle 1,0\rangle\) and \(\vec{w}=\langle 1,1\rangle\).
Medium
Formula: \(\cos\theta = \dfrac{\vec{u}\cdot\vec{w}}{\|\vec{u}\|\,\|\vec{w}\|}\)
πŸ“– EXPLANATION
\(\vec{u}\cdot\vec{w} = 1\cdot1 + 0\cdot1 = 1\)
\(\|\vec{u}\| = 1,\quad \|\vec{w}\| = \sqrt{1^2+1^2} = \sqrt{2}\)
\(\cos\theta = \dfrac{1}{1\cdot\sqrt{2}} = \dfrac{1}{\sqrt{2}}\) β†’ \(\theta = 45Β°\) βœ“
10
Given \(\vec{p}=\langle 3,4\rangle\), find the unit vector \(\hat{p}\).
Medium
πŸ“– EXPLANATION
Step 1: magnitude \(= \sqrt{9+16} = 5\)
Step 2: divide each component by 5: \(\hat{p} = \left\langle\frac{3}{5},\frac{4}{5}\right\rangle\)
Check: \(\left(\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2 = \frac{9}{25}+\frac{16}{25} = 1\) βœ“ (unit vector always has magnitude 1)
UNIT vector = divide by magnitude β†’ length becomes exactly 1
πŸ”² Unit 3 β€” Matrices
🧠 Matrix Quick Reference
MultiplyRow Γ— Column, sizes: (mΓ—n)(nΓ—p)=(mΓ—p)
Determinant 2Γ—2\(\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc\)
Inverse 2Γ—2\(A^{-1}=\dfrac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}\)
No inverse if...\(\det(A)=0\) β†’ singular matrix
Identity matrix\(I=\begin{pmatrix}1&0\\0&1\end{pmatrix}\), \(AI=IA=A\)
"2Γ—2 det: Ad minus Bc β€” A-D diagonal MINUS B-C diagonal"
Inverse: SWAP main diagonal, NEGATE off-diagonal, DIVIDE by det
11
Find the determinant: \(\det\begin{pmatrix}5 & 2 \\ 3 & 4\end{pmatrix}\)
Easy
⚠️ TRAP: It's (5)(4) βˆ’ (2)(3), NOT (5)(4) + (2)(3)!
πŸ“– EXPLANATION
\(\det = (5)(4) - (2)(3) = 20 - 6 = 14\)
Remember: main diagonal product MINUS off-diagonal product
Option A (26) = 20+6 (wrong: added instead of subtracted)
12
Multiply: \(\begin{pmatrix}1&2\\3&4\end{pmatrix}\begin{pmatrix}5\\6\end{pmatrix}\)
Medium
⚠️ TRAP: Matrix multiplication is NOT element-by-element!
πŸ“– EXPLANATION
Row 1 Γ— Column: \(1\cdot5 + 2\cdot6 = 5+12 = 17\)
Row 2 Γ— Column: \(3\cdot5 + 4\cdot6 = 15+24 = 39\)
Result is a 2Γ—1 matrix: \(\begin{pmatrix}17\\39\end{pmatrix}\)
Sizes: (2Γ—2)(2Γ—1) = (2Γ—1) βœ“
ROW Γ— COLUMN β†’ each entry: sum of (row element Γ— column element)
13
Does \(A = \begin{pmatrix}2&4\\1&2\end{pmatrix}\) have an inverse?
Tricky ⚠️
⚠️ TRAP: Check det FIRST before finding inverse!
πŸ“– EXPLANATION
\(\det(A) = (2)(2) - (4)(1) = 4 - 4 = 0\)
When \(\det = 0\), the matrix is singular β€” NO inverse exists!
Notice row 2 is exactly Β½ of row 1 β†’ linearly dependent rows β†’ always det=0
det = 0 β†’ SINGULAR β†’ NO inverse (like dividing by zero!)
14
For \(A=\begin{pmatrix}3&1\\2&1\end{pmatrix}\), what is \(A^{-1}\)?
Tricky ⚠️
⚠️ TRAP: SWAP the main diagonal, NEGATE the off-diagonal!
πŸ“– EXPLANATION
det \(= 3\cdot1 - 1\cdot2 = 1\)
\(A^{-1} = \dfrac{1}{1}\begin{pmatrix}1&-1\\-2&3\end{pmatrix} = \begin{pmatrix}1&-1\\-2&3\end{pmatrix}\)
Verify: \(A\cdot A^{-1} = \begin{pmatrix}3&1\\2&1\end{pmatrix}\begin{pmatrix}1&-1\\-2&3\end{pmatrix} = \begin{pmatrix}1&0\\0&1\end{pmatrix}\) βœ“
πŸŒ€ Unit 4 β€” Polar Coordinates & Graphs
🧠 Polar Quick Reference
Polar β†’ Rect\(x = r\cos\theta,\quad y = r\sin\theta\)
Rect β†’ Polar\(r = \sqrt{x^2+y^2},\quad \theta = \arctan\!\left(\frac{y}{x}\right)\)
Circle: \(r=a\)Circle centered at origin, radius \(a\)
Rose: \(r=a\cos(n\theta)\)n petals if n odd; 2n petals if n even
Limacon: \(r=a+b\sin\theta\)Inner loop if \(|b|>|a|\)
Lemniscate\(r^2=a^2\cos(2\theta)\) β€” figure-8 shape
"r=aΒ±b: if a<b β†’ inner loop, if a=b β†’ cardioid, if a>b β†’ dimpled or convex"
Rose petals: odd n β†’ n petals; even n β†’ 2n petals
15
Convert polar \((r,\theta) = \left(4, \dfrac{\pi}{3}\right)\) to rectangular \((x,y)\).
Easy
πŸ“– EXPLANATION
\(x = r\cos\theta = 4\cos\!\left(\frac{\pi}{3}\right) = 4\cdot\frac{1}{2} = 2\)
\(y = r\sin\theta = 4\sin\!\left(\frac{\pi}{3}\right) = 4\cdot\frac{\sqrt{3}}{2} = 2\sqrt{3}\)
∴ \((x,y) = (2,\ 2\sqrt{3})\) βœ“
⚠️ Option B has x and y swapped β€” common mistake!
16
Convert rectangular \((-\sqrt{3},\ 1)\) to polar form \((r,\theta)\) where \(r>0\) and \(0\le\theta<2\pi\).
Tricky ⚠️
⚠️ TRAP: arctan gives wrong quadrant β€” check (x,y) signs first!
πŸ“– EXPLANATION
\(r = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{3+1} = 2\)
Reference angle: \(\arctan\!\left(\frac{1}{\sqrt{3}}\right) = 30Β° = \frac{\pi}{6}\)
Point \((-\sqrt{3}, 1)\): x is negative, y is positive β†’ Quadrant II
\(\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}\) βœ“
Always CHECK which quadrant (x,y) is in BEFORE adjusting arctan!
17
How many petals does the polar rose \(r = 3\cos(4\theta)\) have?
Tricky ⚠️
⚠️ TRAP: When n is EVEN β†’ number of petals = 2n, not n!
πŸ“– EXPLANATION
For \(r = a\cos(n\theta)\) or \(r = a\sin(n\theta)\):
β€’ If n is odd β†’ n petals
β€’ If n is even β†’ 2n petals
Here n = 4 (even) β†’ 2Γ—4 = 8 petals βœ“
ODD n β†’ n petals | EVEN n β†’ DOUBLE (2n) petals
18
Identify the curve \(r = 2 + 3\sin\theta\). What type is it?
Medium
Form: \(r = a + b\sin\theta\), here \(a=2, b=3\)
⚠️ TRAP: Compare |a| vs |b| to determine the type!
πŸ“– EXPLANATION
Here \(|b| = 3 > |a| = 2\), so the limaçon has an inner loop.
Classification of \(r = a + b\sin\theta\):
β€’ \(a = b\) β†’ cardioid (heart shape)
β€’ \(|b| > |a|\) β†’ inner loop
β€’ \(|a| \geq 2|b|\) β†’ convex (no dimple)
‒ Otherwise → dimpled limaçon
19
Which rectangular equation is equivalent to \(r = 4\sin\theta\)?
Tricky ⚠️
Hint: Multiply both sides by \(r\), then substitute \(r^2 = x^2+y^2\) and \(y = r\sin\theta\)
πŸ“– EXPLANATION
Step 1: Multiply both sides by r: \(r^2 = 4r\sin\theta\)
Step 2: Substitute: \(x^2 + y^2 = 4y\) (Option D is also correct as an intermediate step!)
Step 3: Complete the square: \(x^2 + y^2 - 4y = 0\) β†’ \(x^2 + (y-2)^2 = 4\)
This is a circle centered at (0, 2) with radius 2 βœ“
20
πŸ”₯ CHALLENGE: A vector \(\vec{v}\) has magnitude 6 and direction angle 150Β°. Express in rectangular form \(a\mathbf{i}+b\mathbf{j}\), then find \(\vec{v}\cdot\vec{w}\) where \(\vec{w}=\langle\sqrt{3},-1\rangle\).
Tricky ⚠️
This connects vectors + trig + dot product all in one!
πŸ“– EXPLANATION
Step 1: Convert to rectangular form:
\(v_x = 6\cos(150Β°) = 6\cdot\left(-\frac{\sqrt{3}}{2}\right) = -3\sqrt{3}\)
\(v_y = 6\sin(150Β°) = 6\cdot\frac{1}{2} = 3\)
So \(\vec{v} = \langle -3\sqrt{3},\ 3\rangle\)

Step 2: Dot product with \(\vec{w}=\langle\sqrt{3},-1\rangle\):
\(\vec{v}\cdot\vec{w} = (-3\sqrt{3})(\sqrt{3}) + (3)(-1) = -9 - 3 = -12\) βœ“
150Β° is in Q2 β†’ x-component NEGATIVE, y-component POSITIVE (sin 150Β° = sin 30Β° = Β½)
πŸ“Š Your Final Score
β€”