Self-Study Edition · 2025

PreCalculus
Core Problems

Vectors · Matrices · Polar Graphs · Trigonometry
20 carefully selected high-mistake problems

0 / 20 answered
✓ Correct 0
✗ Incorrect 0
Chapter 1 · Trigonometry Basics
⚡
Memory Key SOHCAHTOA Sin = Opp/Hyp · Cos = Adj/Hyp · Tan = Opp/Adj
Q1
Trig Ratios · Easy
In a right triangle, the opposite side = 3 and hypotenuse = 5.
What is \(\sin\theta\)?
Hint: Which letter in SOHCAHTOA uses Opposite and Hypotenuse?
📘 Explanation Using the Pythagorean theorem, adjacent = \(\sqrt{5^2 - 3^2} = \sqrt{16} = 4\).
\(\sin\theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{3}{5}\).
Watch out: many students accidentally pick \(\frac{4}{5}\) (that's \(\cos\theta\)!).
⚡
Memory Key All Students Take Calculus Quadrant I: All positive · II: Sin · III: Tan · IV: Cos
Q2
Quadrant Signs · Easy
If \(\sin\theta > 0\) and \(\cos\theta < 0\), in which quadrant does \(\theta\) lie?
Hint: Use "All Students Take Calculus" to recall which functions are positive in each quadrant.
📘 Explanation Quadrant II: x is negative (so cos < 0) and y is positive (so sin > 0).
"All Students Take Calculus" → in QII, only Sin is positive. ✓
⚡
Memory Key Pythag Identity \(\sin^2\theta + \cos^2\theta = 1\) — Always true, everywhere!
Q3
Pythagorean Identity · Easy
If \(\sin\theta = \tfrac{5}{13}\), find \(\cos\theta\)
(assume \(\theta\) is in Quadrant I).
\(\sin^2\theta + \cos^2\theta = 1\)
📘 Explanation \(\cos^2\theta = 1 - \sin^2\theta = 1 - \frac{25}{169} = \frac{144}{169}\)
\(\cos\theta = \frac{12}{13}\) (positive in QI).
Classic 5-12-13 Pythagorean triple — memorize it!
⚡
Memory Key Degree → Radian Multiply by \(\pi / 180\). Think: 180° = π radians.
Q4
Angle Conversion · Easy
Convert \(270°\) to radians.
Hint: Multiply by \(\dfrac{\pi}{180}\) and simplify.
📘 Explanation \(270 \times \frac{\pi}{180} = \frac{270\pi}{180} = \frac{3\pi}{2}\).
Key check: 90°=π/2, 180°=π, 270°=3π/2, 360°=2π. These four are must-memorize!
⚡
Memory Key Reference Angle Always positive, always ≤ 90°. Measure from x-axis, not y-axis!
Q5
Reference Angles · Easy
What is the reference angle of \(210°\)?
Hint: 210° is in Quadrant III. Subtract 180° to find the reference angle.
📘 Explanation In Quadrant III: reference angle = \(\theta - 180° = 210° - 180° = 30°\).
Common mistake: students subtract from 360° (that's for QIV!).
Chapter 2 · Polar Graphs
⚡
Memory Key Polar Point \((r, \theta)\): r = distance from origin · θ = angle from positive x-axis
Q6
Polar Coordinates · Easy
Convert polar point \(\left(4,\, \dfrac{\pi}{3}\right)\) to rectangular form \((x, y)\).
\(x = r\cos\theta, \quad y = r\sin\theta\)
📘 Explanation \(\cos\frac{\pi}{3} = \frac{1}{2},\quad \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}\)
\(x = 4 \cdot \frac{1}{2} = 2, \quad y = 4 \cdot \frac{\sqrt{3}}{2} = 2\sqrt{3}\)
Common mix-up: swapping sin and cos. Remember: x uses cos, y uses sin.
⚡
Memory Key Cardioid Shape \(r = a \pm a\cos\theta\) or \(r = a \pm a\sin\theta\) → Heart shape. "Cardio" = heart!
Q7
Polar Curve ID · Easy
What shape does \(r = 2 + 2\cos\theta\) produce?
Hint: When a = b in r = a + b·cosθ, you get a special name.
📘 Explanation When \(a = b\) in \(r = a + b\cos\theta\), the result is a cardioid (heart-shaped curve).
It touches the origin (r = 0) when \(\cos\theta = -1\), i.e., \(\theta = \pi\).
If \(a \neq b\), it becomes a limaçon.
⚡
Memory Key Rose Petals \(r = a\cos(n\theta)\): n odd → n petals · n even → 2n petals
Q8
Rose Curves · Easy
How many petals does \(r = 3\cos(4\theta)\) have?
Hint: n = 4 is even. Apply the even rule.
📘 Explanation For \(r = a\cos(n\theta)\): n is even → \(2n\) petals.
\(n = 4\) (even) → \(2 \times 4 = 8\) petals. ✓
The trap: many guess just "4". Remember to double when n is even!
⚡
Memory Key Circle Polar \(r = a\) → circle with radius |a|. \(r = a\cos\theta\) → circle shifted on x-axis.
Q9
Polar Circles · Easy
What is the rectangular form of \(r = 6\cos\theta\)?
Multiply both sides by \(r\): \(\;r^2 = 6r\cos\theta\)
Recall: \(r^2 = x^2 + y^2\) and \(r\cos\theta = x\)
📘 Explanation \(r^2 = 6r\cos\theta \Rightarrow x^2+y^2 = 6x\)
Complete the square: \(x^2 - 6x + 9 + y^2 = 9\)
\((x-3)^2 + y^2 = 9\) — a circle centered at (3, 0) with radius 3.
⚡
Memory Key Negative r Negative r = go the OPPOSITE direction. \((-3, 60°)\) = \((3, 240°)\)
Q10
Negative Radius · Tricky!
Which point is equivalent to \(\left(-2,\; \dfrac{\pi}{4}\right)\) in polar form?
📘 Explanation Negative r: flip direction → add π to angle.
\(\left(-2, \frac{\pi}{4}\right) = \left(2, \frac{\pi}{4} + \pi\right) = \left(2, \frac{5\pi}{4}\right)\) ✓
This is the single most common mistake on polar exams!
Chapter 3 · Vectors
⚡
Memory Key Magnitude Formula \(|\vec{v}| = \sqrt{a^2 + b^2}\) — Just Pythagorean theorem on components!
Q11
Vector Magnitude · Easy
Find \(|\vec{v}|\) for \(\vec{v} = \langle -5, 12 \rangle\).
📘 Explanation \(|\vec{v}| = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\)
Recognize the 5-12-13 Pythagorean triple!
⚡
Memory Key Dot Product Sign Dot > 0 → acute · Dot = 0 → perpendicular · Dot < 0 → obtuse
Q12
Dot Product · Easy
Are \(\vec{u} = \langle 3, -4 \rangle\) and \(\vec{v} = \langle 8, 6 \rangle\) perpendicular?
\(\vec{u} \cdot \vec{v} = u_1 v_1 + u_2 v_2\)
📘 Explanation \(\vec{u}\cdot\vec{v} = (3)(8) + (-4)(6) = 24 - 24 = 0\)
Dot product = 0 → vectors are perpendicular. ✓
Don't confuse "same magnitude" with perpendicularity — those are unrelated!
⚡
Memory Key Unit Vector Divide by magnitude: \(\hat{u} = \dfrac{\vec{v}}{|\vec{v}|}\)
Q13
Unit Vectors · Easy
Find the unit vector in the direction of \(\vec{v} = \langle 6, 8 \rangle\).
📘 Explanation \(|\vec{v}| = \sqrt{36+64} = \sqrt{100} = 10\)
\(\hat{u} = \frac{1}{10}\langle 6,8\rangle = \left\langle \frac{3}{5}, \frac{4}{5}\right\rangle\)
Verify: \(\left(\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2 = \frac{9}{25}+\frac{16}{25} = 1\) ✓
⚡
Memory Key Angle Between Vectors \(\cos\theta = \dfrac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}|}\) — dot over product of magnitudes
Q14
Angle Between Vectors · Tricky!
Find the angle between \(\vec{u} = \langle 1, 0 \rangle\) and \(\vec{v} = \langle 1, 1 \rangle\).
Hint: \(\langle 1,0 \rangle\) is the x-axis unit vector. What angle does \(\langle 1,1 \rangle\) make?
📘 Explanation \(\vec{u}\cdot\vec{v} = 1, \quad |\vec{u}| = 1, \quad |\vec{v}| = \sqrt{2}\)
\(\cos\theta = \frac{1}{1\cdot\sqrt{2}} = \frac{1}{\sqrt{2}} \Rightarrow \theta = 45°\) ✓
Intuitively, \(\langle1,1\rangle\) points diagonally — 45° from horizontal.
⚡
Memory Key Vector Component Form \(\vec{v} = \langle r\cos\theta,\; r\sin\theta \rangle\) — same as polar-to-rectangular!
Q15
Vector Direction · Connects Trig + Vectors
A vector has magnitude 10 and direction angle 150°. Find its component form.
Hint: Use \(\vec{v} = \langle 10\cos 150°,\; 10\sin 150° \rangle\)
📘 Explanation \(\cos 150° = -\frac{\sqrt{3}}{2},\quad \sin 150° = \frac{1}{2}\)
\(\vec{v} = \langle 10\cdot(-\frac{\sqrt{3}}{2}),\; 10\cdot\frac{1}{2}\rangle = \langle -5\sqrt{3},\; 5\rangle\) ✓
150° is in QII: x negative, y positive.
Chapter 4 · Matrices
⚡
Memory Key Row × Column Matrix mult: (m×n)(n×p) = m×p. Inner dimensions MUST match!
Q16
Matrix Multiplication · Easy
Find entry \(c_{11}\) of \(C = AB\) where:
\(A = \begin{bmatrix}2 & 3\\1 & 0\end{bmatrix}, \quad B = \begin{bmatrix}1 & 4\\2 & -1\end{bmatrix}\)
Row 1 of A · Column 1 of B
📘 Explanation \(c_{11} = (2)(1) + (3)(2) = 2 + 6 = 8\)
Row 1 of A = [2, 3] · Column 1 of B = [1, 2]
Multiply corresponding entries and add. Don't forget: AB ≠ BA in general!
⚡
Memory Key 2×2 Determinant \(\det\begin{bmatrix}a&b\\c&d\end{bmatrix} = ad - bc\) — "diagonal minus anti-diagonal"
Q17
Determinant · Easy
Find the determinant:
\(\det\begin{bmatrix}5 & -2\\3 & 4\end{bmatrix}\)
📘 Explanation \(\det = (5)(4) - (-2)(3) = 20 - (-6) = 20 + 6 = 26\)
Classic mistake: forgetting the negative sign gives \(20 - 6 = 14\). Always mind the signs!
⚡
Memory Key Inverse 2×2 \(A^{-1} = \dfrac{1}{\det A}\begin{bmatrix}d & -b\\-c & a\end{bmatrix}\) — swap diagonal, negate off-diagonal
Q18
Matrix Inverse · Tricky!
When does a 2×2 matrix NOT have an inverse?
Think about what you divide by when computing the inverse.
📘 Explanation The inverse formula divides by \(\det(A)\). If \(\det(A) = 0\), division by zero → no inverse.
Such a matrix is called singular.
Fun fact: det = 0 means the two row vectors are parallel (linearly dependent).
⚡
Memory Key Identity Matrix \(AI = IA = A\). Like multiplying by 1. Diagonal = 1s, rest = 0s.
Q19
Identity Matrix · Easy
Which matrix is the \(2\times 2\) identity matrix \(I\)?
📘 Explanation The identity matrix has 1s on the main diagonal and 0s everywhere else.
For any matrix A: \(AI = IA = A\).
Option A is the exchange matrix (swaps rows). Don't confuse them!
⚡
Memory Key Cramer's Rule Solve \(Ax = b\) using determinants: \(x = \frac{\det(A_x)}{\det(A)}\)
Q20
Systems with Matrices · Connects All Topics!
The system below has:
\(\begin{bmatrix}2 & 1\\4 & 2\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}6\\12\end{bmatrix}\)
How many solutions does this system have?
Hint: Compute the determinant of the coefficient matrix first!
📘 Explanation \(\det = (2)(2) - (1)(4) = 4 - 4 = 0\) → singular matrix, so we can't get a unique solution.
But notice Row 2 = 2 × Row 1, so the equations are dependent → infinitely many solutions!
The complete set: \(x = \frac{6-y}{2}\) for any value of y.
🎓

Well Done!