π Unit 1 Β· Trigonometry Fundamentals
The Sine, Cosine & Beyond
Where most students first trip up β and build back stronger.
01
In a right triangle, the angle \(\theta\) is opposite a side of length 5 and adjacent to a side of length 12. What is \(\sin\theta\)?
π‘ Recall
SOH: \(\sin\theta = \dfrac{\text{Opposite}}{\text{Hypotenuse}}\) | First, find the hypotenuse using the Pythagorean Theorem: \(a^2 + b^2 = c^2\).
SOH: \(\sin\theta = \dfrac{\text{Opposite}}{\text{Hypotenuse}}\) | First, find the hypotenuse using the Pythagorean Theorem: \(a^2 + b^2 = c^2\).
Hypotenuse = longest side. It's always across from the right angle, never touching it.
Correct! Beautiful work.
1Hypotenuse: \(c = \sqrt{5^2 + 12^2} = \sqrt{25+144} = \sqrt{169} = 13\)
2\(\sin\theta = \dfrac{\text{opp}}{\text{hyp}} = \dfrac{5}{13}\) β
3Common trap: \(\dfrac{5}{12}\) uses the adjacent instead of hypotenuse β that's \(\tan\theta\)!
02
Which of the following is the exact value of \(\cos(150Β°)\)?
π‘ Key Idea
\(150Β° = 180Β° - 30Β°\) β it's in Quadrant II, where cosine is negative. Reference angle is \(30Β°\).
\(150Β° = 180Β° - 30Β°\) β it's in Quadrant II, where cosine is negative. Reference angle is \(30Β°\).
CAST: All positive (Q1) Β· Sine positive (Q2) Β· Tan positive (Q3) Β· Cos positive (Q4). Think: "All Students Take Calculus"
Correct!
1Reference angle: \(180Β°-150Β°=30Β°\), so \(\cos 30Β° = \dfrac{\sqrt{3}}{2}\)
2Quadrant II β cosine is negative β \(\cos 150Β° = -\dfrac{\sqrt{3}}{2}\)
03
If \(\sin\theta = \dfrac{3}{5}\) and \(\theta\) is in Quadrant I, find \(\tan\theta\).
π‘ Strategy
Use \(\sin^2\theta + \cos^2\theta = 1\) to find \(\cos\theta\), then \(\tan\theta = \dfrac{\sin\theta}{\cos\theta}\).
Use \(\sin^2\theta + \cos^2\theta = 1\) to find \(\cos\theta\), then \(\tan\theta = \dfrac{\sin\theta}{\cos\theta}\).
tan = sin Γ· cos. Once you know two of the three, you know all six trig functions.
Correct!
1\(\cos^2\theta = 1 - \sin^2\theta = 1 - \tfrac{9}{25} = \tfrac{16}{25}\), so \(\cos\theta = \tfrac{4}{5}\) (Q1 β positive)
2\(\tan\theta = \dfrac{3/5}{4/5} = \dfrac{3}{4}\)
04
Convert \(225Β°\) to radians. Which answer is correct?
π‘ Formula
Multiply by \(\dfrac{\pi}{180Β°}\). Then simplify the fraction fully.
Multiply by \(\dfrac{\pi}{180Β°}\). Then simplify the fraction fully.
180Β° = Ο radians. So 1Β° = Ο/180. Always simplify: cancel common factors between the degree number and 180.
Correct!
1\(225Β° \times \dfrac{\pi}{180} = \dfrac{225\pi}{180}\)
2GCF(225, 180) = 45 β \(\dfrac{225}{180} = \dfrac{5}{4}\)
3Answer: \(\dfrac{5\pi}{4}\) β (this is in Quadrant III)
05
What is the period of the function \(f(x) = 3\sin(2x - \pi)\)?
π‘ General Form
\(f(x) = A\sin(Bx - C) + D\)
Period \(= \dfrac{2\pi}{|B|}\), Amplitude \(= |A|\), Phase Shift \(= \dfrac{C}{B}\)
\(f(x) = A\sin(Bx - C) + D\)
Period \(= \dfrac{2\pi}{|B|}\), Amplitude \(= |A|\), Phase Shift \(= \dfrac{C}{B}\)
Period = 2Ο Γ· B. Most students accidentally put 3 (the amplitude) as B. The number INSIDE with x is B!
Correct!
1Here, \(B = 2\), \(A = 3\), \(C = \pi\)
2Period \(= \dfrac{2\pi}{2} = \pi\)
3Phase Shift \(= \dfrac{\pi}{2}\) (shifted right), Amplitude = 3. These are separate!
β Unit 2 Β· Vectors
Magnitude, Direction & Operations
Vectors have direction. Don't lose it.
06
Find the magnitude of vector \(\vec{v} = \langle 3, -4 \rangle\).
π‘ Formula
\(|\vec{v}| = \sqrt{v_x^2 + v_y^2}\) β this is just the Pythagorean theorem in disguise!
\(|\vec{v}| = \sqrt{v_x^2 + v_y^2}\) β this is just the Pythagorean theorem in disguise!
Magnitude is always positive. Even if components are negative, squaring them makes them positive.
Correct!
1\(|\vec{v}| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5\)
2Classic 3-4-5 Pythagorean triple! Memorize: (3,4,5), (5,12,13), (8,15,17)
07
Given \(\vec{a} = \langle 1, 2 \rangle\) and \(\vec{b} = \langle 3, -1 \rangle\), calculate the dot product \(\vec{a} \cdot \vec{b}\).
π‘ Formula
\(\vec{a} \cdot \vec{b} = a_x b_x + a_y b_y\)
Multiply matching components, then add. Result is a scalar (just a number, not a vector).
\(\vec{a} \cdot \vec{b} = a_x b_x + a_y b_y\)
Multiply matching components, then add. Result is a scalar (just a number, not a vector).
Dot product is a NUMBER, not a vector. If dot product = 0, the vectors are perpendicular (90Β° angle).
Correct!
1\(\vec{a} \cdot \vec{b} = (1)(3) + (2)(-1) = 3 + (-2) = 1\)
2Since the result β 0, these vectors are NOT perpendicular.
3Watch out for D β the dot product is never a vector!
08
What is the unit vector in the direction of \(\vec{v} = \langle 0, -5 \rangle\)?
π‘ Formula
\(\hat{u} = \dfrac{\vec{v}}{|\vec{v}|}\) β divide each component by the magnitude. The result always has magnitude exactly 1.
\(\hat{u} = \dfrac{\vec{v}}{|\vec{v}|}\) β divide each component by the magnitude. The result always has magnitude exactly 1.
Unit vector has magnitude = 1. It ONLY carries direction info. Think of it as a "direction arrow" with fixed length 1.
Correct!
1\(|\vec{v}| = \sqrt{0^2 + (-5)^2} = 5\)
2\(\hat{u} = \dfrac{\langle 0,-5\rangle}{5} = \langle 0, -1\rangle\)
3Check: \(|\langle 0,-1\rangle| = \sqrt{0+1} = 1\) β It points straight down.
09
Find the angle between \(\vec{u} = \langle 1, 0 \rangle\) and \(\vec{w} = \langle 1, 1 \rangle\).
π‘ Formula
\(\cos\theta = \dfrac{\vec{u}\cdot\vec{w}}{|\vec{u}||\vec{w}|}\), then solve for \(\theta = \cos^{-1}(\ldots)\)
\(\cos\theta = \dfrac{\vec{u}\cdot\vec{w}}{|\vec{u}||\vec{w}|}\), then solve for \(\theta = \cos^{-1}(\ldots)\)
This formula is the ONLY way to find the exact angle between two vectors. Memorize the structure: dot product over product of magnitudes.
Correct!
1\(\vec{u}\cdot\vec{w} = (1)(1)+(0)(1)=1\)
2\(|\vec{u}|=1,\quad |\vec{w}|=\sqrt{2}\)
3\(\cos\theta = \dfrac{1}{1\cdot\sqrt{2}} = \dfrac{1}{\sqrt{2}}\), so \(\theta = \cos^{-1}\!\left(\tfrac{1}{\sqrt{2}}\right)=45Β°\)
10
If \(\vec{p} = \langle 2, -3 \rangle\) and \(\vec{q} = \langle -1, 5 \rangle\), find \(2\vec{p} - \vec{q}\).
π‘ Rules
Scalar multiplication: multiply each component. Subtraction: subtract component by component. Do NOT mix x and y components.
Scalar multiplication: multiply each component. Subtraction: subtract component by component. Do NOT mix x and y components.
Scalar Γ vector: multiply every component. Vector Β± vector: operate component by component. x stays with x, y stays with y. Always.
Correct!
1\(2\vec{p} = 2\langle 2,-3\rangle = \langle 4,-6\rangle\)
2\(2\vec{p}-\vec{q} = \langle 4,-6\rangle - \langle -1,5\rangle = \langle 4-(-1),\,-6-5\rangle = \langle 5,-11\rangle\)
β¦ Unit 3 Β· Matrices
Rows, Columns & Multiplication
Matrix multiplication is NOT commutative. That's the #1 trap.
11
If \(A = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}\) and \(B = \begin{pmatrix} 1 \\ 4 \end{pmatrix}\), find \(AB\).
π‘ Matrix Multiplication
Each entry = (row of A) Β· (column of B) as a dot product.
Rows of A Γ Columns of B. Result size: rows of A Γ columns of B.
Each entry = (row of A) Β· (column of B) as a dot product.
Rows of A Γ Columns of B. Result size: rows of A Γ columns of B.
To multiply AΓB: columns of A MUST equal rows of B. Result has rows of A and columns of B. Think (mΓn)(nΓp) = (mΓp).
Correct!
1Row 1 of A: \([2,1]\cdot[1,4] = 2(1)+1(4) = 6\)
2Row 2 of A: \([0,3]\cdot[1,4] = 0(1)+3(4) = 12\)
3Result: \(\begin{pmatrix}6\\12\end{pmatrix}\)
12
Find the determinant of \(M = \begin{pmatrix} 5 & 2 \\ 3 & 4 \end{pmatrix}\).
π‘ Formula
\(\det\begin{pmatrix}a & b\\c & d\end{pmatrix} = ad - bc\)
Diagonal down-left minus diagonal down-right.
\(\det\begin{pmatrix}a & b\\c & d\end{pmatrix} = ad - bc\)
Diagonal down-left minus diagonal down-right.
det = "main diagonal product" MINUS "anti-diagonal product". If det = 0, the matrix has no inverse (singular).
Correct!
1\(\det(M) = (5)(4) - (2)(3) = 20 - 6 = 14\)
2Common error: computing \(5\times4 + 2\times3 = 26\) β wrong! It must be subtraction.
13
Let \(A = \begin{pmatrix}1&2\\0&1\end{pmatrix}\) and \(B = \begin{pmatrix}1&0\\1&1\end{pmatrix}\). Which statement is true?
π‘ Non-Commutativity
Matrix multiplication is usually not commutative: \(AB \neq BA\). Compute both and compare!
Matrix multiplication is usually not commutative: \(AB \neq BA\). Compute both and compare!
AB β BA is the #1 matrix trap. Unlike regular numbers, ORDER MATTERS. Always say "left multiply" or "right multiply."
Correct! This is one of the most important matrix facts.
1\(AB = \begin{pmatrix}1+2&0+2\\0+1&0+1\end{pmatrix}=\begin{pmatrix}3&2\\1&1\end{pmatrix}\)
2\(BA = \begin{pmatrix}1&2\\1&3\end{pmatrix}\)
3\(AB \neq BA\) β matrix multiplication is NOT commutative in general.
14
Find the inverse of \(N = \begin{pmatrix}3&1\\5&2\end{pmatrix}\).
π‘ 2Γ2 Inverse Formula
\(N^{-1} = \dfrac{1}{\det(N)}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}\) where \(\det = ad-bc\)
\(N^{-1} = \dfrac{1}{\det(N)}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}\) where \(\det = ad-bc\)
Steps: β find det β‘ swap a and d β’ negate b and c β£ divide all by det. The order of steps matters!
Correct!
1\(\det(N)=3(2)-1(5)=6-5=1\)
2\(N^{-1}=\dfrac{1}{1}\begin{pmatrix}2&-1\\-5&3\end{pmatrix}=\begin{pmatrix}2&-1\\-5&3\end{pmatrix}\)
3Verify: \(N\cdot N^{-1}=I\) β the identity matrix. Always check!
π Unit 4 Β· Polar Coordinates & Graphs
Roses, Spirals & the Circle of Truth
Polar graphs are beautiful once you see the pattern.
15
Convert the polar point \((4, \frac{\pi}{3})\) to rectangular (Cartesian) coordinates.
π‘ Conversion Formulas
\(x = r\cos\theta\), \(\quad y = r\sin\theta\)
Remember: \(\cos\!\left(\frac{\pi}{3}\right)=\frac{1}{2}\), \(\sin\!\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}\)
\(x = r\cos\theta\), \(\quad y = r\sin\theta\)
Remember: \(\cos\!\left(\frac{\pi}{3}\right)=\frac{1}{2}\), \(\sin\!\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}\)
30-60-90 triangle: sin(30Β°)=Β½, cos(30Β°)=β3/2, sin(60Β°)=β3/2, cos(60Β°)=Β½. Flip sin and cos between 30Β° and 60Β°!
Correct!
1\(x = 4\cos\!\left(\frac{\pi}{3}\right) = 4\cdot\frac{1}{2} = 2\)
2\(y = 4\sin\!\left(\frac{\pi}{3}\right) = 4\cdot\frac{\sqrt{3}}{2} = 2\sqrt{3}\)
3Rectangular: \((2,\; 2\sqrt{3})\) β
16
What shape does the polar equation \(r = 2 + 2\cos\theta\) produce?
π‘ Polar Graph ID Guide
β’ \(r = a\cos(n\theta)\) β Rose curve (n petals if n odd, 2n if n even)
β’ \(r = a + b\cos\theta\) β LimaΓ§on (heart-shape if a=b: cardioid)
β’ \(r = a\) β Circle Β· \(r = a\theta\) β Spiral
β’ \(r = a\cos(n\theta)\) β Rose curve (n petals if n odd, 2n if n even)
β’ \(r = a + b\cos\theta\) β LimaΓ§on (heart-shape if a=b: cardioid)
β’ \(r = a\) β Circle Β· \(r = a\theta\) β Spiral
Cardioid = limaçon where |a|=|b|. The word "cardioid" comes from Greek for heart (kardia). When a=b it makes a perfect heart shape.
Correct! A cardioid β one of the most beautiful curves in math.
1Form: \(r = a + b\cos\theta\) with \(a=2, b=2\). Since \(a=b\), this is a cardioid.
2Cardioids have maximum \(r = 2a = 4\) (at \(\theta=0\)) and pass through the pole (\(r=0\) at \(\theta=\pi\)).
3When \(a \neq b\), same form but gives a limaçon with/without inner loop.
17
How many petals does the rose curve \(r = 3\cos(4\theta)\) have?
π‘ Petal Count Rule
\(r = a\cos(n\theta)\) or \(r = a\sin(n\theta)\):
β If \(n\) is odd: \(n\) petals
β If \(n\) is even: \(2n\) petals
\(r = a\cos(n\theta)\) or \(r = a\sin(n\theta)\):
β If \(n\) is odd: \(n\) petals
β If \(n\) is even: \(2n\) petals
ODD n = n petals. EVEN n = 2n petals. This is because even-n curves trace petals twice! Common wrong answer: "4 petals" β the even rule doubles it.
Correct! Most students say 4 β you got the even-n rule right.
1\(r = 3\cos(4\theta)\): here \(n=4\), which is even.
2Even rule: petals \(= 2n = 2\times4 = 8\)
3Why? The curve completes two full passes over \([0, 2\pi]\), creating each petal twice β but 8 visible petals total.
18
Convert the rectangular point \((-\sqrt{3},\; 1)\) to polar form \((r, \theta)\) with \(r > 0\) and \(0 \le \theta < 2\pi\).
π‘ Formulas
\(r = \sqrt{x^2+y^2}\), \(\quad \tan\theta = \dfrac{y}{x}\)
β οΈ Always check which quadrant the point is in before finding \(\theta\)!
\(r = \sqrt{x^2+y^2}\), \(\quad \tan\theta = \dfrac{y}{x}\)
β οΈ Always check which quadrant the point is in before finding \(\theta\)!
x negative, y positive β Quadrant II. Your calculator gives a Q4 angle for arctan β add Ο to land in Q2. ALWAYS check the quadrant!
Correct! Nailed the quadrant check.
1\(r = \sqrt{(-\sqrt{3})^2+1^2}=\sqrt{3+1}=2\)
2\(\tan\theta=\dfrac{1}{-\sqrt{3}}\Rightarrow\) reference angle \(=\dfrac{\pi}{6}\)
3Q2 (x<0, y>0): \(\theta=\pi-\dfrac{\pi}{6}=\dfrac{5\pi}{6}\). Answer: \(\!\left(2,\dfrac{5\pi}{6}\right)\)
19
A vector \(\vec{v}\) has magnitude 6 and makes an angle of \(120Β°\) with the positive x-axis. Express \(\vec{v}\) in component form \(\langle v_x, v_y \rangle\).
π‘ Vector from Angle
\(\vec{v} = \langle r\cos\theta,\; r\sin\theta \rangle\)
This connects polar coordinates directly to vector components!
\(\vec{v} = \langle r\cos\theta,\; r\sin\theta \rangle\)
This connects polar coordinates directly to vector components!
Polar β Vector is the same formula as Polar β Rectangular. cos(120Β°) = βΒ½ because 120Β° is in Q2 (cosine is negative). sin(120Β°) = +β3/2 (sine is positive in Q2).
Correct! Linking polar and vectors like a pro.
1\(v_x = 6\cos(120Β°) = 6\cdot(-\tfrac{1}{2}) = -3\)
2\(v_y = 6\sin(120Β°) = 6\cdot\tfrac{\sqrt{3}}{2} = 3\sqrt{3}\)
3This bridges three units: trig values + CAST rule + vector components. All in one!
20
The rotation matrix \(R = \begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{pmatrix}\) rotates a vector by angle \(\theta\). Apply a \(90Β°\) rotation to \(\vec{v} = \langle 1, 0 \rangle\). What is the result?
π‘ Rotation Matrix
At \(\theta = 90Β°\): \(\cos 90Β°=0\), \(\sin 90Β°=1\)
So \(R = \begin{pmatrix}0&-1\\1&0\end{pmatrix}\). Multiply by \(\vec{v}\) as a column vector.
At \(\theta = 90Β°\): \(\cos 90Β°=0\), \(\sin 90Β°=1\)
So \(R = \begin{pmatrix}0&-1\\1&0\end{pmatrix}\). Multiply by \(\vec{v}\) as a column vector.
Rotating (1,0) by 90Β° counterclockwise β (0,1). This is intuitive: the x-axis pointing right rotates to the y-axis pointing up. Trust your geometry!
Perfect. You've connected it all β trig values, matrix multiplication, and vector geometry!
1\(R = \begin{pmatrix}0&-1\\1&0\end{pmatrix}\) at \(\theta=90Β°\)
2\(R\vec{v}=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}0(1)+(-1)(0)\\1(1)+0(0)\end{pmatrix}=\begin{pmatrix}0\\1\end{pmatrix}\)
3Geometrically: rotating the positive x-axis 90Β° counterclockwise gives the positive y-axis. Beautiful! π
π
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