Trigonometry

Trigonometry studies the relationships between angles and sides of triangles. Right-triangle definitions extend to all real angles via the unit circle, giving the six trigonometric functions, laws of sines and cosines, and the area formula.

● Right-triangle definitions

$\sin\theta=\frac{\text{opp}}{hyp},\quad\cos\theta=\frac{\text{adj}}{hyp},\quad\tan\theta=\frac{\text{opp}}{\text{adj}},$ $\csc\theta=\frac{1}{\sin\theta},\quad\sec\theta=\frac{1}{\cos\theta},\quad\cot\theta=\frac{1}{\tan\theta}.$

Memory aid for signs by quadrant (ASTC): Q1 All positive; Q2 Sin positive; Q3 Tan positive; Q4 Cos positive.

● Degrees, radians, and arc length

$180°=\pi\text{ rad}.\quad\text{Arc length: }s=r\theta\;(\theta\text{ in radians}).\quad\text{Sector area: }A=\tfrac{1}{2}r^2\theta.$

Degrees	Radians	$\sin$	$\cos$	$\tan$
0°	0	0	1	0
30°	$\pi/6$	$\frac{1}{2}$	$\frac{\sqrt3}{2}$	$\frac{1}{\sqrt3}$
45°	$\pi/4$	$\frac{\sqrt2}{2}$	$\frac{\sqrt2}{2}$	1
60°	$\pi/3$	$\frac{\sqrt3}{2}$	$\frac{1}{2}$	$\sqrt3$
90°	$\pi/2$	1	0	undef

● Law of Sines

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}.$

Use when: AAS, ASA, or SSA (ambiguous case — may give 0, 1, or 2 triangles).

SSA ambiguous case: given $a, b, A$ . If $a < b\sin A$ → no triangle; if $a=b\sin A$ → right triangle; if $b\sin A < a < b$ → two triangles; if $a\ge b$ → one triangle.

⊙ Explorer · The SSA ambiguous caseinteractive

side a: 6.0

side b: 8.0

angle A: 30°

height b sin A4.000

number of triangles2

triangle 1: B, C41.8°, 108.2°

triangle 2: B, C138.2°, 11.8°

The dashed circle has radius a and centre C. Each point where it meets the base ray is a possible vertex B. With a = 6, b = 8, A = 30° (Worked Example 1.3), b sin A = 4 < a < b, so the circle cuts the ray twice: two triangles. Slide a below 4 (no triangle), to exactly 4 (one right triangle), or to 8 or more (one triangle).

● Law of Cosines & Area

$c^2=a^2+b^2-2ab\cos C;\qquad\cos C=\frac{a^2+b^2-c^2}{2ab}.$ $\text{Area}=\tfrac{1}{2}ab\sin C.\quad\text{Heron's: }A=\sqrt{s(s-a)(s-b)(s-c)},\;s=\tfrac{a+b+c}{2}.$

Use Law of Cosines when: SAS (two sides + included angle) or SSS (three sides).

Worked Example 1.1 — Law of Sines AAS

Problem. In $\triangle ABC$ , $A=40°$ , $B=75°$ , $b=12$ . Find $a$ and the area.

$C=180-40-75=65°.\quad\dfrac{a}{\sin40°}=\dfrac{12}{\sin75°}.$

$a=\dfrac{12\sin40°}{\sin75°}=\dfrac{12(0.6428)}{0.9659}\approx7.99.$

Area $=\dfrac{1}{2}ab\sin C=\dfrac{1}{2}(7.99)(12)\sin65°\approx43.5.$

Worked Example 1.2 — Law of Cosines SSS

Problem. A triangle has sides $a=7, b=9, c=5$ . Find angle $B$ .

$\cos B=\dfrac{a^2+c^2-b^2}{2ac}=\dfrac{49+25-81}{2(7)(5)}=\dfrac{-7}{70}=-0.1.$

$B=\arccos(-0.1)\approx95.7°.$

Worked Example 1.3 — Ambiguous SSA

Problem. Given $a=6,\;b=8,\;A=30°$ . Find all possible triangles.

$b\sin A=8\sin30°=4.$ Since $b\sin A=4 < a=6 < b=8$ : two triangles possible.

$\sin B=\dfrac{b\sin A}{a}=\dfrac{8(0.5)}{6}=\dfrac{2}{3}.\quad B_1=\arcsin(2/3)\approx41.8°$ or $B_2=180°-41.8°=138.2°.$

Triangle 1: $B_1\approx41.8°,\;C_1\approx108.2°.$ Triangle 2: $B_2\approx138.2°,\;C_2\approx11.8°.$ Both valid since all angles positive and sum to 180°.

Worked Example 1.4 — Arc length and sector area

Problem. A sector has radius 10 cm and central angle 120°. Find the arc length and area.

$\theta=120°=\dfrac{2\pi}{3}$ rad. Arc length: $s=r\theta=10\cdot\dfrac{2\pi}{3}=\dfrac{20\pi}{3}\approx20.9$ cm.

Area: $A=\dfrac{1}{2}r^2\theta=\dfrac{1}{2}(100)\dfrac{2\pi}{3}=\dfrac{100\pi}{3}\approx104.7\text{ cm}^2.$

Oblique

\triangle ABC

illustrating side-angle relationships for the Law of Sines and Law of Cosines.

▲ Pre-Calculus Traps & Commandments — Trigonometry & Triangle Laws

Radian vs. Degree Mode: Always verify calculator mode before computing. Arc length $s = r\theta$ and sector area $A = \frac{1}{2}r^2\theta$ strictly require $\theta$ in radians.
Ambiguous SSA Case: When using the Law of Sines with SSA, check the altitude $h = b\sin A$ . If $h < a < b$ , two distinct valid triangles exist: $B_1 = \arcsin(b\sin A / a)$ (acute) and $B_2 = 180° - B_1$ (obtuse), provided $A + B_2 < 180°$ .
Law of Cosines Angle Selection: In SSS triangles, solve for the largest angle first (opposite the longest side) using the Law of Cosines. Because $\arccos$ outputs values across $[0°, 180°]$ , it automatically identifies obtuse angles, avoiding the ambiguity of $\arcsin$ .
Area Formulas: Use $K = \frac{1}{2}ab\sin C$ when two sides and the included angle are known (SAS). Use Heron's formula $K = \sqrt{s(s-a)(s-b)(s-c)}$ with semi-perimeter $s = (a+b+c)/2$ when all three sides are known (SSS).

In

\triangle ABC

,

a = 8

,

b = 11

, and

A = 35°

. Determine the number of possible triangles and find the measure of angle

B

for all valid cases to the nearest tenth of a degree.

Calculate the altitude $h = b\sin A = 11\sin 35° \approx 11(0.5736) = 6.31$ .

Since $h < a < b$ ( $6.31 < 8 < 11$ ), there are two possible triangles.

By the Law of Sines: $\sin B = \frac{b\sin A}{a} = \frac{6.31}{8} \approx 0.7887.$

First solution (acute): $B_1 = \arcsin(0.7887) \approx 52.1°$ .

Second solution (obtuse): $B_2 = 180° - 52.1° = 127.9°$ .

Check validity: $A + B_2 = 35° + 127.9° = 162.9° < 180°$ , which allows $C_2 = 17.1° > 0$ . Both $B_1 \approx 52.1°$ and $B_2 \approx 127.9°$ are valid.

A surveyor measures distances from an observation point

C

to two landmarks

A

and

B

:

AC = 140\text{ m}

and

BC = 210\text{ m}

, with included angle

\angle ACB = 78°

. Find the distance

AB

between the landmarks to the nearest tenth of a meter, and compute the area of

\triangle ABC

to the nearest square meter.

Apply the Law of Cosines to find side $c = AB$ :

c^2 = a^2 + b^2 - 2ab\cos C = 210^2 + 140^2 - 2(210)(140)\cos 78°.

c^2 = 44100 + 19600 - 58800(0.2079) = 63700 - 12225 = 51475.

c = \sqrt{51475} \approx 226.9\text{ m}.

Compute the area:

\text{Area} = \frac{1}{2}ab\sin C = \frac{1}{2}(210)(140)\sin 78° = 14700(0.9781) \approx 14379\text{ m}^2.

📝 Chapter Quiz

Questions Level

2

Unit Circle & Trigonometric Identities

The unit circle extends trigonometry to all real numbers. Trigonometric identities are equations true for all values of the variable — they allow simplification, equation solving, and calculus manipulations.

● Unit circle and periodicity

For angle $\theta$ in standard position, terminal point on unit circle: $(\cos\theta,\sin\theta).$

\sin(\theta+2\pi)=\sin\theta,\quad\cos(\theta+2\pi)=\cos\theta,\quad\tan(\theta+\pi)=\tan\theta.

Even/Odd: $\cos(-\theta)=\cos\theta$ (even); $\sin(-\theta)=-\sin\theta$ (odd); $\tan(-\theta)=-\tan\theta$ (odd).

● Fundamental identities

$\sin^2\theta+\cos^2\theta=1,\quad1+\tan^2\theta=\sec^2\theta,\quad1+\cot^2\theta=\csc^2\theta.$ $\tan\theta=\frac{\sin\theta}{\cos\theta},\quad\cot\theta=\frac{\cos\theta}{\sin\theta}.$

⊙ Explorer · The unit circleinteractive

angle θ: 40

cos θ0.766

sin θ0.643

tan θ0.839

quadrantI

The point $P=(\cos\theta,\sin\theta)$ always lies on $x^2+y^2=1$ — that's exactly the Pythagorean identity $\sin^2\theta+\cos^2\theta=1$ above.

● Sum & difference formulas

$\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B,$ $\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B,$ $\tan(A\pm B)=\frac{\tan A\pm\tan B}{1\mp\tan A\tan B}.$

● Double-angle & half-angle formulas

$\sin2\theta=2\sin\theta\cos\theta,$ $\cos2\theta=\cos^2\theta-\sin^2\theta=1-2\sin^2\theta=2\cos^2\theta-1,$ $\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}.$ $\sin\frac{\theta}{2}=\pm\sqrt{\frac{1-\cos\theta}{2}},\quad\cos\frac{\theta}{2}=\pm\sqrt{\frac{1+\cos\theta}{2}},\quad\tan\frac{\theta}{2}=\frac{1-\cos\theta}{\sin\theta}.$

● Product-to-sum & sum-to-product formulas

$\sin A\cos B=\tfrac{1}{2}[\sin(A+B)+\sin(A-B)],$ $\cos A\cos B=\tfrac{1}{2}[\cos(A-B)+\cos(A+B)],$ $\sin A+\sin B=2\sin\!\left(\frac{A+B}{2}\right)\cos\!\left(\frac{A-B}{2}\right).$

Worked Example 2.1 — Exact value

Problem. Find the exact value of $\cos(75°)$ .

$\cos75°=\cos(45°+30°)=\cos45°\cos30°-\sin45°\sin30°$

$=\dfrac{\sqrt2}{2}\cdot\dfrac{\sqrt3}{2}-\dfrac{\sqrt2}{2}\cdot\dfrac{1}{2}=\dfrac{\sqrt6-\sqrt2}{4}.$

Worked Example 2.2 — Identity proof

Problem. Prove: $\dfrac{\sin2\theta}{1+\cos2\theta}=\tan\theta.$

LHS: $\dfrac{2\sin\theta\cos\theta}{1+(2\cos^2\theta-1)}=\dfrac{2\sin\theta\cos\theta}{2\cos^2\theta}=\dfrac{\sin\theta}{\cos\theta}=\tan\theta=$ RHS. $\checkmark$

Worked Example 2.3 — Solve trig equation

Problem. Solve $\cos2\theta+\cos\theta=0$ on $[0°,360°)$ .

Replace $\cos2\theta=2\cos^2\theta-1$ : $2\cos^2\theta-1+\cos\theta=0.$

Factor: $(2\cos\theta-1)(\cos\theta+1)=0.$

$\cos\theta=\frac{1}{2}\Rightarrow\theta=60°,300°.\quad\cos\theta=-1\Rightarrow\theta=180°.$

Worked Example 2.4 — Given value, find related

Problem. If $\sin\theta=\frac{5}{13}$ and $\theta$ is in Q2, find $\sin2\theta$ and $\cos\frac{\theta}{2}.$

Q2: $\cos\theta=-\frac{12}{13}.\quad\sin2\theta=2\sin\theta\cos\theta=2\!\left(\frac{5}{13}\right)\!\left(-\frac{12}{13}\right)=-\frac{120}{169}.$

$\cos\frac{\theta}{2}=+\sqrt{\frac{1+\cos\theta}{2}}=\sqrt{\frac{1-12/13}{2}}=\sqrt{\frac{1/13}{2}}=\sqrt{\frac{1}{26}}=\frac{1}{\sqrt{26}}.$ (Q2 $\Rightarrow\theta/2\in$ Q1, so cosine positive.)

Terminal point on the unit circle at

\theta = 150°

(

5\pi/6

rad), showing coordinates

(-\sqrt{3}/2, 1/2)

with reference triangle in Quadrant II.

▲ Pre-Calculus Traps & Commandments — Unit Circle & Trigonometric Identities

ASTC Quadrant Signs: Remember All Students Take Calculus: Q1 (All positive), Q2 (Sine positive, Cos/Tan negative), Q3 (Tangent positive, Sin/Cos negative), Q4 (Cosine positive, Sin/Tan negative).
Pythagorean Identity Sign Determination: When solving $\cos\theta = \pm\sqrt{1 - \sin^2\theta}$ , actively select the sign based on the quadrant of $\theta$ . Never leave $\pm$ in an evaluated expression.
Never Divide by a Variable Trig Function: In equations like $\sin(2\theta) = \sin\theta$ , never divide both sides by $\sin\theta$ . Doing so eliminates roots where $\sin\theta = 0$ ! Instead, expand $2\sin\theta\cos\theta - \sin\theta = 0$ and factor out $\sin\theta$ .
Double-Angle Cosine Selection: $\cos(2\theta)$ has three interchangeable identities: $\cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$ . Choose the identity that matches the other trigonometric functions in the equation to form a solvable single-variable quadratic.

Solve the trigonometric equation

\cos(2\theta) + 3\sin\theta = 2

for all solutions

\theta \in [0, 2\pi)

.

Substitute $\cos(2\theta) = 1 - 2\sin^2\theta$ to express the entire equation in terms of $\sin\theta$ :

(1 - 2\sin^2\theta) + 3\sin\theta = 2.

-2\sin^2\theta + 3\sin\theta - 1 = 0 \implies 2\sin^2\theta - 3\sin\theta + 1 = 0.

Factor the quadratic in $\sin\theta$ :

(2\sin\theta - 1)(\sin\theta - 1) = 0.

Case 1: $2\sin\theta - 1 = 0 \implies \sin\theta = \frac{1}{2}$ . On $[0, 2\pi)$ , $\theta = \frac{\pi}{6}$ or $\theta = \frac{5\pi}{6}$ .

Case 2: $\sin\theta - 1 = 0 \implies \sin\theta = 1$ . On $[0, 2\pi)$ , $\theta = \frac{\pi}{2}$ .

Combining all solutions: $\theta \in \left\{\frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}\right\}$ .

Verify the trigonometric identity:

\frac{\sin(2x)}{\sin x} - \frac{\cos(2x)}{\cos x} = \sec x.

Expand the double-angle expressions on the left-hand side:

\text{LHS} = \frac{2\sin x\cos x}{\sin x} - \frac{2\cos^2 x - 1}{\cos x}.

Cancel $\sin x$ in the first term and combine over a common denominator of $\cos x$ :

= 2\cos x - \frac{2\cos^2 x - 1}{\cos x} = \frac{2\cos^2 x - (2\cos^2 x - 1)}{\cos x}.

= \frac{2\cos^2 x - 2\cos^2 x + 1}{\cos x} = \frac{1}{\cos x} = \sec x = \text{RHS}. \quad \checkmark

📝 Chapter Quiz

Questions Level

3

Conic Sections

Conic sections are the curves formed by intersecting a double cone with a plane. Each conic has a focus–directrix geometric definition and appears in physics (planetary orbits, optics, projectile paths).

● Standard forms (centred at origin)

Conic	Standard equation	Key features
Circle	$x^2+y^2=r^2$	Centre $(0,0)$ , radius $r$
Parabola (opens up)	$x^2=4py$	Focus $(0,p)$ , directrix $y=-p$
Ellipse ( $a>b>0$ )	$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$	Vertices $(\pm a,0)$ , foci $(\pm c,0)$ , $c^2=a^2-b^2$
Hyperbola (horiz.)	$\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$	Vertices $(\pm a,0)$ , foci $(\pm c,0)$ , $c^2=a^2+b^2$ , asymptotes $y=\pm\frac{b}{a}x$

● Shifted conics (centre (h,k))

Replace $x$ with $(x-h)$ and $y$ with $(y-k)$ in the standard form:

\text{Circle: }(x-h)^2+(y-k)^2=r^2.

\text{Ellipse: }\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1.

\text{Parabola: }(x-h)^2=4p(y-k)\;(\text{opens up if }p>0).

To find centre: complete the square in both $x$ and $y$ .

● Eccentricity

$e=\frac{c}{a}.$

Circle: $e=0.$ Ellipse: $0<e<1.$ Parabola: $e=1.$ Hyperbola: $e>1.$ Higher eccentricity → more "stretched."

⊙ Explorer · Eccentricity — one family, four shapesinteractive

eccentricity e: 0

shapecircle

c (focus distance)0.000

Same $a=3$ throughout. $e=0$ is a circle, $0<e<1$ an ellipse, $e=1$ a parabola, $e>1$ a hyperbola — exactly the definition above, live.

Worked Example 3.1 — Circle from general form

Problem. Write $x^2+y^2-6x+8y+9=0$ in standard form. Find centre, radius.

Group: $(x^2-6x)+(y^2+8y)=-9.$ Complete the square:

$(x^2-6x+9)+(y^2+8y+16)=-9+9+16=16.$

$(x-3)^2+(y+4)^2=16.$ Centre $(3,-4)$ , radius 4.

Worked Example 3.2 — Ellipse

Problem. Find the foci and eccentricity of $\dfrac{(x-1)^2}{25}+\dfrac{(y+2)^2}{16}=1.$

$a^2=25,\;b^2=16,\;c^2=25-16=9,\;c=3.$ Centre $(1,-2).$

Foci: $(1\pm3,\,-2)=(4,-2)$ and $(-2,-2).$ $e=c/a=3/5.$

Worked Example 3.3 — Parabola

Problem. Find the focus and directrix of $(y-3)^2=-12(x+1).$

Form: $(y-k)^2=4p(x-h)$ with $4p=-12\Rightarrow p=-3.$ Vertex $(-1,3).$

Opens left (since $p<0$ ). Focus: $(-1+(-3),3)=(-4,3).$ Directrix: $x=-1-(-3)=2.$

Worked Example 3.4 — Hyperbola

Problem. Write $4x^2-9y^2-8x+36y-68=0$ in standard form and identify foci.

$4(x^2-2x)-9(y^2-4y)=68.$ Complete squares: $4(x-1)^2-4-9(y-2)^2+36=68.$

$4(x-1)^2-9(y-2)^2=36.$ Divide by 36: $\dfrac{(x-1)^2}{9}-\dfrac{(y-2)^2}{4}=1.$

$a^2=9,\;b^2=4,\;c^2=13,\;c=\sqrt{13}.$ Centre $(1,2).$ Foci: $(1\pm\sqrt{13},\,2).$

Horizontal hyperbola centered at

(1, 2)

with vertices at

(-2, 2)

and

(4, 2)

, foci at

(1 \pm \sqrt{13}, 2)

, and asymptotes

y - 2 = \pm\frac{2}{3}(x - 1)

.

▲ Pre-Calculus Traps & Commandments — Conic Sections

Ellipse vs. Hyperbola Focal Distances: For an ellipse, $c^2 = a^2 - b^2$ (foci are inside, $c < a$ ). For a hyperbola, $c^2 = a^2 + b^2$ (foci are outside the vertices, $c > a$ ). Never mix up these signs!
Identifying Conics from General Quadratic: In $Ax^2 + Cy^2 + Dx + Ey + F = 0$ : Circle if $A = C$ ; Ellipse if $A \neq C$ but $AC > 0$ (same sign); Hyperbola if $AC < 0$ (opposite signs); Parabola if either $A = 0$ or $C = 0$ (one variable squared).
Parabola Directrix vs. Focus: In standard forms $(y-k)^2 = 4p(x-h)$ and $(x-h)^2 = 4p(y-k)$ , the focus and directrix lie at equal distance $|p|$ from the vertex. The parabola wraps around the focus and curves away from the directrix.
Hyperbola Asymptotes: Find asymptotes directly by replacing 1 with 0 in standard form: $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 0 \implies y - k = \pm\frac{b}{a}(x - h)$ . For vertical hyperbolas, the slopes are $\pm\frac{a}{b}$ .

Find the vertex, focus, and directrix of the parabola given by the general equation

y^2 + 8x - 4y + 28 = 0

.

Complete the square on the $y$ -terms:

y^2 - 4y = -8x - 28 \implies (y^2 - 4y + 4) = -8x - 28 + 4.

(y - 2)^2 = -8x - 24 = -8(x + 3).

This matches $(y - k)^2 = 4p(x - h)$ with:

Vertex: $(h, k) = (-3, 2)$
Focal parameter: $4p = -8 \implies p = -2$ (opens horizontally to the left)
Focus: $(h + p, k) = (-3 + (-2), 2) = (-5, 2)$
Directrix: $x = h - p = -3 - (-2) \implies x = -1$

An ellipse has vertices at

(1, 6)

and

(1, -4)

, and a focus at

(1, 4)

. Determine its standard equation and compute its eccentricity.

The vertices share $x = 1$ , so the major axis is vertical.

Center is the midpoint of the vertices: $(h, k) = \left(1, \frac{6 + (-4)}{2}\right) = (1, 1)$ .

Semi-major axis: $a = 6 - 1 = 5$ (so $a^2 = 25$ ).

Focal distance from center to $(1, 4)$ : $c = 4 - 1 = 3$ .

For an ellipse, $b^2 = a^2 - c^2 = 25 - 9 = 16 \implies b = 4$ .

Standard equation: $\frac{(x - 1)^2}{16} + \frac{(y - 1)^2}{25} = 1.$

Eccentricity: $e = \frac{c}{a} = \frac{3}{5} = 0.6.$

📝 Chapter Quiz

Questions Level

4

Vectors & Parametric Equations

A vector has both magnitude and direction. Vectors model force, velocity, and displacement. Parametric equations describe motion and curves by expressing both coordinates as functions of a parameter $t$ .

● Vector fundamentals

For $\mathbf{u}=\langle u_1,u_2\rangle$ and $\mathbf{v}=\langle v_1,v_2\rangle$ , scalar $k$ :

\mathbf{u}+\mathbf{v}=\langle u_1+v_1,u_2+v_2\rangle,\quad k\mathbf{u}=\langle ku_1,ku_2\rangle,

|\mathbf{u}|=\sqrt{u_1^2+u_2^2},\quad\hat{\mathbf{u}}=\frac{\mathbf{u}}{|\mathbf{u}|}\quad(\text{unit vector}).

Component form from magnitude/angle: $\mathbf{v}=\langle|\mathbf{v}|\cos\theta,\;|\mathbf{v}|\sin\theta\rangle.$

● Dot product & angle

$\mathbf{u}\cdot\mathbf{v}=u_1v_1+u_2v_2=|\mathbf{u}||\mathbf{v}|\cos\theta.$

Perpendicular: $\mathbf{u}\cdot\mathbf{v}=0.\quad$ Parallel: $\mathbf{v}=k\mathbf{u}.$

Vector projection of $\mathbf{u}$ onto $\mathbf{v}$ :

\text{proj}_{\mathbf{v}}\mathbf{u}=\frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{v}|^2}\,\mathbf{v},\qquad\text{scalar projection}=\frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{v}|}.

⊙ Explorer · Dot product and projectioninteractive

u₁: 6

u₂: 2

v₁: 3

v₂: 4

u · v26

angle θ34.7°

proj_v u⟨3.120, 4.160⟩

scalar projection5.200

Defaults to Worked Example 4.2: u = ⟨6, 2⟩, v = ⟨3, 4⟩, u · v = 26, |v|² = 25, so proj_v u = (26/25)⟨3, 4⟩ = ⟨78/25, 104/25⟩. The dashed segment from u to its projection is always perpendicular to v. Make u · v = 0 and the projection collapses to the zero vector; make it negative and the projection points against v (θ > 90°).

● Parametric equations & eliminating the parameter

Curve: $x=f(t),\;y=g(t).$

Method 1: solve for $t$ from one equation, substitute into the other.

Method 2 (trig): if $x=a\cos t$ and $y=b\sin t$ , use $\cos^2t+\sin^2t=1\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1.$

Projectile motion: $x=v_0\cos\alpha\cdot t,\quad y=v_0\sin\alpha\cdot t-\tfrac{1}{2}g t^2.$

Worked Example 4.1 — Dot product and angle

Problem. Find the angle between $\mathbf{u}=\langle4,-3\rangle$ and $\mathbf{v}=\langle2,5\rangle$ .

$\mathbf{u}\cdot\mathbf{v}=4(2)+(-3)(5)=8-15=-7.$

$|\mathbf{u}|=\sqrt{16+9}=5,\quad|\mathbf{v}|=\sqrt{4+25}=\sqrt{29}.$

$\cos\theta=\dfrac{-7}{5\sqrt{29}}\approx\dfrac{-7}{26.93}\approx-0.260.\quad\theta\approx105.1°.$

Worked Example 4.2 — Projection

Problem. Project $\mathbf{u}=\langle6,2\rangle$ onto $\mathbf{v}=\langle3,4\rangle$ .

$\mathbf{u}\cdot\mathbf{v}=18+8=26.\quad|\mathbf{v}|^2=9+16=25.$

$\text{proj}_{\mathbf{v}}\mathbf{u}=\dfrac{26}{25}\langle3,4\rangle=\left\langle\dfrac{78}{25},\dfrac{104}{25}\right\rangle.$

Worked Example 4.3 — Eliminate parameter

Problem. Convert $x=3\cos t,\;y=5\sin t,\;t\in[0,2\pi]$ to Cartesian form.

$\dfrac{x}{3}=\cos t,\quad\dfrac{y}{5}=\sin t.\quad\cos^2t+\sin^2t=1\Rightarrow\dfrac{x^2}{9}+\dfrac{y^2}{25}=1.$

An ellipse with semi-axes $a=5$ (vertical), $b=3$ (horizontal).

Worked Example 4.4 — Projectile

Problem. A ball is kicked at 20 m/s at 60° to the horizontal. Find: (a) max height, (b) range, (c) time of flight.

$v_x=20\cos60°=10\text{ m/s},\quad v_y=20\sin60°=10\sqrt3\text{ m/s}.$

(a) At max height $v_y=0$ : $t_{\text{top}}=10\sqrt3/9.8\approx1.77\text{ s}.\quad H=10\sqrt3(1.77)-\frac12(9.8)(1.77)^2\approx15.3\text{ m}.$

(b) Time of flight: $T=2t_{\text{top}}\approx3.53\text{ s}.\quad\text{Range}=10(3.53)\approx35.3\text{ m}.$

Parametric trajectory traced counterclockwise by

x(t) = 3\cos t

,

y(t) = 5\sin t

for

t \in [0, 2\pi]

, forming the ellipse

x^2/9 + y^2/25 = 1

.

▲ Pre-Calculus Traps & Commandments — Vectors & Parametric Equations

Dot Product Output: The dot product $\mathbf{u}\cdot\mathbf{v} = u_1 v_1 + u_2 v_2$ yields a scalar (a single number), never a vector. Writing component brackets around the dot product is a major conceptual error.
Vector vs. Scalar Projection: Scalar projection is the signed magnitude $\text{comp}_{\mathbf{v}}\mathbf{u} = \frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{v}|}$ . Vector projection incorporates direction: $\text{proj}_{\mathbf{v}}\mathbf{u} = \frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{v}|^2}\,\mathbf{v}$ . Notice that the denominator is $|\mathbf{v}|^2$ in the vector projection.
Parametric Domain Restrictions: When eliminating $t$ to find a Cartesian equation, always record the natural domain and range restricted by the parameter. For instance, $x = 2\cos t$ limits $x$ to $[-2, 2]$ .
Orthogonal vs. Parallel Conditions: Two nonzero vectors are orthogonal (perpendicular) if and only if $\mathbf{u}\cdot\mathbf{v} = 0$ . They are parallel if and only if $\mathbf{u} = k\mathbf{v}$ (their components are strictly proportional).

Given vectors

\mathbf{a} = 2\mathbf{i} - 5\mathbf{j}

and

\mathbf{b} = -3\mathbf{i} + 4\mathbf{j}

, find the scalar projection of

\mathbf{a}

onto

\mathbf{b}

and the angle

\theta

between them to the nearest tenth of a degree.

Calculate the dot product:

\mathbf{a}\cdot\mathbf{b} = (2)(-3) + (-5)(4) = -6 - 20 = -26.

Calculate the magnitudes:

|\mathbf{a}| = \sqrt{2^2 + (-5)^2} = \sqrt{4 + 25} = \sqrt{29} \approx 5.385,

|\mathbf{b}| = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5.

Scalar projection of $\mathbf{a}$ onto $\mathbf{b}$ :

\text{comp}_{\mathbf{b}}\mathbf{a} = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|} = \frac{-26}{5} = -5.2.

Angle $\theta$ between vectors:

\cos\theta = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|} = \frac{-26}{5\sqrt{29}} \approx -0.9656 \implies \theta = \arccos(-0.9656) \approx 164.9°.

Eliminate the parameter to determine the Cartesian equation of the curve

x = 1 + 2\sec t

,

y = -2 + 3\tan t

. State the shape, center, and vertices.

Isolate the trigonometric terms:

\sec t = \frac{x - 1}{2}, \qquad \tan t = \frac{y + 2}{3}.

Apply the fundamental Pythagorean identity $\sec^2 t - \tan^2 t = 1$ :

\left(\frac{x - 1}{2}\right)^2 - \left(\frac{y + 2}{3}\right)^2 = 1 \implies \frac{(x - 1)^2}{4} - \frac{(y + 2)^2}{9} = 1.

This is a horizontal hyperbola with:

Center: $(h, k) = (1, -2)$
Transverse semi-axis: $a = 2$
Vertices: $(h \pm a, k) = (1 \pm 2, -2) = (-1, -2)$ and $(3, -2)$

📝 Chapter Quiz

Questions Level

5

Polar Coordinates

Polar coordinates $(r,\theta)$ locate a point by distance from the origin and angle from the positive $x$ -axis. Many curves that are complicated in Cartesian form have elegant polar equations.

● Conversion between polar and Cartesian

$x=r\cos\theta,\quad y=r\sin\theta,\quad r^2=x^2+y^2,\quad\theta=\arctan\!\left(\frac{y}{x}\right)\text{ (adjust for quadrant)}.$

Note: A point has infinitely many polar representations: $(r,\theta)\equiv(r,\theta+2\pi k)\equiv(-r,\theta+\pi).$

● Common polar curves

Name	Equation	Description
Circle at origin	$r=a$	Radius $\|a\|$
Circle through origin	$r=2a\cos\theta$	Centre $(a,0)$ , radius $\|a\|$
Cardioid	$r=a(1+\cos\theta)$	Heart-shaped; passes through origin
Limaçon	$r=a+b\cos\theta$	Inner loop if $a<b$ ; no loop if $a>b$
Rose ( $n$ odd)	$r=a\cos(n\theta)$	$n$ petals
Rose ( $n$ even)	$r=a\cos(n\theta)$	$2n$ petals
Lemniscate	$r^2=a^2\cos2\theta$	Figure-eight shape
Spiral	$r=a\theta$	Archimedean spiral

⊙ Explorer · Limaçons r = a + b cos θinteractive

a: 2.0

b: 2.0

trace to θ: 360°

shapecardioid (a = b)

max r (θ = 0)4.00

r at θ = π0.00

current point (r, θ)(4.00, 360°)

Defaults to Worked Example 5.3, r = 2 + 2 cos θ: a cardioid with maximum r = 4 at θ = 0 and r = 0 at θ = π. Make a < b to see the inner loop (r turns negative), a > b for no loop, and b = 0 for a circle. Replacing θ with −θ leaves cos θ unchanged, so every curve here is symmetric about the polar axis.

● Symmetry in polar graphs

About the polar axis (x-axis): replace $\theta$ with $-\theta$ — equation unchanged.

About $\theta=\pi/2$ (y-axis): replace $\theta$ with $\pi-\theta$ — equation unchanged.

About the pole (origin): replace $r$ with $-r$ or $\theta$ with $\theta+\pi$ — equation unchanged.

Worked Example 5.1 — Converting to Cartesian

Problem. Convert $r=4\cos\theta$ to Cartesian and identify the curve.

Multiply both sides by $r$ : $r^2=4r\cos\theta\Rightarrow x^2+y^2=4x.$

Complete the square: $(x-2)^2+y^2=4.$ Circle with centre $(2,0)$ , radius 2.

Worked Example 5.2 — Polar to Cartesian equation

Problem. Convert $r=\dfrac{3}{2-\cos\theta}$ and identify the conic.

$r(2-\cos\theta)=3\Rightarrow2r-r\cos\theta=3\Rightarrow2\sqrt{x^2+y^2}-x=3.$

$2\sqrt{x^2+y^2}=x+3\Rightarrow4(x^2+y^2)=(x+3)^2=x^2+6x+9.$

$3x^2-6x+4y^2=9.$ Divide: $\dfrac{(x-1)^2}{4}+\dfrac{y^2}{3}=1.$ Ellipse.

Worked Example 5.3 — Graphing a cardioid

Problem. Sketch $r=2+2\cos\theta$ . Find max $r$ , min $r$ , and symmetry.

Max: $\theta=0\Rightarrow r=4.$ Min: $\theta=\pi\Rightarrow r=0.$ Midline: $\theta=\pi/2$ or $3\pi/2\Rightarrow r=2.$

Symmetric about the polar axis (replacing $\theta$ with $-\theta$ gives same equation). This is a cardioid pointing right.

Polar circle

r = 4\cos\theta

, equivalent in rectangular coordinates to

(x - 2)^2 + y^2 = 4

, with radius

r = 2

passing through the pole

(0, 0)

.

▲ Pre-Calculus Traps & Commandments — Polar Coordinates

Multiple Polar Representations: Unlike Cartesian coordinates, every polar point $(r, \theta)$ possesses infinitely many equivalent coordinates: $(r, \theta + 2\pi k)$ and $(-r, \theta + (2k+1)\pi)$ . When comparing points or finding intersections, always test both positive and negative radius forms!
Quadrant Adjustment for Angle $\theta$ : The standard conversion $\theta = \arctan(y/x)$ only returns principal angles in $(-\pi/2, \pi/2)$ (Quadrants I and IV). When converting points with $x < 0$ (Quadrants II and III), you must add $\pi$ (or $180°$ ) to place the angle in the correct quadrant.
Interpreting Negative Radius $r < 0$ : To plot a point with $r < 0$ , locate the terminal ray at angle $\theta$ , then travel backward through the origin (pole) along the opposite ray by distance $|r|$ .
Rose Curve Petal Counts: For curves $r = a\cos(n\theta)$ and $r = a\sin(n\theta)$ : if $n$ is odd, the graph contains exactly $n$ petals (traced once over $[0, \pi)$ ); if $n$ is even, the graph contains $2n$ petals (traced over $[0, 2\pi)$ ).

Convert the Cartesian equation

(x^2 + y^2)^2 = 4(x^2 - y^2)

into polar form and identify the name and key features of the resulting curve.

Substitute standard Cartesian-to-polar relations $x^2 + y^2 = r^2$ , $x = r\cos\theta$ , and $y = r\sin\theta$ :

(r^2)^2 = 4(r^2\cos^2\theta - r^2\sin^2\theta) \implies r^4 = 4r^2(\cos^2\theta - \sin^2\theta).

Apply the double-angle identity $\cos^2\theta - \sin^2\theta = \cos(2\theta)$ :

r^4 = 4r^2\cos(2\theta).

Divide through by $r^2$ (for $r \neq 0$ ):

r^2 = 4\cos(2\theta).

This is a lemniscate (figure-eight curve) centered at the pole with loops oriented along the polar axis and maximum reach $r = 2$ at $\theta = 0$ and $\theta = \pi$ .

Find all distinct points of intersection between the two polar curves

r = 2\sin\theta

and

r = 2\cos\theta

on

0 \le \theta < 2\pi

.

Equate the two equations:

2\sin\theta = 2\cos\theta \implies \tan\theta = 1.

On $[0, 2\pi)$ , $\theta = \frac{\pi}{4}$ or $\theta = \frac{5\pi}{4}$ .

For $\theta = \frac{\pi}{4}$ : $r = 2\sin\left(\frac{\pi}{4}\right) = \sqrt{2}$ , giving point $\left(\sqrt{2}, \frac{\pi}{4}\right)$ .
For $\theta = \frac{5\pi}{4}$ : $r = 2\sin\left(\frac{5\pi}{4}\right) = -\sqrt{2}$ . However, $\left(-\sqrt{2}, \frac{5\pi}{4}\right)$ represents the identical geometric point as $\left(\sqrt{2}, \frac{\pi}{4}\right)$ .

Next, check the pole (origin):

$r = 2\sin\theta = 0$ when $\theta = 0$ or $\pi$ .
$r = 2\cos\theta = 0$ when $\theta = \frac{\pi}{2}$ or $\frac{3\pi}{2}$ .

Because both curves pass through the origin, the pole $(0, 0)$ is a valid intersection point.

Therefore, the distinct intersection points are the pole $(0, 0)$ and $\left(\sqrt{2}, \frac{\pi}{4}\right)$ .

📝 Chapter Quiz

Questions Level

6

Exponential & Logarithmic Functions

Exponential functions model multiplicative change — growth and decay. Logarithms are their inverses and appear naturally in measurement scales (Richter, decibel, pH) and in solving exponential equations.

● Properties of exponential functions

f(x)=ab^x

Domain: $\mathbb{R}$ . Range: $(0,\infty)$ when $a>0.$ Horizontal asymptote: $y=0.$

b^m\cdot b^n=b^{m+n},\quad\frac{b^m}{b^n}=b^{m-n},\quad(b^m)^n=b^{mn},\quad b^0=1,\quad b^{-n}=\frac{1}{b^n}.

Natural exponential: $f(x)=e^x,\;e\approx2.71828.$

● Logarithm properties (

\log_b x=y\iff b^y=x

)

$\log_b(MN)=\log_b M+\log_b N,\quad\log_b\!\left(\frac{M}{N}\right)=\log_b M-\log_b N,$ $\log_b(M^p)=p\log_b M,\quad\log_b x=\frac{\log x}{\log b}=\frac{\ln x}{\ln b}.$

$\log_b 1=0,\;\log_b b=1,\;b^{\log_b x}=x,\;\log_b(b^x)=x.$

● Growth and decay models

$P(t)=P_0\,b^t,\quad P(t)=P_0\,e^{rt}\quad(r>0:\text{growth};\;r<0:\text{decay}).$ $\text{Half-life: }A(t)=A_0\!\left(\tfrac{1}{2}\right)^{t/h}.\quad\text{Compound: }A=P\!\left(1+\frac{r}{n}\right)^{nt}.$

Logistic growth: $P(t)=\dfrac{K}{1+Ae^{-bt}}$ where $K$ is the carrying capacity.

⊙ Explorer · Exponential growth & decayinteractive

rate r: 8

modelgrowth

P(10) / P₀2.226

doubling time8.66

$P(t)=P_0e^{rt}$ . Same shape as Worked Example 6.3 (half-life) when $r<0$ , and 6.4 (compound growth) when $r>0$ .

● Solving exponential and logarithmic equations

Exponential: (1) isolate base power; (2) apply $\ln$ both sides; (3) solve.

Logarithmic: (1) combine using log properties; (2) exponentiate; (3) check domain.

Warning: always verify solutions — $\log$ of a negative number is undefined.

Worked Example 6.1 — Exponential equation

Problem. Solve $5^{x+1}=3^{2x-1}$ . Give exact and decimal answers.

Take $\ln$ : $(x+1)\ln5=(2x-1)\ln3.$

$x\ln5+\ln5=2x\ln3-\ln3.\quad x(\ln5-2\ln3)=-\ln3-\ln5.$

$x=\dfrac{-\ln3-\ln5}{\ln5-2\ln3}=\dfrac{-\ln15}{\ln5-\ln9}=\dfrac{\ln15}{\ln9-\ln5}=\dfrac{\ln15}{\ln(9/5)}\approx\dfrac{2.708}{0.588}\approx4.61.$

Worked Example 6.2 — Logarithmic equation

Problem. Solve $\log(x+4)-\log(x-1)=1.$

$\log\!\left(\dfrac{x+4}{x-1}\right)=1\Rightarrow\dfrac{x+4}{x-1}=10\Rightarrow x+4=10x-10\Rightarrow9x=14\Rightarrow x=\dfrac{14}{9}.$

Check: $x=14/9>1$ ✓, and $x+4>0$ ✓.

Worked Example 6.3 — Half-life

Problem. Strontium-90 has a half-life of 28.8 years. How long until only 15% of an initial sample remains?

$A(t)=A_0(0.5)^{t/28.8}.$ Set $A=0.15\,A_0:\quad0.5^{t/28.8}=0.15.$

$\dfrac{t}{28.8}\ln0.5=\ln0.15\Rightarrow t=28.8\cdot\dfrac{\ln0.15}{\ln0.5}=28.8\cdot\dfrac{-1.897}{-0.693}\approx78.8\text{ years}.$

Worked Example 6.4 — Compound interest

Problem. \$5000 is invested at 6% p.a. compounded monthly. Find the amount after 10 years, and compare with continuous compounding.

Monthly: $A=5000\!\left(1+\dfrac{0.06}{12}\right)^{120}=5000(1.005)^{120}\approx5000(1.8194)\approx\$9097.$

Continuous: $A=5000\,e^{0.06\times10}=5000\,e^{0.6}\approx5000(1.8221)\approx\$9110.$ Difference: \$13 over 10 years.

Mutual reflection of

y = e^x

and

y = \ln x

across the identity line

y = x

. Note the horizontal asymptote

y = 0

for

e^x

and vertical asymptote

x = 0

for

\ln x

.

▲ Pre-Calculus Traps & Commandments — Exponentials & Logarithms

Log Domain & Extraneous Roots: The argument of any logarithm must be strictly positive ( $x > 0$ ). When solving equations such as $\log(x) + \log(x-3) = 1$ by condensing to $\log(x(x-3)) = 1$ , the resulting quadratic may produce extraneous solutions that render the original individual arguments negative or zero. Always verify candidates in the uncombined equation!
Common False Log Rules: Remember that $\log(A + B) \neq \log A + \log B$ , $\frac{\log A}{\log B} \neq \log(A - B)$ , and $(\log A)^p \neq p\log A$ . The power rule $\log(A^p) = p\log A$ applies only when the exponent is attached directly to the inner argument.
Change of Base Order: $\log_b a = \frac{\ln a}{\ln b} = \frac{\log_{10} a}{\log_{10} b}$ . The argument $a$ goes in the numerator, and the old base $b$ goes in the denominator.
Continuous vs. Compounded Interest: Use $A = Pe^{rt}$ strictly for continuous compounding. For compounding $n$ times per year (e.g. monthly $n=12$ , quarterly $n=4$ ), use the discrete formula $A = P\left(1 + \frac{r}{n}\right)^{nt}$ .

Solve the logarithmic equation

\log_3(x + 6) + \log_3(x) = 3

and eliminate any extraneous solutions.

Combine the logarithms on the left-hand side using the product rule:

\log_3[x(x + 6)] = 3.

Convert to exponential form:

x(x + 6) = 3^3 = 27 \implies x^2 + 6x - 27 = 0.

Factor the quadratic:

(x + 9)(x - 3) = 0 \implies x = -9 \text{ or } x = 3.

Check the domain restrictions in the original equation:

For $x = -9$ : $\log_3(-9)$ is undefined, so $x = -9$ is an extraneous solution.
For $x = 3$ : $\log_3(9) + \log_3(3) = 2 + 1 = 3$ . Valid $\checkmark$ .

Therefore, the unique solution is $x = 3$ .

A radioactive substance follows the decay model

M(t) = M_0 e^{-kt}

. If 20% of the sample decays in 15 days, find the exact decay constant

k

and the substance's half-life in days to the nearest tenth.

If 20% decays in 15 days, then 80% remains at $t = 15$ :

0.80 M_0 = M_0 e^{-15k} \implies e^{-15k} = 0.80.

Take the natural logarithm of both sides:

-15k = \ln(0.80) \implies k = -\frac{\ln(0.80)}{15} = \frac{\ln(1.25)}{15} \approx 0.01488\text{ day}^{-1}.

The half-life $t_{1/2}$ satisfies $e^{-k t_{1/2}} = 0.5$ :

t_{1/2} = \frac{\ln 2}{k} = \frac{15\ln 2}{-\ln(0.80)} \approx \frac{15(0.6931)}{0.2231} \approx 46.6\text{ days}.

📝 Chapter Quiz

Questions Level

7

Rational Functions & Introduction to Limits

Rational functions are quotients of polynomials. Their graphs exhibit asymptotic behaviour — the conceptual bridge to AP Calculus. Limits describe what a function approaches, even where it is not defined.

● Asymptotes and holes

For $f(x)=\dfrac{p(x)}{q(x)}$ (fully reduced), $n=\deg p,\;d=\deg q$ :

Vertical asymptote: $x=a$ where $q(a)=0$ after cancelling common factors.

Hole (removable discontinuity): at $x=a$ where a factor $(x-a)$ cancels from both $p$ and $q.$

Horizontal asymptote:

n<d:\;y=0.\quad n=d:\;y=\frac{\text{leading coeff of }p}{\text{leading coeff of }q}.\quad n>d:\;\text{oblique (polynomial long division)}.

⊙ Explorer · Asymptotes and holesinteractive

leading coefficient a: 2.0

zero at x = z: 2

denominator factor x − v: 3

common factor x − h: -2

f(x) = 2(x − 2)(x + 2) / ((x − 3)(x + 2))

hole(s)(-2, 1.600)

vertical asymptotex = 3

horizontal asymptotey = 2

x-interceptx = 2

Defaults to Worked Example 7.1, f(x) = (2x² − 8)/(x² − x − 6) = 2(x − 2)(x + 2)/((x − 3)(x + 2)). The factor (x + 2) cancels, so x = −2 is a hole at (−2, 8/5), not an asymptote; x = 3 is the vertical asymptote; numerator and denominator have equal degree, so the horizontal asymptote is y = 2. Set z = v and that factor cancels too.

● Introduction to limits

$\displaystyle\lim_{x\to a}f(x)=L$ means $f(x)$ can be made arbitrarily close to $L$ by taking $x$ close (not equal) to $a.$

One-sided limits: $\displaystyle\lim_{x\to a^-}f(x)$ (from left), $\displaystyle\lim_{x\to a^+}f(x)$ (from right). The two-sided limit exists iff both one-sided limits are equal.

Limits at infinity: $\displaystyle\lim_{x\to\pm\infty}f(x)=$ horizontal asymptote (if finite).

● Graphing rational functions — procedure

1. Find domain (exclude zeros of $q$ ). 2. Factorise; cancel; identify holes. 3. Find vertical asymptotes. 4. Find horizontal/oblique asymptote. 5. Find intercepts. 6. Analyse sign; sketch.

Worked Example 7.1 — Full analysis

Problem. Analyse $f(x)=\dfrac{2x^2-8}{x^2-x-6}.$

Factor: $\dfrac{2(x-2)(x+2)}{(x-3)(x+2)}.$ Cancel $(x+2)$ : hole at $x=-2,\;y=\frac{2(-2-2)}{-2-3}=\frac{-8}{-5}=\frac{8}{5}.$

Simplified: $\dfrac{2(x-2)}{x-3},\;x\neq-2.$ VA: $x=3.$ HA: $n=d\Rightarrow y=2.$ $x$ -intercept: $x=2.$ $y$ -intercept: $f(0)=\frac{-8}{-6}=\frac{4}{3}.$

Worked Example 7.2 — Oblique asymptote

Problem. Find all asymptotes of $f(x)=\dfrac{x^2+2x+3}{x-1}.$

VA: $x=1.$ $n=2>d=1$ : oblique asymptote — divide:

$x^2+2x+3=(x-1)(x+3)+6.$ So oblique asymptote: $y=x+3.$

Worked Example 7.3 — Evaluating limits

Problem. Find $\displaystyle\lim_{x\to2}\frac{x^2-4}{x-2}$ and $\displaystyle\lim_{x\to\infty}\frac{3x^2-1}{2x^2+5}.$

(a) Factor: $\dfrac{(x-2)(x+2)}{x-2}=x+2\to4$ as $x\to2.$ (Removable discontinuity.)

(b) Divide numerator and denominator by $x^2$ : $\dfrac{3-1/x^2}{2+5/x^2}\to\dfrac{3}{2}$ as $x\to\infty.$

Worked Example 7.4 — Solving rational inequalities

Problem. Solve $\dfrac{x+1}{x-2}>0.$

Critical values: $x=-1$ (numerator zero) and $x=2$ (denominator zero). Test sign in each interval:

$(-\infty,-1)$ : both negative → positive ✓. $(-1,2)$ : $+/-$ → negative ✗. $(2,\infty)$ : both positive ✓.

Solution: $x\in(-\infty,-1)\cup(2,\infty).$

Rational function

f(x) = \frac{x^2+2x+3}{x-1} = x + 3 + \frac{6}{x-1}

displaying vertical asymptote

x = 1

and slant (oblique) asymptote

y = x + 3

.

▲ Pre-Calculus Traps & Commandments — Rational Functions & Limits

Holes vs. Vertical Asymptotes: A factor $(x - c)$ that cancels completely from both numerator and denominator produces a removable discontinuity (hole) at $(c, g(c))$ , where $g(x)$ is the simplified function. Denominator roots that remain after canceling yield true vertical asymptotes.
Asymptote Crossing: A curve can never intersect a vertical asymptote (the function is strictly undefined there). However, a rational curve can intersect its horizontal or slant asymptote at finite values of $x$ . Set $f(x) = y_{\text{asymptote}}$ to locate intersection coordinates.
Slant Asymptote Existence: A slant (oblique) asymptote exists if and only if the degree of the numerator is exactly 1 higher than the denominator ( $\deg p = \deg q + 1$ ). The asymptote equation $y = mx + b$ is the quotient polynomial from long division, ignoring the remainder.
Rational Inequality Sign Charts: Never multiply both sides of a rational inequality by a variable denominator expression (e.g. $x - 2$ ) because its sign is undetermined. Move all terms to one side to compare with 0, find all partition points (zeros of numerator and denominator), and test intervals on a sign chart.

Find all vertical and horizontal asymptotes, the exact coordinates of any holes, and all intercepts of the rational function

R(x) = \dfrac{2x^2 + 5x - 3}{x^2 + 4x + 3}

.

Factor numerator and denominator completely:

R(x) = \frac{(2x - 1)(x + 3)}{(x + 1)(x + 3)}.

The factor $(x + 3)$ cancels for $x \neq -3$ . Evaluate the simplified form $g(x) = \frac{2x - 1}{x + 1}$ at $x = -3$ :

g(-3) = \frac{2(-3) - 1}{-3 + 1} = \frac{-7}{-2} = 3.5.

Key features:

Hole: $(-3, 3.5)$
Vertical asymptote: $x + 1 = 0 \implies x = -1$
Horizontal asymptote: Equal degrees (2 and 2) $\implies y = 2/1 = 2$
$x$ -intercept: Numerator root $2x - 1 = 0 \implies (0.5, 0)$
$y$ -intercept: $R(0) = -3/3 = -1 \implies (0, -1)$

Determine the slant asymptote of

g(x) = \dfrac{2x^3 - 3x^2 + 5}{x^2 + 1}

and find the coordinates of any point where the curve intersects this asymptote.

Since $\deg(\text{num}) = 3$ and $\deg(\text{denom}) = 2$ with $3 - 2 = 1$ , a slant asymptote exists. Perform polynomial long division:

(2x^3 - 3x^2 + 5) \div (x^2 + 1) = (2x - 3) + \frac{-2x + 8}{x^2 + 1}.

The linear quotient gives the slant asymptote: $y = 2x - 3$ .

To find where the graph intersects this asymptote, set the remainder equal to zero:

\frac{-2x + 8}{x^2 + 1} = 0 \implies -2x + 8 = 0 \implies x = 4.

Substitute $x = 4$ into the asymptote equation:

y = 2(4) - 3 = 5.

The curve intersects its slant asymptote at the point $(4, 5)$ .

📝 Chapter Quiz

Questions Level

8

Sequences & Series

A sequence is an ordered list of numbers; a series is the sum of a sequence's terms. Arithmetic and geometric types appear in financial models, while infinite series connect to calculus through limits.

● Arithmetic sequences & series

Common difference $d$ ; $n$ -th term and sum:

a_n=a_1+(n-1)d,\quad S_n=\frac{n}{2}(a_1+a_n)=\frac{n}{2}[2a_1+(n-1)d].

● Geometric sequences & series

Common ratio $r$ ; $n$ -th term, finite sum, and infinite sum:

a_n=a_1\,r^{n-1},\quad S_n=\frac{a_1(1-r^n)}{1-r}\;(r\neq1),\quad S_\infty=\frac{a_1}{1-r}\;(|r|<1).

⊙ Explorer · Geometric partial sumsinteractive

first term a₁: 1.20

ratio r: 0.40

terms n: 10

Sₙ = a₁(1 − rⁿ)/(1 − r)1.99979

S∞ = a₁/(1 − r)2.00000

S∞ − Sₙ0.00021

Defaults to Worked Example 8.2: a₁ = 6/5, r = 2/5, so S∞ = (6/5)/(3/5) = 2. The dots climb toward the dashed line and the gap S∞ − Sₙ = a₁rⁿ/(1 − r) shrinks by a factor r each step. Push |r| to 1 or beyond and the dashed line disappears: the series diverges.

● Sigma notation

$\sum_{k=1}^{n}k=\frac{n(n+1)}{2},\quad\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6},\quad\sum_{k=1}^{n}k^3=\left[\frac{n(n+1)}{2}\right]^2.$ $\sum_{k=0}^{n}ar^k=\frac{a(1-r^{n+1})}{1-r}\quad(\text{geometric partial sum}).$

● Binomial theorem

$(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k,\quad\binom{n}{k}=\frac{n!}{k!(n-k)!}.$

Pascal's triangle rows give binomial coefficients for small $n.$

Finding a specific term: the $(r+1)$ -th term of $(a+b)^n$ is $\dbinom{n}{r}a^{n-r}b^r.$

● Mathematical induction (overview)

To prove a statement $P(n)$ for all $n\ge1$ : (1) Base case: verify $P(1)$ . (2) Inductive step: assume $P(k)$ true, prove $P(k+1)$ follows.

Worked Example 8.1 — Arithmetic series

Problem. Find the sum of the arithmetic series $5+9+13+\cdots+101.$

$d=4,\;a_1=5,\;a_n=101.\quad n=\dfrac{101-5}{4}+1=25.$

$S_{25}=\dfrac{25}{2}(5+101)=\dfrac{25}{2}(106)=1325.$

Worked Example 8.2 — Infinite geometric series

Problem. Find $\sum_{n=1}^{\infty}3\!\left(\dfrac{2}{5}\right)^n.$

$a_1=3(2/5)=6/5,\;r=2/5,\;|r|<1.\quad S_\infty=\dfrac{6/5}{1-2/5}=\dfrac{6/5}{3/5}=\dfrac{6}{3}=2.$

Worked Example 8.3 — Binomial expansion

Problem. Find the term containing $x^3$ in the expansion of $\left(2x-\dfrac{1}{x}\right)^7.$

General term: $\binom{7}{k}(2x)^{7-k}\!\left(-\dfrac{1}{x}\right)^k=\binom{7}{k}2^{7-k}(-1)^k\,x^{7-k-k}=\binom{7}{k}2^{7-k}(-1)^k\,x^{7-2k}.$

Set $7-2k=3\Rightarrow k=2.$ Term: $\binom{7}{2}2^5(-1)^2x^3=21\cdot32\cdot x^3=672\,x^3.$

Worked Example 8.4 — Proof by induction

Problem. Prove $1+2+3+\cdots+n=\dfrac{n(n+1)}{2}$ for all $n\ge1.$

Base case $n=1$ : LHS $=1$ , RHS $=\dfrac{1\cdot2}{2}=1.\;\checkmark$

Inductive step: Assume true for $n=k$ : $1+\cdots+k=\dfrac{k(k+1)}{2}.$

For $n=k+1$ : $1+\cdots+k+(k+1)=\dfrac{k(k+1)}{2}+(k+1)=(k+1)\!\left(\dfrac{k}{2}+1\right)=\dfrac{(k+1)(k+2)}{2}.\;\checkmark$

Convergence of geometric partial sums

S_n = \sum_{k=1}^n (0.5)^{k-1}

with

a_1 = 1

and

r = 0.5

, approaching the horizontal limit

S_\infty = \frac{1}{1 - 0.5} = 2

.

▲ Pre-Calculus Traps & Commandments — Sequences, Series & Binomial Theorem

Lower Index Pitfalls in Geometric Series: In $\sum_{n=1}^\infty a r^{n-1}$ , the first term is $a r^0 = a$ . But in $\sum_{n=1}^\infty a r^n$ , the first term is $a_1 = ar$ ! Always evaluate the true first term by substituting the initial index value rather than blindly copying the leading coefficient.
Infinite Series Convergence Condition: The formula $S_\infty = \frac{a_1}{1 - r}$ is valid strictly when $|r| < 1$ . If $|r| \ge 1$ , the terms do not approach 0 and the series diverges. Plugging $r = 2$ into the formula produces an invalid result.
Binomial Expansion Negative Signs: When expanding $(A - B)^n$ , the negative sign belongs to the second term: $\binom{n}{k} A^{n-k} (-B)^k = (-1)^k \binom{n}{k} A^{n-k} B^k$ . Terms alternate signs, with odd powers of $k$ being negative.
Strict Induction Logic: A proof by mathematical induction requires two distinct parts: (1) verifying the base case (usually $n = 1$ ), and (2) explicitly using the inductive hypothesis $P(k)$ to prove $P(k+1)$ . You must show where the assumed formula for the first $k$ terms is substituted.

Find the 6th term in the expansion of

(3x - 2y^2)^8

in fully simplified form.

In the binomial theorem expansion $\sum_{k=0}^n \binom{n}{k} A^{n-k} B^k$ , the 1st term has $k = 0$ , so the 6th term has $k = 5$ .

Identify terms: $n = 8$ , $A = 3x$ , $B = -2y^2$ :

T_6 = \binom{8}{5} (3x)^{8-5} (-2y^2)^5 = \binom{8}{3} (3x)^3 (-2y^2)^5.

Compute each factor:

$\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$
$(3x)^3 = 27x^3$
$(-2y^2)^5 = -32y^{10}$

Multiply the coefficients:

T_6 = 56 \times 27 \times (-32) x^3 y^{10} = 56 \times (-864) x^3 y^{10} = -48384 x^3 y^{10}.

An infinite geometric series has common ratio

r = -0.4

and converges to an infinite sum of

S_\infty = 25

. Find the first term

a_1

, and compute the exact value of the partial sum

S_4

.

Use the infinite sum formula $S_\infty = \frac{a_1}{1 - r}$ :

25 = \frac{a_1}{1 - (-0.4)} = \frac{a_1}{1.4} \implies a_1 = 25 \times 1.4 = 35.

Compute the 4th partial sum $S_4 = \frac{a_1(1 - r^4)}{1 - r}$ :

(-0.4)^4 = 0.0256.

S_4 = \frac{35(1 - 0.0256)}{1.4} = \frac{35(0.9744)}{1.4} = 25 \times 0.9744 = 24.36.

📝 Chapter Quiz

Questions Level

§

Practice Set

1Medium

If $\sin\theta=\dfrac{3}{5}$ and $\theta$ is in Quadrant I, find $\cos\theta$ , $\tan\theta$ , and $\sin 2\theta$ .

$\cos\theta=\sqrt{1-9/25}=4/5.\quad\tan\theta=(3/5)/(4/5)=3/4.\quad\sin2\theta=2(3/5)(4/5)=24/25.$

2Medium

Write the equation of the ellipse with foci at $(\pm 3,0)$ and vertices at $(\pm 5,0)$ .

$c=3,\;a=5,\;b^2=a^2-c^2=16.\quad\dfrac{x^2}{25}+\dfrac{y^2}{16}=1.$

3Hard

Convert $r=4\cos\theta$ to Cartesian form and identify the curve.

$r^2=4r\cos\theta\Rightarrow x^2+y^2=4x\Rightarrow(x-2)^2+y^2=4.$ Circle, centre $(2,0)$ , radius 2.

4Medium

Find the sum of the geometric series $12+8+\tfrac{16}{3}+\cdots$

$r=2/3,\;|r|<1.\quad S_\infty=12/(1-2/3)=36.$

5Easy

Solve $\log_2(x+3)+\log_2(x-3)=4$ .

$\log_2[(x+3)(x-3)]=4\Rightarrow x^2-9=16\Rightarrow x=5.$ (Reject $-5$ : $\log$ undefined.)

6Hard

Find the vertical and horizontal asymptotes of $f(x)=\dfrac{3x^2+1}{x^2-4}$ and describe end behaviour.

Vertical: $x=\pm2.$ Horizontal: degrees equal, $y=3/1=3.$ As $x\to\pm\infty$ , $f(x)\to 3$ from above (since numerator grows slightly faster for large $|x|$ ).

7Medium

Find the angle between $\mathbf{u}=\langle 1,\sqrt{3}\rangle$ and $\mathbf{v}=\langle 2,0\rangle$ .

$\mathbf{u}\cdot\mathbf{v}=2.\;|\mathbf{u}|=2,\;|\mathbf{v}|=2.\;\cos\theta=2/4=1/2.\;\theta=60°.$

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