Reference sheet

Precalculus reference sheet

The main definitions, formulas, identities and conditions of the precalculus course on one page, unit by unit, with links to the lessons that explain each topic in full.

For Precalculus

This sheet follows the ten units of the course in order. Each unit starts with the big idea in a few sentences, then gives the definitions, formulas and rules you will look up most often, with the conditions they need. When you want the full explanation or more examples, follow the lesson links under each topic.

Angles are in radians unless they are marked in degrees. A formula holds wherever both sides are defined unless a condition says otherwise.

Functions, Relations, and Graphs

A function assigns exactly one output to each input. This unit sets up the language for talking about functions, from intervals and domains to graphs and rates of change. It then builds new functions from old ones by shifting, combining, composing and inverting them.

Sets and intervals

A set is a collection of distinct elements, and x∈Ax \in A means that xx belongs to AA. The empty set ∅\varnothing has no elements, so it is not the same as {0}\{0\}, which has one. Set-builder notation describes a set by a condition, as in {x∣x>2}\{x \mid x > 2\}.

IntervalInequalityEndpoints included
(a,b)(a, b)a<x<ba < x < bneither
[a,b][a, b]a≤x≤ba \le x \le bboth
[a,b)[a, b)a≤x<ba \le x < bonly aa
(a,b](a, b]a<x≤ba < x \le bonly bb
(a,∞)(a, \infty)x>ax > anone
(−∞,b](-\infty, b]x≤bx \le bonly bb
(−∞,∞)(-\infty, \infty)every real xxnone

Infinity is not a number, so it always gets a parenthesis. The union A∪BA \cup B holds the elements in AA or BB or both, and the intersection A∩BA \cap B holds the elements in both. For example [0,5)∩(2,8]=(2,5)[0, 5) \cap (2, 8] = (2, 5).

Adding the same number to both sides of an inequality, or multiplying both sides by the same positive number, keeps its direction. Multiplying or dividing by a negative number reverses it, so −2x<6-2x < 6 gives x>−3x > -3. A compound inequality such as −1≤2x+3<7-1 \le 2x + 3 < 7 is solved on all three parts at once, which gives −2≤x<2-2 \le x < 2. Testing one value from your answer in the original inequality catches a missed reversal.

Lessons Sets, intervals and inequalities

Functions, domain and range

A relation is a set of ordered pairs (x,y)(x, y). It is a function when each input xx is paired with exactly one output yy, written y=f(x)y = f(x). The input is the independent variable and the output is the dependent variable. Several inputs may share an output, but one input may not have two. A curve in the plane is the graph of a function of xx exactly when no vertical line meets it more than once.

  • Notation. The symbol f(x)f(x) is the output at the input xx, not ff times xx. To find f(a+h)f(a + h), replace every xx with a+ha + h. For f(x)=x2f(x) = x^2 this gives a2+2ah+h2a^2 + 2ah + h^2, which in general is neither f(a)+hf(a) + h nor f(a)+f(h)f(a) + f(h).
  • Natural domain. When only a formula is given, the domain is every real xx for which the formula gives a real number. Leave out zeros of denominators, keep the radicand of an even root at least 00, and keep the argument of a logarithm positive. Odd roots accept every real number. A domain can also be stated outright or set by the context, such as a length that must be positive.
  • Range. The range is the set of outputs the function actually takes. It sits inside the codomain, the set where outputs are allowed to lie. On a graph it is the projection of the graph onto the yy-axis.

For example the function

f(x)=x+3x−1f(x) = \frac{\sqrt{x + 3}}{x - 1}

has domain [−3,1)∪(1,∞)[-3, 1) \cup (1, \infty). Restrictions in the original formula stay even after simplifying, so x2−4x−2\frac{x^2 - 4}{x - 2} is undefined at x=2x = 2 although it simplifies to x+2x + 2. The function x2x^2 has range [0,∞)[0, \infty) on all of R\mathbb{R}, but range [0,4][0, 4] when its domain is [−1,2][-1, 2].

Lessons Relations and functions, Function notation, Domains, Range and codomain

Lines, graphs and rates of change

The slope of the line through (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2), with x1≠x2x_1 \ne x_2, is

m=y2−y1x2−x1.m = \frac{y_2 - y_1}{x_2 - x_1}.
FormEquationNotes
Slope-intercepty=mx+by = mx + bslope mm, yy-intercept (0,b)(0, b)
Point-slopey−y1=m(x−x1)y - y_1 = m(x - x_1)slope mm through (x1,y1)(x_1, y_1)
Horizontal liney=by = bslope 00
Vertical linex=ax = aundefined slope, not a function of xx

A linear function changes at a constant rate, its slope, measured in output units per input unit. When b=0b = 0 and m≠0m \ne 0 the output is proportional to the input. Two distinct nonvertical lines are parallel exactly when their slopes are equal, and two nonvertical lines are perpendicular exactly when their slopes multiply to −1-1.

The average rate of change of ff from x=ax = a to x=bx = b, with a≠ba \ne b, is

f(b)−f(a)b−a,\frac{f(b) - f(a)}{b - a},

the slope of the secant line through (a,f(a))(a, f(a)) and (b,f(b))(b, f(b)). For f(x)=x2f(x) = x^2 from 11 to 33 it is 9−13−1=4\frac{9 - 1}{3 - 1} = 4. Its sign says whether the output rose or fell overall.

To read a graph, the xx-intercepts are the solutions of f(x)=0f(x) = 0 and the yy-intercept is f(0)f(0) when 00 is in the domain. A function is increasing on an interval when x1<x2x_1 < x_2 gives f(x1)<f(x2)f(x_1) < f(x_2) there, decreasing when it gives f(x1)>f(x2)f(x_1) > f(x_2), and constant when it gives f(x1)=f(x2)f(x_1) = f(x_2). A local maximum value f(c)f(c) is at least as large as f(x)f(x) for every xx near cc, and a global maximum value is at least as large as every value of ff. Minimum values work the same way with the inequality reversed. Values read from a graph are estimates until the formula confirms them.

Lessons Linear functions, Reading graphs, Average rate of change

Piecewise, absolute value and symmetric functions

A piecewise function uses different formulas on different parts of its domain. The branch conditions should not overlap, and an input that meets none of them is not in the domain. On the graph a closed dot marks an endpoint that belongs to its branch and an open dot marks one that does not. The absolute value is the piecewise function

∣x∣={xif x≥0,−xif x<0,\lvert x \rvert = \begin{cases} x & \text{if } x \ge 0, \\ -x & \text{if } x < 0, \end{cases}

which measures the distance from xx to 00. In the same way ∣x−a∣\lvert x - a \rvert is the distance from xx to aa. For c>0c > 0,

Equation or inequalitySolution
∣u∣=c\lvert u \rvert = cu=cu = c or u=−cu = -c
∣u∣<c\lvert u \rvert < c−c<u<c-c < u < c, the inside of an interval
∣u∣>c\lvert u \rvert > cu<−cu < -c or u>cu > c, the outside of an interval

The same holds with ≤\le and ≥\ge, endpoints included. If c=0c = 0, the equation ∣u∣=0\lvert u \rvert = 0 means u=0u = 0, the inequality ∣u∣<0\lvert u \rvert < 0 has no solutions and ∣u∣>0\lvert u \rvert > 0 holds for every u≠0u \ne 0. If c<0c < 0, the equation ∣u∣=c\lvert u \rvert = c and the inequality ∣u∣<c\lvert u \rvert < c have no solutions, while ∣u∣>c\lvert u \rvert > c always holds. When the right side contains the variable, as in ∣x−1∣=2x\lvert x - 1 \rvert = 2x, check each candidate in the original equation. There the candidates are −1-1 and 13\frac{1}{3}, and only 13\frac{1}{3} works.

A function is even when f(−x)=f(x)f(-x) = f(x) and odd when f(−x)=−f(x)f(-x) = -f(x) for every xx in its domain, which must contain −x-x whenever it contains xx. An even graph is symmetric about the yy-axis and an odd graph is symmetric about the origin. For example x2x^2 is even, x3x^3 is odd and x2+xx^2 + x is neither. The zero function is both.

Lessons Piecewise functions, Absolute value, Even and odd functions

Transformations of graphs

Start from the graph of y=f(x)y = f(x) and let c>0c > 0.

EquationEffect on the graph
y=f(x)+cy = f(x) + c or y=f(x)−cy = f(x) - cshift up or down by cc
y=f(x−c)y = f(x - c) or y=f(x+c)y = f(x + c)shift right or left by cc
y=−f(x)y = -f(x)reflect across the xx-axis
y=f(−x)y = f(-x)reflect across the yy-axis
y=c f(x)y = c\,f(x)multiply yy-coordinates by cc, a stretch if c>1c > 1 and a compression if c<1c < 1
y=f(cx)y = f(cx)multiply xx-coordinates by 1c\dfrac{1}{c}, a compression if c>1c > 1 and a stretch if c<1c < 1

Changes inside the function act on xx and work the opposite way to how they look. In general, for nonzero aa and bb the graph of y=a f(b(x−h))+ky = a\,f\big(b(x - h)\big) + k takes each point (p,q)(p, q) of the graph of ff to

(pb+h, aq+k).\left(\frac{p}{b} + h,\ aq + k\right).

For example y=2x−3+1y = 2\sqrt{x - 3} + 1 takes the point (4,2)(4, 2) of y=xy = \sqrt{x} to (7,5)(7, 5).

Lessons Shifts and reflections, Stretches and compressions

Combining and inverting functions

OperationDefinitionDomain
Sum, difference, product(f±g)(x)=f(x)±g(x)(f \pm g)(x) = f(x) \pm g(x) and (fg)(x)=f(x) g(x)(fg)(x) = f(x)\,g(x)inputs in both domains
Quotient(fg)(x)=f(x)g(x)\left(\dfrac{f}{g}\right)(x) = \dfrac{f(x)}{g(x)}inputs in both domains with g(x)≠0g(x) \ne 0
Composition(f∘g)(x)=f(g(x))(f \circ g)(x) = f(g(x))inputs xx of gg with g(x)g(x) in the domain of ff

A composition applies gg first, and the order matters. With f(x)=xf(x) = \sqrt{x} and g(x)=x−2g(x) = x - 2, the composition f(g(x))=x−2f(g(x)) = \sqrt{x - 2} has domain [2,∞)[2, \infty), while g(f(x))=x−2g(f(x)) = \sqrt{x} - 2 has domain [0,∞)[0, \infty).

A function is one-to-one when different inputs always give different outputs, so f(x1)=f(x2)f(x_1) = f(x_2) forces x1=x2x_1 = x_2. Its graph passes the horizontal line test, and a function that is strictly increasing or strictly decreasing is always one-to-one. A one-to-one ff has an inverse f−1f^{-1} defined by

f−1(y)=x  ⟺  f(x)=y.f^{-1}(y) = x \iff f(x) = y.

Then f−1(f(x))=xf^{-1}(f(x)) = x for every xx in the domain of ff and f(f−1(y))=yf(f^{-1}(y)) = y for every yy in the range of ff. Domain and range swap, and the graph of f−1f^{-1} is the reflection of the graph of ff across the line y=xy = x. To find a formula, solve y=f(x)y = f(x) for xx and then swap the letters, so f(x)=2x+3f(x) = 2x + 3 has f−1(x)=x−32f^{-1}(x) = \frac{x - 3}{2}. The inverse is not the reciprocal, since f−1(x)f^{-1}(x) is generally not 1f(x)\frac{1}{f(x)}. A function that is not one-to-one gets an inverse once its domain is restricted, as x2x^2 on [0,∞)[0, \infty) has the inverse x\sqrt{x}.

Lessons Algebra of functions, Composition, One-to-one functions, Inverse functions

Polynomial, Rational, and Algebraic Functions

Polynomials are built from constants and a variable using only addition and multiplication, and rational functions divide one polynomial by another. Their graphs are read from zeros, end behavior and asymptotes. Complex numbers make it possible to factor every polynomial completely.

Quadratics, distance and circles

A quadratic function f(x)=ax2+bx+cf(x) = ax^2 + bx + c with a≠0a \ne 0 has a parabola as its graph, opening up when a>0a > 0 and down when a<0a < 0. Completing the square gives the vertex form

f(x)=a(x−h)2+k,h=−b2a,k=f(h),\begin{gathered} f(x) = a(x - h)^2 + k, \\[0.8em] h = -\frac{b}{2a}, \qquad k = f(h), \end{gathered}

with vertex (h,k)(h, k) and axis of symmetry x=hx = h. For example x2+6x+5=(x+3)2−4x^2 + 6x + 5 = (x + 3)^2 - 4 has vertex (−3,−4)(-3, -4). The zeros of ax2+bx+cax^2 + bx + c are

x=−b±b2−4ac2a.x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.

With real coefficients the discriminant says what kind of zeros to expect.

Discriminant b2−4acb^2 - 4acZeros
positivetwo real zeros
zeroone repeated real zero, −b2a-\dfrac{b}{2a}
negativeno real zeros, two complex conjugate zeros

For points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) the distance and the midpoint are

d=(x2−x1)2+(y2−y1)2,M=(x1+x22,y1+y22).\begin{aligned} d &= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}, \\[0.6em] M &= \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right). \end{aligned}

The circle with center (h,k)(h, k) and radius r>0r > 0 is

(x−h)2+(y−k)2=r2.(x - h)^2 + (y - k)^2 = r^2.

A general equation x2+y2+Dx+Ey+F=0x^2 + y^2 + Dx + Ey + F = 0 is put in this form by completing the square in xx and in yy. For example x2+y2−4x+6y−12=0x^2 + y^2 - 4x + 6y - 12 = 0 becomes (x−2)2+(y+3)2=25(x - 2)^2 + (y + 3)^2 = 25, the circle with center (2,−3)(2, -3) and radius 55. If the right side comes out 00 the graph is a single point, and if it comes out negative there is no graph.

Lessons Quadratic functions, Circles, distance and midpoints

Powers, roots and equations

A power function has the form f(x)=xpf(x) = x^p for a constant pp. For an integer n≥2n \ge 2, the principal root an\sqrt[n]{a} is the real number whose nnth power is aa, taking the nonnegative one when nn is even. Even roots need a≥0a \ge 0, and odd roots accept every real aa.

RuleCondition
am/n=(an)m=amna^{m/n} = \left(\sqrt[n]{a}\right)^m = \sqrt[n]{a^m}integers mm and n≥1n \ge 1 with mn\dfrac{m}{n} in lowest terms, a≥0a \ge 0 when nn is even, a≠0a \ne 0 when m≤0m \le 0
a−p=1apa^{-p} = \dfrac{1}{a^p}a>0a > 0, or a≠0a \ne 0 when pp is an integer
x2=∣x∣\sqrt{x^2} = \lvert x \rvertevery real xx
ab=a b\sqrt{ab} = \sqrt{a}\,\sqrt{b}a≥0a \ge 0 and b≥0b \ge 0

On [0,∞)[0, \infty) the functions xn\sqrt[n]{x} and xnx^n undo each other, and for odd nn they do so on all of R\mathbb{R}.

To solve a rational equation, note the excluded values, multiply both sides by the least common denominator and solve. To solve a radical equation, isolate the radical and raise both sides to the power that removes it. Both steps can produce candidates that fail the original equation, so check every one. For example x+2=x\sqrt{x + 2} = x leads to x+2=x2x + 2 = x^2, so x=2x = 2 or x=−1x = -1. Only x=2x = 2 works, since 1≠−1\sqrt{1} \ne -1.

Lessons Power and root functions, Rational and radical equations

Polynomials, zeros and division

A polynomial of degree nn has the form P(x)=anxn+⋯+a1x+a0P(x) = a_n x^n + \cdots + a_1 x + a_0 with an≠0a_n \ne 0. For n≥1n \ge 1 its end behavior follows the leading term anxna_n x^n.

Degree nnan>0a_n > 0an<0a_n < 0
evenrises at both endsfalls at both ends
oddfalls on the left, rises on the rightrises on the left, falls on the right
  • Zeros. A number cc is a zero of multiplicity mm when (x−c)m(x - c)^m is a factor of P(x)P(x) and (x−c)m+1(x - c)^{m+1} is not. At a real zero of odd multiplicity the graph crosses the xx-axis, and at one of even multiplicity it touches the axis and turns back. A multiplicity of 22 or more flattens the graph against the axis. A polynomial of degree n≥1n \ge 1 has at most nn real zeros and at most n−1n - 1 turning points.
  • Division. For polynomials PP and DD with D≠0D \ne 0 there are unique polynomials QQ and RR with P(x)=D(x) Q(x)+R(x)P(x) = D(x)\,Q(x) + R(x), where R=0R = 0 or deg⁡R<deg⁡D\deg R < \deg D. Synthetic division is a shortcut when D(x)=x−cD(x) = x - c.
  • Remainder and factor theorems. The remainder when P(x)P(x) is divided by x−cx - c is P(c)P(c). So x−cx - c is a factor of P(x)P(x) exactly when P(c)=0P(c) = 0.
  • Rational zeros theorem. If PP has integer coefficients and a0≠0a_0 \ne 0, every rational zero, written in lowest terms as pq\frac{p}{q}, has pp dividing a0a_0 and qq dividing ana_n. The theorem only lists candidates, and some zeros may be irrational or complex.

For example the candidates for P(x)=x3−6x2+11x−6P(x) = x^3 - 6x^2 + 11x - 6 are ±1\pm 1, ±2\pm 2, ±3\pm 3 and ±6\pm 6. Since P(1)=0P(1) = 0, synthetic division by x−1x - 1 gives

x3−6x2+11x−6=(x−1)(x2−5x+6)=(x−1)(x−2)(x−3).\begin{aligned} &x^3 - 6x^2 + 11x - 6 \\[0.4em] &\quad = (x - 1)(x^2 - 5x + 6) \\[0.4em] &\quad = (x - 1)(x - 2)(x - 3). \end{aligned}

Lessons End behavior, Zeros and multiplicity, Polynomial division, Remainder and factor theorems, Rational zeros

Complex numbers and complete factorization

The imaginary unit ii satisfies i2=−1i^2 = -1, and a complex number z=a+biz = a + bi has real part aa and imaginary part bb. Plotted as the point (a,b)(a, b) in the complex plane, it lies at distance ∣z∣=a2+b2\lvert z \rvert = \sqrt{a^2 + b^2} from the origin.

OperationRule
Sum(a+bi)+(c+di)=(a+c)+(b+d)i(a + bi) + (c + di) = (a + c) + (b + d)i
Product(a+bi)(c+di)=(ac−bd)+(ad+bc)i(a + bi)(c + di) = (ac - bd) + (ad + bc)i
Conjugatea+bi‾=a−bi\overline{a + bi} = a - bi, and z z‾=a2+b2z\,\overline{z} = a^2 + b^2
Quotientmultiply top and bottom by the conjugate of the bottom, which must be nonzero

For example

3+i1−i=(3+i)(1+i)(1−i)(1+i)=2+4i2=1+2i.\begin{aligned} \frac{3 + i}{1 - i} &= \frac{(3 + i)(1 + i)}{(1 - i)(1 + i)} \\[0.8em] &= \frac{2 + 4i}{2} = 1 + 2i. \end{aligned}

For a>0a > 0 we write −a=ia\sqrt{-a} = i\sqrt{a}, but the rule a b=ab\sqrt{a}\,\sqrt{b} = \sqrt{ab} fails when both numbers are negative, since −1 −1=−1\sqrt{-1}\,\sqrt{-1} = -1.

Fundamental theorem of algebra. Every polynomial of degree n≥1n \ge 1 with complex coefficients has a complex zero. As a result it factors as

P(x)=an(x−c1)(x−c2)⋯(x−cn)P(x) = a_n (x - c_1)(x - c_2) \cdots (x - c_n)

and has exactly nn zeros counted with multiplicity. When the coefficients are real, nonreal zeros come in conjugate pairs a±bia \pm bi, so PP factors over the reals into linear factors and irreducible quadratic factors with negative discriminant. For example x4−1=(x−1)(x+1)(x2+1)=(x−1)(x+1)(x−i)(x+i)x^4 - 1 = (x - 1)(x + 1)(x^2 + 1) = (x - 1)(x + 1)(x - i)(x + i).

Lessons Complex numbers, The fundamental theorem of algebra

Rational functions and asymptotes

A rational function is a quotient f(x)=P(x)Q(x)f(x) = \frac{P(x)}{Q(x)} of polynomials, with QQ not the zero polynomial. Its domain leaves out every zero of QQ in the original formula, and canceling a factor does not put that point back. Factor the top and bottom and cancel common factors.

  • A zero cc of the denominator that cancels completely leaves a hole at x=cx = c, at the height the simplified formula gives there.
  • A zero cc of the denominator that remains gives a vertical asymptote x=cx = c, where ∣f(x)∣\lvert f(x) \rvert grows without bound as xx approaches cc from either side.

With leading terms anxna_n x^n on top and bmxmb_m x^m below, the behavior as x→∞x \to \infty and as x→−∞x \to -\infty is

DegreesAsymptote
n<mn < mhorizontal, y=0y = 0
n=mn = mhorizontal, y=anbmy = \dfrac{a_n}{b_m}
n=m+1n = m + 1slant, the line y=s(x)y = s(x) where ss is the quotient from polynomial division
n>m+1n > m + 1no horizontal or slant asymptote, and the quotient s(x)s(x) still gives the end behavior

The graph can cross a horizontal or slant asymptote, but never a vertical one. For example the function

f(x)=x2−4x2−x−2=(x−2)(x+2)(x−2)(x+1)f(x) = \frac{x^2 - 4}{x^2 - x - 2} = \frac{(x - 2)(x + 2)}{(x - 2)(x + 1)}

has a hole at (2,43)\left(2, \frac{4}{3}\right), the vertical asymptote x=−1x = -1 and the horizontal asymptote y=1y = 1. Division gives x2+1x−1=x+1+2x−1\frac{x^2 + 1}{x - 1} = x + 1 + \frac{2}{x - 1}, so the line y=x+1y = x + 1 is a slant asymptote of that function.

To graph a rational function, find its domain, holes, intercepts and asymptotes. Then make a sign chart between the zeros and excluded values, and draw each piece approaching its asymptotes.

Lessons Rational functions, Vertical asymptotes and holes, Horizontal asymptotes, Slant asymptotes, Graphing

Polynomial and rational inequalities

  1. Move every term to one side so the other side is 00, combine into one fraction if needed, and factor.
  2. Mark the critical values on a number line. These are the zeros of the numerator and of the denominator.
  3. Test one point in each interval to find the sign of the expression there.
  4. Keep the intervals with the sign you need. Include zeros of the numerator for ≤\le or ≥\ge, and never include zeros of the denominator.

Never multiply both sides by an expression whose sign you do not know. For example x2−x−6>0x^2 - x - 6 > 0 factors as (x−3)(x+2)>0(x - 3)(x + 2) > 0, with solution (−∞,−2)∪(3,∞)(-\infty, -2) \cup (3, \infty). The inequality

x−3x+1≤0\frac{x - 3}{x + 1} \le 0

has solution (−1,3](-1, 3].

Lessons Polynomial inequalities, Rational inequalities

Exponential and Logarithmic Functions

An exponential function multiplies its output by the same factor over equal steps of input. A logarithm undoes it by returning the exponent, so every fact about exponentials has a logarithmic twin. Together they model growth, decay and quantities measured on logarithmic scales.

Exponential functions and the number e

An exponential function has the form f(x)=a bxf(x) = a\,b^x with a≠0a \ne 0 and a base b>0b > 0, b≠1b \ne 1. Each step of 11 in xx multiplies the output by bb, so equal steps in the input give equal ratios of outputs. For a>0a > 0 it shows growth when b>1b > 1 and decay when 0<b<10 < b < 1. Its domain is R\mathbb{R}, its range is (0,∞)(0, \infty) when a>0a > 0, and the line y=0y = 0 is a horizontal asymptote. For b>0b > 0 and all real xx and yy,

bxby=bx+y,bxby=bx−y,(bx)y=bxy,b−x=1bx.\begin{aligned} b^x b^y &= b^{x + y}, & \frac{b^x}{b^y} &= b^{x - y}, \\[0.6em] (b^x)^y &= b^{xy}, & b^{-x} &= \frac{1}{b^x}. \end{aligned}

The number e≈2.71828e \approx 2.71828 is the limit

e=lim⁡n→∞(1+1n)n.e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n.

It comes from compound interest. A principal PP at annual rate rr, compounded nn times a year, grows to P(1+rn)ntP\left(1 + \frac{r}{n}\right)^{nt} after tt years. As nn grows this approaches PertPe^{rt}, which is continuous compounding. A table of values suggests the limit, but numbers alone do not prove it.

Lessons Exponential functions, The number e

Logarithms

For a base b>0b > 0 with b≠1b \ne 1 and an argument x>0x > 0,

y=log⁡bx  ⟺  by=x.y = \log_b x \iff b^y = x.

A logarithm is an exponent. The common logarithm is log⁡x=log⁡10x\log x = \log_{10} x and the natural logarithm is ln⁡x=log⁡ex\ln x = \log_e x. The values log⁡b1=0\log_b 1 = 0 and log⁡bb=1\log_b b = 1 come from b0=1b^0 = 1 and b1=bb^1 = b. The inverse rules are log⁡b(bx)=x\log_b(b^x) = x for every real xx and blog⁡bx=xb^{\log_b x} = x for every x>0x > 0.

Featurey=bxy = b^xy=log⁡bxy = \log_b x
Domain(−∞,∞)(-\infty, \infty)(0,∞)(0, \infty)
Range(0,∞)(0, \infty)(−∞,∞)(-\infty, \infty)
Asymptotey=0y = 0x=0x = 0
Points(0,1)(0, 1) and (1,b)(1, b)(1,0)(1, 0) and (b,1)(b, 1)
Directionincreasing if b>1b > 1, decreasing if 0<b<10 < b < 1the same

The two graphs are reflections of each other across the line y=xy = x. Shifts and stretches work as for any function, so y=ln⁡(x−2)+1y = \ln(x - 2) + 1 has domain (2,∞)(2, \infty) and vertical asymptote x=2x = 2.

Lessons Logarithmic functions, Graphs

Properties of logarithms

For a base b>0b > 0 with b≠1b \ne 1, numbers M>0M > 0 and N>0N > 0, and any real pp,

PropertyFormula
Productlog⁡b(MN)=log⁡bM+log⁡bN\log_b(MN) = \log_b M + \log_b N
Quotientlog⁡bMN=log⁡bM−log⁡bN\log_b \dfrac{M}{N} = \log_b M - \log_b N
Powerlog⁡b(Mp)=plog⁡bM\log_b(M^p) = p\log_b M
Change of baselog⁡bM=ln⁡Mln⁡b=log⁡aMlog⁡ab\log_b M = \dfrac{\ln M}{\ln b} = \dfrac{\log_a M}{\log_a b} for any base a>0a > 0, a≠1a \ne 1

There is no rule for log⁡b(M+N)\log_b(M + N). Watch the conditions when you expand. The expression ln⁡(x2)\ln(x^2) is defined for every x≠0x \ne 0 and equals 2ln⁡∣x∣2\ln\lvert x \rvert, while 2ln⁡x2\ln x needs x>0x > 0. Change of base gives log⁡210=ln⁡10ln⁡2≈3.3219\log_2 10 = \frac{\ln 10}{\ln 2} \approx 3.3219. Keep exact forms like ln⁡10ln⁡2\frac{\ln 10}{\ln 2} until the last step and round only the final answer.

Lessons Product and quotient properties, Power rule and change of base

Solving exponential and logarithmic equations

  • Exponential equations. Write both sides with the same base and set the exponents equal, or isolate the exponential and take a logarithm of both sides. From 3e2x=123e^{2x} = 12 you get e2x=4e^{2x} = 4, so x=ln⁡42=ln⁡2x = \frac{\ln 4}{2} = \ln 2.
  • Logarithmic equations. Combine into a single logarithm, rewrite in exponential form and solve. Then keep only the candidates that make every logarithm in the original equation have a positive argument. From log⁡2x+log⁡2(x−2)=3\log_2 x + \log_2(x - 2) = 3 you get x(x−2)=8x(x - 2) = 8, so x=4x = 4 or x=−2x = -2, and only x=4x = 4 works.

Lessons Exponential equations, Logarithmic equations

Growth, decay and models

ModelFormulaConditions and notes
Percent rate per periodA=A0(1+r)tA = A_0(1 + r)^tgrowth factor 1+r1 + r, growth if r>0r > 0, decay if −1<r<0-1 < r < 0
Continuous rateA=A0ektA = A_0 e^{kt}growth if k>0k > 0, decay if k<0k < 0
Compound interestA=P(1+rn)ntA = P\left(1 + \dfrac{r}{n}\right)^{nt}rate rr compounded nn times per year
Continuous compoundingA=PertA = Pe^{rt}the limit of the row above as n→∞n \to \infty
Half-life TTA=A0(12)t/TA = A_0 \left(\dfrac{1}{2}\right)^{t/T}for A0ektA_0e^{kt} with k<0k < 0, T=ln⁡2∣k∣T = \dfrac{\ln 2}{\lvert k \rvert}
Doubling time TTA=A0 2t/TA = A_0\,2^{t/T}for A0ektA_0e^{kt} with k>0k > 0, T=ln⁡2kT = \dfrac{\ln 2}{k}
LogisticP(t)=M1+Ae−ktP(t) = \dfrac{M}{1 + Ae^{-kt}}M>0M > 0, k>0k > 0, P0>0P_0 > 0 and A=M−P0P0A = \dfrac{M - P_0}{P_0}

A percent rate and a continuous rate describe the same model when 1+r=ek1 + r = e^k, that is when k=ln⁡(1+r)k = \ln(1 + r).

A logistic model grows almost exponentially at first and then levels off toward the carrying capacity MM. When 0<P0<M0 < P_0 < M it grows fastest at P=M2P = \frac{M}{2}, reached at t=ln⁡Akt = \frac{\ln A}{k} (a negative time when P0>M2P_0 > \frac{M}{2}), and the curve is symmetric about that point.

To choose a model, look at equal steps in the input. Equal differences in the output suggest a linear model and equal ratios suggest an exponential one. A residual is the observed value minus the predicted value, and residuals with a clear pattern mean the model is missing something. Predictions inside the range of the data are safer than predictions beyond it. A logarithmic scale records a constant multiple of the logarithm of a quantity, as pH records −log⁡10-\log_{10} of the concentration of hydrogen ions. Equal steps on such a scale are equal factors in the quantity, so a drop of 11 in pH means ten times the concentration.

Lessons Growth and decay models, Choosing a model, Logistic growth

Trigonometric Functions

Trigonometry starts with ratios of sides in right triangles and grows into functions defined for every angle by the unit circle. A full turn brings you back to the same point, so these functions repeat, which makes them the natural models for waves and oscillations.

Angles and arcs

An angle in standard position has its vertex at the origin and its initial side along the positive xx-axis. It is positive when measured counterclockwise. A full turn is 360∘360^\circ, or 2π2\pi radians, so to convert degrees to radians multiply by π180\frac{\pi}{180}.

Degrees0∘0^\circ30∘30^\circ45∘45^\circ60∘60^\circ90∘90^\circ180∘180^\circ270∘270^\circ360∘360^\circ
Radians00π6\dfrac{\pi}{6}π4\dfrac{\pi}{4}π3\dfrac{\pi}{3}π2\dfrac{\pi}{2}π\pi3π2\dfrac{3\pi}{2}2π2\pi

Angles that differ by whole turns, θ+2kπ\theta + 2k\pi or θ+360∘k\theta + 360^\circ k for an integer kk, are coterminal. For a circle of radius rr and a central angle θ\theta in radians,

QuantityFormula
Arc lengths=rθs = r\theta
Sector areaA=12r2θA = \dfrac{1}{2}r^2\theta
Angular speedω=θt\omega = \dfrac{\theta}{t}
Linear speedv=st=rωv = \dfrac{s}{t} = r\omega

These formulas need radians, so convert degrees first. A wheel of radius 0.30.3 meters turning at 1010 radians per second moves a point on its rim at 33 meters per second.

Lessons Angles and radians, Arc length and sector area

Right triangles and special angles

For an acute angle θ\theta in a right triangle,

sin⁡θ=oppositehypotenuse,cos⁡θ=adjacenthypotenuse,tan⁡θ=oppositeadjacent,\begin{aligned} \sin\theta &= \frac{\text{opposite}}{\text{hypotenuse}}, \\[0.6em] \cos\theta &= \frac{\text{adjacent}}{\text{hypotenuse}}, \\[0.6em] \tan\theta &= \frac{\text{opposite}}{\text{adjacent}}, \end{aligned}

and the reciprocal ratios are csc⁡θ=1sin⁡θ\csc\theta = \frac{1}{\sin\theta}, sec⁡θ=1cos⁡θ\sec\theta = \frac{1}{\cos\theta} and cot⁡θ=1tan⁡θ\cot\theta = \frac{1}{\tan\theta}. Similar triangles have the same ratios, so the ratios depend only on the angle. Angles of elevation and depression are measured from the horizontal.

The isosceles right triangle with legs 11 has hypotenuse 2\sqrt{2}. The triangle with angles 30∘30^\circ, 60∘60^\circ and 90∘90^\circ has sides 11, 3\sqrt{3} and 22 opposite them. These give the exact values below.

θ\theta00π6\dfrac{\pi}{6}π4\dfrac{\pi}{4}π3\dfrac{\pi}{3}π2\dfrac{\pi}{2}
sin⁡θ\sin\theta0012\dfrac{1}{2}22\dfrac{\sqrt{2}}{2}32\dfrac{\sqrt{3}}{2}11
cos⁡θ\cos\theta1132\dfrac{\sqrt{3}}{2}22\dfrac{\sqrt{2}}{2}12\dfrac{1}{2}00
tan⁡θ\tan\theta0033\dfrac{\sqrt{3}}{3}113\sqrt{3}undefined

Lessons Right triangle trigonometry, Special right triangles

Any angle and the unit circle

If (x,y)(x, y) is a point other than the origin on the terminal side of θ\theta and r=x2+y2r = \sqrt{x^2 + y^2}, then

FunctionValueDefined when
sin⁡θ\sin\thetayr\dfrac{y}{r}always
cos⁡θ\cos\thetaxr\dfrac{x}{r}always
tan⁡θ\tan\thetayx\dfrac{y}{x}x≠0x \ne 0
csc⁡θ\csc\thetary\dfrac{r}{y}y≠0y \ne 0
sec⁡θ\sec\thetarx\dfrac{r}{x}x≠0x \ne 0
cot⁡θ\cot\thetaxy\dfrac{x}{y}y≠0y \ne 0

On the unit circle r=1r = 1, so the point at angle θ\theta is (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta). You reach it from (1,0)(1, 0) by traveling a distance ∣θ∣\lvert \theta \rvert along the circle, counterclockwise when θ>0\theta > 0 and clockwise when θ<0\theta < 0. A full turn returns to the same point, so sin⁡(θ+2π)=sin⁡θ\sin(\theta + 2\pi) = \sin\theta and cos⁡(θ+2π)=cos⁡θ\cos(\theta + 2\pi) = \cos\theta.

The reference angle is the acute angle between the terminal side and the xx-axis. A function’s value at θ\theta equals its value at the reference angle, with the sign it has in that quadrant.

QuadrantIIIIIIIV
Reference angle for 0<θ<2π0 < \theta < 2\piθ\thetaπ−θ\pi - \thetaθ−π\theta - \pi2π−θ2\pi - \theta
Positive functionsall sixsin⁡\sin and csc⁡\csctan⁡\tan and cot⁡\cotcos⁡\cos and sec⁡\sec

For example 4π3\frac{4\pi}{3} lies in quadrant III with reference angle π3\frac{\pi}{3}, so sin⁡4π3=−32\sin\frac{4\pi}{3} = -\frac{\sqrt{3}}{2}.

Lessons General and reference angles, The unit circle

Graphs of the six functions

Here kk stands for any integer.

FunctionDomainRangePeriodVertical asymptotes
sin⁡x\sin xall xx[−1,1][-1, 1]2π2\pinone
cos⁡x\cos xall xx[−1,1][-1, 1]2π2\pinone
tan⁡x\tan xx≠π2+kπx \ne \dfrac{\pi}{2} + k\piall realsπ\pix=π2+kπx = \dfrac{\pi}{2} + k\pi
cot⁡x\cot xx≠kπx \ne k\piall realsπ\pix=kπx = k\pi
sec⁡x\sec xx≠π2+kπx \ne \dfrac{\pi}{2} + k\pi(−∞,−1]∪[1,∞)(-\infty, -1] \cup [1, \infty)2π2\pix=π2+kπx = \dfrac{\pi}{2} + k\pi
csc⁡x\csc xx≠kπx \ne k\pi(−∞,−1]∪[1,∞)(-\infty, -1] \cup [1, \infty)2π2\pix=kπx = k\pi

Sine and cosine are smooth waves between −1-1 and 11, and the cosine graph is the sine graph shifted left by π2\frac{\pi}{2}. Tangent rises from one asymptote to the next and cotangent falls. Secant and cosecant are U-shaped branches that touch the cosine and sine graphs where those equal 11 or −1-1 and open away from the xx-axis.

Lessons Sine and cosine, Tangent and cotangent, Secant and cosecant

Sinusoids and harmonic motion

For A≠0A \ne 0 and B>0B > 0, the graphs of

y=Asin⁡(B(x−C))+Dandy=Acos⁡(B(x−C))+D\begin{gathered} y = A\sin\big(B(x - C)\big) + D \\[0.4em] \text{and} \quad y = A\cos\big(B(x - C)\big) + D \end{gathered}

have these features.

FeatureValue
Amplitude∣A∣\lvert A \rvert
Period2πB\dfrac{2\pi}{B}
FrequencyB2π\dfrac{B}{2\pi} cycles per unit of xx
Phase shiftCC, to the right when C>0C > 0
Midliney=Dy = D
Range[D−∣A∣, D+∣A∣]\big[D - \lvert A \rvert,\ D + \lvert A \rvert\big]

A negative AA reflects the wave across its midline, and the tangent and cotangent versions have period πB\frac{\pi}{B}. The graph of y=3sin⁡(2(x−π4))+1y = 3\sin\big(2(x - \frac{\pi}{4})\big) + 1 has amplitude 33, period π\pi, a shift of π4\frac{\pi}{4} to the right and range [−2,4][-2, 4]. To fit a wave to data with highest value MM and lowest value mm, take ∣A∣=M−m2\lvert A \rvert = \frac{M - m}{2}, D=M+m2D = \frac{M + m}{2} and B=2πperiodB = \frac{2\pi}{\text{period}}.

In simple harmonic motion the displacement from the equilibrium position is d(t)=asin⁡(ωt)d(t) = a\sin(\omega t) or d(t)=acos⁡(ωt)d(t) = a\cos(\omega t) with ω>0\omega > 0. The amplitude is ∣a∣\lvert a \rvert, the period is 2πω\frac{2\pi}{\omega} and the frequency is ω2π\frac{\omega}{2\pi}. A phase constant φ\varphi shifts the starting point, as in asin⁡(ωt+φ)a\sin(\omega t + \varphi). With damping, d(t)=ae−ctsin⁡(ωt)d(t) = ae^{-ct}\sin(\omega t) for c>0c > 0 swings between the curves ±ae−ct\pm ae^{-ct}, which shrink toward 00.

Lessons Amplitude, period and phase shift, Harmonic motion

Analytic Trigonometry and Triangles

Identities are equations that hold for every angle where both sides are defined, and they let you rewrite a trigonometric expression in whatever form a problem needs. Inverse functions answer which angle has a given value. The laws of sines and cosines then solve triangles of any shape.

Fundamental identities

TypeIdentities
Reciprocalcsc⁡θ=1sin⁡θ\csc\theta = \dfrac{1}{\sin\theta}, sec⁡θ=1cos⁡θ\sec\theta = \dfrac{1}{\cos\theta}, cot⁡θ=1tan⁡θ\cot\theta = \dfrac{1}{\tan\theta}
Quotienttan⁡θ=sin⁡θcos⁡θ\tan\theta = \dfrac{\sin\theta}{\cos\theta}, cot⁡θ=cos⁡θsin⁡θ\cot\theta = \dfrac{\cos\theta}{\sin\theta}
Pythagoreansin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1, 1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta, 1+cot⁡2θ=csc⁡2θ1 + \cot^2\theta = \csc^2\theta
Even and oddcos⁡(−θ)=cos⁡θ\cos(-\theta) = \cos\theta, sin⁡(−θ)=−sin⁡θ\sin(-\theta) = -\sin\theta, tan⁡(−θ)=−tan⁡θ\tan(-\theta) = -\tan\theta
Cofunctionsin⁡(π2−θ)=cos⁡θ\sin\left(\dfrac{\pi}{2} - \theta\right) = \cos\theta, cos⁡(π2−θ)=sin⁡θ\cos\left(\dfrac{\pi}{2} - \theta\right) = \sin\theta

Each identity holds for every θ\theta where both sides are defined. The first Pythagorean identity is x2+y2=1x^2 + y^2 = 1 on the unit circle, and the other two follow by dividing it by cos⁡2θ\cos^2\theta or by sin⁡2θ\sin^2\theta.

To verify an identity, start from one side and rewrite it until it matches the other. Do not treat the identity as an equation and change both sides at once. The usual moves are rewriting in sines and cosines, factoring, combining fractions and multiplying by a conjugate. For every θ\theta with sin⁡θ≠0\sin\theta \ne 0,

sin⁡θ1−cos⁡θ=sin⁡θ (1+cos⁡θ)1−cos⁡2θ=1+cos⁡θsin⁡θ.\frac{\sin\theta}{1 - \cos\theta} = \frac{\sin\theta\,(1 + \cos\theta)}{1 - \cos^2\theta} = \frac{1 + \cos\theta}{\sin\theta}.

Checking a few values can show that a claimed identity is false, but it never proves one true.

Lessons Fundamental identities, Verifying identities

Inverse trigonometric functions

Each inverse function undoes a trigonometric function restricted to a principal branch, a set of angles where the function is one-to-one and still takes all of its values.

FunctionDomainPrincipal range
arcsin⁡x\arcsin x[−1,1][-1, 1][−π2,π2]\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]
arccos⁡x\arccos x[−1,1][-1, 1][0,π][0, \pi]
arctan⁡x\arctan xall xx(−π2,π2)\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)
arccot⁡x\operatorname{arccot} xall xx(0,π)(0, \pi)
arcsec⁡x\operatorname{arcsec} x∣x∣≥1\lvert x \rvert \ge 1[0,π2)∪(π2,π]\left[0, \dfrac{\pi}{2}\right) \cup \left(\dfrac{\pi}{2}, \pi\right]
arccsc⁡x\operatorname{arccsc} x∣x∣≥1\lvert x \rvert \ge 1[−π2,0)∪(0,π2]\left[-\dfrac{\pi}{2}, 0\right) \cup \left(0, \dfrac{\pi}{2}\right]

The last three rows are the branches this course uses, and with them arcsec⁡x=arccos⁡1x\operatorname{arcsec} x = \arccos\frac{1}{x} and arccsc⁡x=arcsin⁡1x\operatorname{arccsc} x = \arcsin\frac{1}{x} for ∣x∣≥1\lvert x \rvert \ge 1. Some books choose other branches for these three, so check which one a text uses. The notation sin⁡−1x\sin^{-1} x means arcsin⁡x\arcsin x, not 1sin⁡x\frac{1}{\sin x}.

Composing in one order always works, as sin⁡(arcsin⁡x)=x\sin(\arcsin x) = x for every xx in [−1,1][-1, 1]. In the other order, arcsin⁡(sin⁡x)=x\arcsin(\sin x) = x only for xx in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right], so arcsin⁡(sin⁡5π6)=π6\arcsin\left(\sin\frac{5\pi}{6}\right) = \frac{\pi}{6}. A reference triangle simplifies the mixed compositions, and the principal range decides the signs.

ExpressionEqualsHolds for
cos⁡(arcsin⁡x)\cos(\arcsin x) and sin⁡(arccos⁡x)\sin(\arccos x)1−x2\sqrt{1 - x^2}−1≤x≤1-1 \le x \le 1
tan⁡(arcsin⁡x)\tan(\arcsin x)x1−x2\dfrac{x}{\sqrt{1 - x^2}}−1<x<1-1 < x < 1
sin⁡(arctan⁡x)\sin(\arctan x)x1+x2\dfrac{x}{\sqrt{1 + x^2}}all xx
cos⁡(arctan⁡x)\cos(\arctan x)11+x2\dfrac{1}{\sqrt{1 + x^2}}all xx
sin⁡(arcsec⁡x)\sin(\operatorname{arcsec} x)x2−1∣x∣\dfrac{\sqrt{x^2 - 1}}{\lvert x \rvert}∣x∣≥1\lvert x \rvert \ge 1

Lessons Inverse sine and cosine, Inverse tangent and the others, Compositions

Sum, difference and double-angle formulas

Take the upper signs together or the lower signs together.

sin⁡(α±β)=sin⁡αcos⁡β±cos⁡αsin⁡βcos⁡(α±β)=cos⁡αcos⁡β∓sin⁡αsin⁡βtan⁡(α±β)=tan⁡α±tan⁡β1∓tan⁡αtan⁡β\begin{aligned} \sin(\alpha \pm \beta) &= \sin\alpha\cos\beta \pm \cos\alpha\sin\beta \\[0.4em] \cos(\alpha \pm \beta) &= \cos\alpha\cos\beta \mp \sin\alpha\sin\beta \\[0.4em] \tan(\alpha \pm \beta) &= \frac{\tan\alpha \pm \tan\beta}{1 \mp \tan\alpha\tan\beta} \end{aligned}

For example

cos⁡π12=cos⁡(π3−π4)=12⋅22+32⋅22=6+24.\begin{aligned} \cos\frac{\pi}{12} &= \cos\left(\frac{\pi}{3} - \frac{\pi}{4}\right) \\[0.6em] &= \frac{1}{2}\cdot\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2} \\[0.6em] &= \frac{\sqrt{6} + \sqrt{2}}{4}. \end{aligned}

Setting α=β=x\alpha = \beta = x gives the double-angle formulas.

FunctionDouble angle
sin⁡2x\sin 2x2sin⁡xcos⁡x2\sin x\cos x
cos⁡2x\cos 2xcos⁡2x−sin⁡2x=2cos⁡2x−1=1−2sin⁡2x\cos^2 x - \sin^2 x = 2\cos^2 x - 1 = 1 - 2\sin^2 x
tan⁡2x\tan 2x2tan⁡x1−tan⁡2x\dfrac{2\tan x}{1 - \tan^2 x}

Lessons Sum and difference formulas, Double-angle formulas

Half-angle and product formulas

Solving the forms of cos⁡2x\cos 2x for the squares gives the power-reducing formulas,

sin⁡2x=1−cos⁡2x2,cos⁡2x=1+cos⁡2x2,tan⁡2x=1−cos⁡2x1+cos⁡2x,\begin{aligned} \sin^2 x &= \frac{1 - \cos 2x}{2}, \\[0.8em] \cos^2 x &= \frac{1 + \cos 2x}{2}, \\[0.8em] \tan^2 x &= \frac{1 - \cos 2x}{1 + \cos 2x}, \end{aligned}

and replacing xx with x2\frac{x}{2} gives the half-angle formulas,

sin⁡x2=±1−cos⁡x2,cos⁡x2=±1+cos⁡x2,tan⁡x2=1−cos⁡xsin⁡x=sin⁡x1+cos⁡x.\begin{aligned} \sin\frac{x}{2} &= \pm\sqrt{\frac{1 - \cos x}{2}}, \\[0.8em] \cos\frac{x}{2} &= \pm\sqrt{\frac{1 + \cos x}{2}}, \\[0.8em] \tan\frac{x}{2} &= \frac{1 - \cos x}{\sin x} = \frac{\sin x}{1 + \cos x}. \end{aligned}

The sign in a half-angle formula is the sign of that function at x2\frac{x}{2}, so it depends on the quadrant of x2\frac{x}{2}, not of xx. The first form for tan⁡x2\tan\frac{x}{2} needs sin⁡x≠0\sin x \ne 0 and the second needs cos⁡x≠−1\cos x \ne -1.

Adding and subtracting the sum and difference formulas gives the product-to-sum and sum-to-product formulas.

ProductSum
sin⁡αcos⁡β\sin\alpha\cos\beta12[sin⁡(α+β)+sin⁡(α−β)]\dfrac{1}{2}\big[\sin(\alpha + \beta) + \sin(\alpha - \beta)\big]
cos⁡αcos⁡β\cos\alpha\cos\beta12[cos⁡(α−β)+cos⁡(α+β)]\dfrac{1}{2}\big[\cos(\alpha - \beta) + \cos(\alpha + \beta)\big]
sin⁡αsin⁡β\sin\alpha\sin\beta12[cos⁡(α−β)−cos⁡(α+β)]\dfrac{1}{2}\big[\cos(\alpha - \beta) - \cos(\alpha + \beta)\big]
SumProduct
sin⁡α+sin⁡β\sin\alpha + \sin\beta2sin⁡α+β2cos⁡α−β22\sin\dfrac{\alpha + \beta}{2}\cos\dfrac{\alpha - \beta}{2}
sin⁡α−sin⁡β\sin\alpha - \sin\beta2cos⁡α+β2sin⁡α−β22\cos\dfrac{\alpha + \beta}{2}\sin\dfrac{\alpha - \beta}{2}
cos⁡α+cos⁡β\cos\alpha + \cos\beta2cos⁡α+β2cos⁡α−β22\cos\dfrac{\alpha + \beta}{2}\cos\dfrac{\alpha - \beta}{2}
cos⁡α−cos⁡β\cos\alpha - \cos\beta−2sin⁡α+β2sin⁡α−β2-2\sin\dfrac{\alpha + \beta}{2}\sin\dfrac{\alpha - \beta}{2}

Two waves with close frequencies add up to a fast wave inside a slow envelope, as in cos⁡11t+cos⁡9t=2cos⁡10t cos⁡t\cos 11t + \cos 9t = 2\cos 10t\,\cos t. That slow swell is the beat you hear from two nearly equal notes.

Lessons Power-reducing and half-angle, Product-to-sum and sum-to-product

Trigonometric equations

Isolate the trigonometric function, find the solutions in one period, then add whole periods. For any integer kk,

EquationSolutionsCondition
sin⁡x=c\sin x = cx=arcsin⁡c+2kπx = \arcsin c + 2k\pi or x=π−arcsin⁡c+2kπx = \pi - \arcsin c + 2k\pi−1≤c≤1-1 \le c \le 1
cos⁡x=c\cos x = cx=±arccos⁡c+2kπx = \pm\arccos c + 2k\pi−1≤c≤1-1 \le c \le 1
tan⁡x=c\tan x = cx=arctan⁡c+kπx = \arctan c + k\piany real cc

When ∣c∣>1\lvert c \rvert > 1 the equations sin⁡x=c\sin x = c and cos⁡x=c\cos x = c have no solution.

  • Quadratic in form. Treat the trigonometric function as the unknown and factor. The equation 2sin⁡2x−sin⁡x−1=02\sin^2 x - \sin x - 1 = 0 factors as (2sin⁡x+1)(sin⁡x−1)=0(2\sin x + 1)(\sin x - 1) = 0, so sin⁡x=−12\sin x = -\frac{1}{2} or sin⁡x=1\sin x = 1, and the solutions in [0,2π)[0, 2\pi) are π2\frac{\pi}{2}, 7π6\frac{7\pi}{6} and 11π6\frac{11\pi}{6}.
  • Multiple angles. To find the solutions in [0,2π)[0, 2\pi) of an equation in nxnx, with nn a positive integer, solve for u=nxu = nx on [0,2nπ)[0, 2n\pi) and then divide by nn. So sin⁡2x=12\sin 2x = \frac{1}{2} has four solutions in [0,2π)[0, 2\pi), namely π12\frac{\pi}{12}, 5π12\frac{5\pi}{12}, 13π12\frac{13\pi}{12} and 17π12\frac{17\pi}{12}.
  • Cautions. Do not divide by an expression such as cos⁡x\cos x that can be zero. Factor it out instead. Squaring both sides can add false solutions, so check every candidate and drop any where a function in the equation is undefined.

Lessons Linear equations, Quadratic equations

Solving triangles

Label a triangle with angles AA, BB and CC and opposite sides aa, bb and cc. The sides must satisfy the triangle inequality, so each one is shorter than the sum of the other two.

GivenUseFormula
two angles and a side (AAS or ASA)law of sinesasin⁡A=bsin⁡B=csin⁡C\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}
two sides and an angle opposite one of them (SSA)law of sines, checking the ambiguous casethe same
two sides and the angle between them (SAS)law of cosinesc2=a2+b2−2abcos⁡Cc^2 = a^2 + b^2 - 2ab\cos C
three sides (SSS)law of cosinescos⁡C=a2+b2−c22ab\cos C = \dfrac{a^2 + b^2 - c^2}{2ab}

In the ambiguous case you know AA, aa and bb. When AA is acute, compare aa with the height h=bsin⁡Ah = b\sin A.

CaseNumber of triangles
a<ha < hnone
a=ha = hone, a right triangle
h<a<bh < a < btwo
a≥ba \ge bone

When AA is right or obtuse there is one triangle if a>ba > b and none if a≤ba \le b.

The area of a triangle is half the product of two sides and the sine of the angle between them. From the three sides alone, Heron’s formula gives it with the semiperimeter s=a+b+c2s = \frac{a + b + c}{2}.

Area=12absin⁡CArea=s(s−a)(s−b)(s−c)\begin{aligned} \text{Area} &= \frac{1}{2}ab\sin C \\[0.8em] \text{Area} &= \sqrt{s(s - a)(s - b)(s - c)} \end{aligned}

For example, with a=5a = 5, b=7b = 7 and C=60∘C = 60^\circ the law of cosines gives c2=25+49−35=39c^2 = 25 + 49 - 35 = 39, so c=39c = \sqrt{39}, and the area is 12(5)(7)sin⁡60∘=3534\frac{1}{2}(5)(7)\sin 60^\circ = \frac{35\sqrt{3}}{4}. In navigation a bearing is usually measured clockwise from north.

Lessons Law of sines, Law of cosines and area

Systems, Vectors, and Matrices

A system asks for values that satisfy several equations at once. Vectors and matrices give a compact language for these problems, and elimination, determinants and inverses solve them in an organized way.

Systems of equations and inequalities

A solution of a system satisfies every equation at once. A linear system has exactly one solution, no solution (inconsistent) or infinitely many solutions (dependent). Two lines in the plane show all three cases, since they cross once, run parallel without meeting, or coincide.

  • Substitution. Solve one equation for a variable and put the result into the others.
  • Elimination. Add multiples of the equations to cancel a variable. With three variables, eliminate the same variable twice to get a system in two.
  • Nonlinear systems. Substitution and elimination still work, but check every candidate in every original equation. A line meets a parabola or a circle in zero, one or two points. A line that meets a circle in exactly one point is tangent to it, and so is a nonvertical line that meets a parabola y=ax2+bx+cy = ax^2 + bx + c exactly once. A vertical line also meets that parabola once, but it crosses the curve instead of touching it. For example y=x2y = x^2 and y=x+2y = x + 2 meet where x2−x−2=0x^2 - x - 2 = 0, at (−1,1)(-1, 1) and (2,4)(2, 4).

For an inequality in two variables, draw the boundary curve solid for ≤\le or ≥\ge and dashed for << or >>. Test a point that is not on the boundary and shade the side where the inequality holds. The solution of a system of inequalities is the overlap of the shaded regions, called the feasible region.

Lessons Linear systems, Nonlinear systems, Inequalities in two variables

Partial fractions

A rational function P(x)Q(x)\frac{P(x)}{Q(x)} is proper when deg⁡P<deg⁡Q\deg P < \deg Q. If it is not, divide first. Then factor QQ over the reals and write one group of terms for each factor.

Factor of Q(x)Q(x)Terms in the decomposition
ax+bax + bAax+b\dfrac{A}{ax + b}
(ax+b)m(ax + b)^mA1ax+b+A2(ax+b)2+⋯+Am(ax+b)m\dfrac{A_1}{ax + b} + \dfrac{A_2}{(ax + b)^2} + \cdots + \dfrac{A_m}{(ax + b)^m}
ax2+bx+cax^2 + bx + c with b2−4ac<0b^2 - 4ac < 0Ax+Bax2+bx+c\dfrac{Ax + B}{ax^2 + bx + c}
(ax2+bx+c)m(ax^2 + bx + c)^m with b2−4ac<0b^2 - 4ac < 0A1x+B1ax2+bx+c+⋯+Amx+Bm(ax2+bx+c)m\dfrac{A_1x + B_1}{ax^2 + bx + c} + \cdots + \dfrac{A_mx + B_m}{(ax^2 + bx + c)^m}

Multiply through by Q(x)Q(x) and find the constants by matching coefficients or by substituting convenient values of xx. For example

x+5(x−1)(x+2)=2x−1−1x+2,3x2+2x(x2+1)=2x+xx2+1,\begin{aligned} \frac{x + 5}{(x - 1)(x + 2)} &= \frac{2}{x - 1} - \frac{1}{x + 2}, \\[0.8em] \frac{3x^2 + 2}{x(x^2 + 1)} &= \frac{2}{x} + \frac{x}{x^2 + 1}, \end{aligned}

where each equation holds wherever its left side is defined.

Lessons Linear factors, Quadratic factors

Vectors

A vector has a magnitude and a direction, and arrows with the same length and direction represent the same vector. The vector from P(x1,y1)P(x_1, y_1) to Q(x2,y2)Q(x_2, y_2) has components ⟨x2−x1,y2−y1⟩\langle x_2 - x_1, y_2 - y_1 \rangle. For v=⟨a,b⟩=ai+bj\mathbf{v} = \langle a, b \rangle = a\mathbf{i} + b\mathbf{j}, where i=⟨1,0⟩\mathbf{i} = \langle 1, 0 \rangle and j=⟨0,1⟩\mathbf{j} = \langle 0, 1 \rangle,

QuantityFormula
Magnitude∥v∥=a2+b2\lVert \mathbf{v} \rVert = \sqrt{a^2 + b^2}
Sum⟨a,b⟩+⟨c,d⟩=⟨a+c,b+d⟩\langle a, b \rangle + \langle c, d \rangle = \langle a + c, b + d \rangle
Scalar multiplek⟨a,b⟩=⟨ka,kb⟩k\langle a, b \rangle = \langle ka, kb \rangle, with length multiplied by ∣k∣\lvert k \rvert and direction reversed if k<0k < 0
Unit vector along v\mathbf{v}v∥v∥\dfrac{\mathbf{v}}{\lVert \mathbf{v} \rVert} for v≠0\mathbf{v} \ne \mathbf{0}
From length and direction angle θ\thetav=∥v∥ ⟨cos⁡θ,sin⁡θ⟩\mathbf{v} = \lVert \mathbf{v} \rVert\,\langle \cos\theta, \sin\theta \rangle

Geometrically, draw u\mathbf{u} and v\mathbf{v} from the same point. Then u+v\mathbf{u} + \mathbf{v} is the diagonal from that point of the parallelogram they span. Equivalently, place the tail of v\mathbf{v} at the head of u\mathbf{u} and draw the arrow from the tail of u\mathbf{u} to the head of v\mathbf{v}.

Lessons Geometric vectors, Component form

Dot product, projections and work

For u=⟨u1,u2⟩\mathbf{u} = \langle u_1, u_2 \rangle and v=⟨v1,v2⟩\mathbf{v} = \langle v_1, v_2 \rangle,

u⋅v=u1v1+u2v2=∥u∥ ∥v∥cos⁡θ,\mathbf{u}\cdot\mathbf{v} = u_1v_1 + u_2v_2 = \lVert \mathbf{u} \rVert\,\lVert \mathbf{v} \rVert\cos\theta,

where θ\theta is the angle between the vectors, with 0≤θ≤π0 \le \theta \le \pi. The angle is defined only when both vectors are nonzero.

  • Orthogonal vectors. The vectors are orthogonal when u⋅v=0\mathbf{u}\cdot\mathbf{v} = 0. The zero vector counts as orthogonal to every vector.
  • Parallel vectors. Nonzero vectors are parallel when one is a scalar multiple of the other, which happens exactly when θ=0\theta = 0 or θ=π\theta = \pi.
  • Projection. For v≠0\mathbf{v} \ne \mathbf{0}, the scalar component and the vector projection of u\mathbf{u} onto v\mathbf{v} are
comp⁡vu=u⋅v∥v∥,proj⁡vu=u⋅v∥v∥2 v.\operatorname{comp}_{\mathbf{v}}\mathbf{u} = \frac{\mathbf{u}\cdot\mathbf{v}}{\lVert \mathbf{v} \rVert}, \qquad \operatorname{proj}_{\mathbf{v}}\mathbf{u} = \frac{\mathbf{u}\cdot\mathbf{v}}{\lVert \mathbf{v} \rVert^2}\,\mathbf{v}.

The rest, u−proj⁡vu\mathbf{u} - \operatorname{proj}_{\mathbf{v}}\mathbf{u}, is orthogonal to v\mathbf{v}, so u\mathbf{u} splits into a part along v\mathbf{v} and a part perpendicular to it. For u=⟨3,1⟩\mathbf{u} = \langle 3, 1 \rangle and v=⟨1,1⟩\mathbf{v} = \langle 1, 1 \rangle the projection is ⟨2,2⟩\langle 2, 2 \rangle and the rest is ⟨1,−1⟩\langle 1, -1 \rangle.

A constant force F\mathbf{F} that moves an object along the displacement D\mathbf{D} does the work

W=F⋅D=∥F∥ ∥D∥cos⁡θ,W = \mathbf{F}\cdot\mathbf{D} = \lVert \mathbf{F} \rVert\,\lVert \mathbf{D} \rVert\cos\theta,

in joules when the force is in newtons and the distance in meters. The work is negative when the force has a component against the motion.

Lessons Dot product, Projections and work

Matrices and elimination

An m×nm \times n matrix has mm rows and nn columns. Matrices of the same size add entry by entry, and a scalar multiplies every entry. The product ABAB is defined when AA is m×nm \times n and BB is n×pn \times p. It is m×pm \times p, and its entry in row ii and column jj is the dot product of row ii of AA with column jj of BB. Applying ABAB to a vector means applying BB first and then AA, as in a composition of functions. So the order matters, and the two products below differ.

[1234][0110]=[2143][0110][1234]=[3412]\begin{aligned} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} &= \begin{bmatrix} 2 & 1 \\ 4 & 3 \end{bmatrix} \\[0.8em] \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} &= \begin{bmatrix} 3 & 4 \\ 1 & 2 \end{bmatrix} \end{aligned}

In general AB≠BAAB \ne BA. The identity matrix II has 11 on the diagonal and 00 elsewhere, and AI=IA=AAI = IA = A for a square matrix AA of the same size. The zero matrix has every entry 00.

Gaussian elimination. Write the system as an augmented matrix and simplify it with the three row operations, which do not change the solutions.

  1. Swap two rows.
  2. Multiply a row by a nonzero constant.
  3. Add a multiple of one row to another row.

In row echelon form any rows of zeros sit at the bottom, each pivot lies to the right of the pivot in the row above it, and every entry below a pivot is 00. From there, back-substitute, or continue to reduced row echelon form, where each pivot is 11 with zeros above and below it.

Final formSolutions
a row that reads 0=c0 = c with c≠0c \ne 0none, the system is inconsistent
consistent, with a pivot in every variable columnexactly one
consistent, with a variable column that has no pivotinfinitely many, with free variables

Lessons Matrices and addition, Matrix multiplication, Gaussian elimination

Determinants and inverse matrices

The determinants of a 2×22 \times 2 matrix and of a 3×33 \times 3 matrix, expanded along the first row, are

∣abcd∣=ad−bc,∣a1a2a3b1b2b3c1c2c3∣=a1∣b2b3c2c3∣−a2∣b1b3c1c3∣+a3∣b1b2c1c2∣.\begin{aligned} \begin{vmatrix} a & b \\ c & d \end{vmatrix} &= ad - bc, \\[1em] \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} &= a_1\begin{vmatrix} b_2 & b_3 \\ c_2 & c_3 \end{vmatrix} - a_2\begin{vmatrix} b_1 & b_3 \\ c_1 & c_3 \end{vmatrix} \\[0.8em] &\quad + a_3\begin{vmatrix} b_1 & b_2 \\ c_1 & c_2 \end{vmatrix}. \end{aligned}

The number ∣det⁡A∣\lvert \det A \rvert is the factor by which AA scales areas in the plane or volumes in space, and a negative determinant means the orientation is reversed. A square matrix AA is invertible exactly when det⁡A≠0\det A \ne 0, and a matrix with det⁡A=0\det A = 0 is called singular. The inverse satisfies AA−1=A−1A=IAA^{-1} = A^{-1}A = I, and for a 2×22 \times 2 matrix with ad−bc≠0ad - bc \ne 0

[abcd]−1=1ad−bc[d−b−ca].\begin{bmatrix} a & b \\ c & d \end{bmatrix}^{-1} = \frac{1}{ad - bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}.

The inverse is not made of the reciprocals of the entries. When AA is invertible, the system AX=BAX = B has the unique solution X=A−1BX = A^{-1}B. When AA is singular, it has no solution or infinitely many. Cramer’s rule writes the unique solution as xi=det⁡Aidet⁡Ax_i = \frac{\det A_i}{\det A}, where AiA_i is AA with column ii replaced by BB. For example the system with 2x+y=52x + y = 5 and x−y=1x - y = 1 has det⁡A=−3\det A = -3 and the solution x=2x = 2, y=1y = 1.

Lessons Determinants, Inverse matrices

Complex Polar Form and Polar Coordinates

Polar coordinates locate a point by its distance from the origin and its angle. Complex numbers written the same way multiply by multiplying lengths and adding angles, which makes powers and roots easy. Some curves also have much simpler equations in polar form.

Polar coordinates

A point with polar coordinates (r,θ)(r, \theta) lies at distance ∣r∣\lvert r \rvert from the pole, along the ray at angle θ\theta from the polar axis when r>0r > 0 and along the opposite ray when r<0r < 0. So (r,θ)(r, \theta), (r,θ+2kπ)(r, \theta + 2k\pi) for any integer kk and (−r,θ+π)(-r, \theta + \pi) all name the same point, and (0,θ)(0, \theta) is the pole for every θ\theta. The two systems are linked by

x=rcos⁡θ,y=rsin⁡θ,r2=x2+y2,\begin{gathered} x = r\cos\theta, \qquad y = r\sin\theta, \\[0.4em] r^2 = x^2 + y^2, \end{gathered}

and tan⁡θ=yx\tan\theta = \frac{y}{x} when x≠0x \ne 0. To convert a point to polar form, take r=x2+y2r = \sqrt{x^2 + y^2} and choose θ\theta in the quadrant of (x,y)(x, y). The arctangent alone returns angles in (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right), which is the wrong quadrant whenever x<0x < 0. For example (−1,3)(-1, \sqrt{3}) has r=2r = 2 and θ=2π3\theta = \frac{2\pi}{3}, not −π3-\frac{\pi}{3}.

Equations convert with the same formulas. Multiplying r=2sin⁡θr = 2\sin\theta by rr gives x2+y2=2yx^2 + y^2 = 2y, the circle with center (0,1)(0, 1) and radius 11. Multiplying by rr can add the pole to the graph, so check whether the original equation already contains it.

Lessons Polar coordinates, Converting coordinates

Complex numbers in polar form

A nonzero complex number z=a+biz = a + bi can be written as

z=r(cos⁡θ+isin⁡θ)=rcis⁡θ,z = r(\cos\theta + i\sin\theta) = r\operatorname{cis}\theta,

where the modulus is r=∣z∣=a2+b2r = \lvert z \rvert = \sqrt{a^2 + b^2} and the argument θ\theta is an angle with cos⁡θ=ar\cos\theta = \frac{a}{r} and sin⁡θ=br\sin\theta = \frac{b}{r}. Arguments of the same number differ by multiples of 2π2\pi. This sheet takes the principal argument in (−π,π](-\pi, \pi], and some texts use [0,2π)[0, 2\pi) instead. The number 00 has modulus 00 and no defined argument. For z1=r1cis⁡θ1z_1 = r_1\operatorname{cis}\theta_1 and z2=r2cis⁡θ2z_2 = r_2\operatorname{cis}\theta_2,

z1z2=r1r2cis⁡(θ1+θ2),z1z2=r1r2cis⁡(θ1−θ2)for z2≠0.\begin{aligned} z_1 z_2 &= r_1 r_2\operatorname{cis}(\theta_1 + \theta_2), \\[0.6em] \frac{z_1}{z_2} &= \frac{r_1}{r_2}\operatorname{cis}(\theta_1 - \theta_2) \quad \text{for } z_2 \ne 0. \end{aligned}

Multiplying by rcis⁡θr\operatorname{cis}\theta scales by rr and rotates by θ\theta, so multiplying by i=cis⁡π2i = \operatorname{cis}\frac{\pi}{2} is a quarter turn counterclockwise.

Lessons Polar form, Products and quotients

Powers and roots

De Moivre’s theorem says that for every integer nn, with r>0r > 0 when nn is negative,

(rcis⁡θ)n=rncis⁡(nθ).\big(r\operatorname{cis}\theta\big)^n = r^n\operatorname{cis}(n\theta).

For example (1+i)8=(2cis⁡π4)8=16cis⁡2π=16(1 + i)^8 = \left(\sqrt{2}\operatorname{cis}\frac{\pi}{4}\right)^8 = 16\operatorname{cis}2\pi = 16. For a positive integer nn, a nonzero number w=rcis⁡θw = r\operatorname{cis}\theta has exactly nn distinct nnth roots,

zk=rn cis⁡θ+2kπn,k=0,1,…,n−1.\begin{gathered} z_k = \sqrt[n]{r}\,\operatorname{cis}\frac{\theta + 2k\pi}{n}, \\[0.8em] k = 0, 1, \ldots, n - 1. \end{gathered}

They lie on the circle of radius rn\sqrt[n]{r}, spaced 2πn\frac{2\pi}{n} apart, and for n≥3n \ge 3 they are the vertices of a regular polygon with nn sides. The nnth roots of unity are the numbers cis⁡2kπn\operatorname{cis}\frac{2k\pi}{n}. For example the cube roots of 88 are 22 and −1±i3-1 \pm i\sqrt{3}.

Lessons De Moivre’s theorem and roots of unity

Polar graphs

Each test below shows a symmetry when the equation does not change. A test that fails does not rule the symmetry out.

ReplaceSymmetric about
θ\theta with −θ-\thetathe polar axis
θ\theta with π−θ\pi - \thetathe line θ=π2\theta = \dfrac{\pi}{2}
rr with −r-rthe pole

The common curves, with a>0a > 0 and b>0b > 0, are

CurveEquationShape
Circler=ar = a, r=2acos⁡θr = 2a\cos\theta or r=2asin⁡θr = 2a\sin\thetaradius aa, centered at the pole, at (a,0)(a, 0) or at (0,a)(0, a) in rectangular coordinates
Limaçonr=b±acos⁡θr = b \pm a\cos\theta or r=b±asin⁡θr = b \pm a\sin\thetainner loop if b<ab < a, cardioid if b=ab = a, dimpled if a<b<2aa < b < 2a, convex if b≥2ab \ge 2a
Roser=acos⁡nθr = a\cos n\theta or r=asin⁡nθr = a\sin n\theta for an integer n≥2n \ge 2petals of length aa, with nn petals if nn is odd and 2n2n petals if nn is even
Lemniscater2=a2cos⁡2θr^2 = a^2\cos 2\theta or r2=a2sin⁡2θr^2 = a^2\sin 2\thetaa figure eight through the pole, drawn only where the right side is at least 00

The largest value of ∣r∣\lvert r \rvert is a+ba + b on a limaçon and aa on a rose. A rose with odd nn is traced once as θ\theta runs over [0,π)[0, \pi), and one with even nn needs [0,2π)[0, 2\pi). Negative values of rr are plotted on the opposite ray, which is how some petals and loops appear.

Lessons Limaçons and cardioids, Roses and lemniscates

Conics, Parametric Equations, and Space

Parabolas, ellipses and hyperbolas are sets of points defined by distances to fixed points and lines, and each has a standard equation. Parametric equations describe a curve as a path traced over time, and three dimensional coordinates carry distance into space.

Parabolas, ellipses and hyperbolas

ConicSet of points
Parabolaequally far from a focus and a directrix line
Ellipsewhose distances to two foci add up to 2a2a
Hyperbolawhose distances to two foci differ by 2a2a in absolute value

In standard position,

ConicEquationKey facts
Parabolax2=4pyx^2 = 4py with p≠0p \ne 0vertex (0,0)(0, 0), focus (0,p)(0, p), directrix y=−py = -p, opens up if p>0p > 0
Parabolay2=4pxy^2 = 4px with p≠0p \ne 0vertex (0,0)(0, 0), focus (p,0)(p, 0), directrix x=−px = -p, opens right if p>0p > 0
Ellipsex2a2+y2b2=1\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 with a>b>0a > b > 0vertices (±a,0)(\pm a, 0), covertices (0,±b)(0, \pm b), foci (±c,0)(\pm c, 0) with c2=a2−b2c^2 = a^2 - b^2
Hyperbolax2a2−y2b2=1\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 with a,b>0a, b > 0vertices (±a,0)(\pm a, 0), foci (±c,0)(\pm c, 0) with c2=a2+b2c^2 = a^2 + b^2, asymptotes y=±baxy = \pm\dfrac{b}{a}x

Swapping xx and yy gives the ellipse with a vertical major axis and the hyperbola that opens up and down. The ellipse has semimajor axis aa and semiminor axis bb, and the hyperbola has a transverse axis of length 2a2a and a conjugate axis of length 2b2b. The latus rectum of a parabola, the chord through the focus parallel to the directrix, has length 4∣p∣4\lvert p \rvert.

The eccentricity e=cae = \frac{c}{a} satisfies 0≤e<10 \le e < 1 for an ellipse and e>1e > 1 for a hyperbola. A circle is the ellipse with a=ba = b, and it has e=0e = 0. For example the ellipse x225+y29=1\frac{x^2}{25} + \frac{y^2}{9} = 1 has c=4c = 4, foci (±4,0)(\pm 4, 0) and e=45e = \frac{4}{5}.

Lessons Parabolas, Ellipses, Hyperbolas

Shifted and rotated conics

Replacing xx with x−hx - h and yy with y−ky - k moves a conic so that its center or vertex is at (h,k)(h, k). To read a conic from an expanded equation, complete the square in each variable. For example

4x2+9y2−16x+18y−11=0  ⟺  (x−2)29+(y+1)24=1,\begin{aligned} 4x^2 + 9y^2 - 16x + 18y - 11 &= 0 \\[0.8em] \iff \frac{(x - 2)^2}{9} + \frac{(y + 1)^2}{4} &= 1, \end{aligned}

an ellipse with center (2,−1)(2, -1), a=3a = 3, b=2b = 2 and foci (2±5,−1)(2 \pm \sqrt{5}, -1).

In the general equation Ax2+Bxy+Cy2+Dx+Ey+F=0Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 with B≠0B \ne 0, rotating the axes through the angle θ\theta with

cot⁡2θ=A−CB,0<θ<π2,\cot 2\theta = \frac{A - C}{B}, \qquad 0 < \theta < \frac{\pi}{2},

removes the xyxy term. The new coordinates XX and YY are related to xx and yy by

x=Xcos⁡θ−Ysin⁡θ,y=Xsin⁡θ+Ycos⁡θ.\begin{aligned} x &= X\cos\theta - Y\sin\theta, \\[0.4em] y &= X\sin\theta + Y\cos\theta. \end{aligned}

For example xy=1xy = 1 has A=CA = C, so θ=π4\theta = \frac{\pi}{4}, and it becomes X22−Y22=1\frac{X^2}{2} - \frac{Y^2}{2} = 1, a hyperbola. The discriminant names the curve without rotating it.

Discriminant B2−4ACB^2 - 4ACNondegenerate conic
negativeellipse or circle
zeroparabola
positivehyperbola

The equation can also describe a degenerate case, such as a single point, one line, two lines or no points at all, so check before naming the curve.

Lessons Shifted conics, Rotation of axes

Polar equations of conics

With a focus at the pole, eccentricity e>0e > 0 and a directrix at distance d>0d > 0 from the pole,

r=ed1±ecos⁡θorr=ed1±esin⁡θ.\begin{gathered} r = \frac{ed}{1 \pm e\cos\theta} \\[0.8em] \text{or} \quad r = \frac{ed}{1 \pm e\sin\theta}. \end{gathered}

The cosine form has the directrix x=±dx = \pm d and the sine form has the directrix y=±dy = \pm d, with the signs matching. The conic is an ellipse when e<1e < 1, a parabola when e=1e = 1 and a hyperbola when e>1e > 1. For example

r=62+cos⁡θ=31+12cos⁡θr = \frac{6}{2 + \cos\theta} = \frac{3}{1 + \frac{1}{2}\cos\theta}

has e=12e = \frac{1}{2} and d=6d = 6, so it is an ellipse with directrix x=6x = 6.

Lessons Polar equations of conics

Parametric equations

Parametric equations x=f(t)x = f(t) and y=g(t)y = g(t) for tt in an interval trace a path, and its orientation is the direction of increasing tt. Different parametrizations can trace the same curve at different speeds or in different directions. Both x=cos⁡tx = \cos t, y=sin⁡ty = \sin t for 0≤t≤2π0 \le t \le 2\pi and x=cos⁡2tx = \cos 2t, y=sin⁡2ty = \sin 2t for 0≤t≤π0 \le t \le \pi go once around the unit circle counterclockwise, the second twice as fast.

CurveParametrization
Line through (x0,y0)(x_0, y_0) with a nonzero direction ⟨a,b⟩\langle a, b \ranglex=x0+atx = x_0 + at, y=y0+bty = y_0 + bt
Circle with center (h,k)(h, k) and radius rrx=h+rcos⁡tx = h + r\cos t, y=k+rsin⁡ty = k + r\sin t, 0≤t<2π0 \le t < 2\pi
Ellipse x2a2+y2b2=1\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1x=acos⁡tx = a\cos t, y=bsin⁡ty = b\sin t, 0≤t<2π0 \le t < 2\pi
Graph of y=f(x)y = f(x)x=tx = t, y=f(t)y = f(t)

To eliminate the parameter, solve one equation for tt and substitute, or use an identity such as cos⁡2t+sin⁡2t=1\cos^2 t + \sin^2 t = 1. The rectangular equation loses the orientation and can describe more than the path, so carry the restrictions along. For example x=tx = \sqrt{t}, y=t+1y = t + 1 for t≥0t \ge 0 gives y=x2+1y = x^2 + 1, but only the half with x≥0x \ge 0.

Lessons Parametric equations, Eliminating the parameter

Three dimensions

Three mutually perpendicular axes meet at the origin, and a point is written (x,y,z)(x, y, z). The three coordinate planes split space into eight octants, and a point with a zero coordinate lies on one of those planes rather than in an octant. In the usual right-handed system, curling the fingers of your right hand from the positive xx-axis toward the positive yy-axis makes your thumb point along the positive zz-axis. The distance between (x1,y1,z1)(x_1, y_1, z_1) and (x2,y2,z2)(x_2, y_2, z_2) and the sphere with center (h,k,l)(h, k, l) and radius r>0r > 0 are

d=(x2−x1)2+(y2−y1)2+(z2−z1)2,r2=(x−h)2+(y−k)2+(z−l)2.\begin{aligned} d &= \sqrt{\begin{aligned} &(x_2 - x_1)^2 + (y_2 - y_1)^2 \\ &\quad + (z_2 - z_1)^2 \end{aligned}}, \\[0.8em] r^2 &= (x - h)^2 + (y - k)^2 + (z - l)^2. \end{aligned}

The midpoint of a segment averages the coordinates of its endpoints. For example the distance from (1,2,3)(1, 2, 3) to (3,−1,9)(3, -1, 9) is 4+9+36=7\sqrt{4 + 9 + 36} = 7.

Lessons 3D coordinates and distance

Sequences, Series, and Induction

A sequence is a function whose inputs are whole numbers, and a series adds up its terms. Arithmetic and geometric sequences have closed formulas for their terms and their sums. Induction proves that a statement holds for every whole number from some starting point on.

Sequences and factorials

A sequence is a function whose domain is a set of integers, usually n=1,2,3,…n = 1, 2, 3, \ldots or starting at 00, and its terms are written ana_n. An explicit formula gives ana_n directly from nn, as in an=2n+1a_n = 2n + 1. A recursive formula gives each term from earlier ones and needs starting values. The Fibonacci sequence 1,1,2,3,5,8,13,…1, 1, 2, 3, 5, 8, 13, \ldots starts from F1=F2=1F_1 = F_2 = 1 and follows Fn=Fn−1+Fn−2F_n = F_{n-1} + F_{n-2} for n≥3n \ge 3. A recursive rule alone does not tell you whether the terms settle toward a limit.

For a positive integer nn the factorial is n!=n(n−1)⋯2⋅1n! = n(n - 1)\cdots 2 \cdot 1, and 0!=10! = 1. Since (n+1)!=(n+1) n!(n + 1)! = (n + 1)\,n!, quotients of factorials simplify quickly, as in 10!8!=10⋅9=90\frac{10!}{8!} = 10 \cdot 9 = 90.

Lessons Sequences and factorials, Recursive sequences

Sigma notation

Sigma notation writes a sum compactly,

∑k=1nak=a1+a2+⋯+an.\sum_{k=1}^{n} a_k = a_1 + a_2 + \cdots + a_n.

The index kk is a dummy variable, so any letter works, and the bounds give its first and last values. The partial sum SnS_n of a sequence is the sum of its first nn terms. A sequence is a list of numbers, while a series is the sum of its terms.

RuleFormula
Sum and difference∑k=1n(ak±bk)=∑k=1nak±∑k=1nbk\displaystyle\sum_{k=1}^{n} (a_k \pm b_k) = \sum_{k=1}^{n} a_k \pm \sum_{k=1}^{n} b_k
Constant multiple∑k=1nc ak=c∑k=1nak\displaystyle\sum_{k=1}^{n} c\,a_k = c\sum_{k=1}^{n} a_k
Constant∑k=1nc=nc\displaystyle\sum_{k=1}^{n} c = nc
First nn integers∑k=1nk=n(n+1)2\displaystyle\sum_{k=1}^{n} k = \frac{n(n + 1)}{2}
First nn squares∑k=1nk2=n(n+1)(2n+1)6\displaystyle\sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6}
First nn cubes∑k=1nk3=(n(n+1)2)2\displaystyle\sum_{k=1}^{n} k^3 = \left(\frac{n(n + 1)}{2}\right)^2

To shift the index, substitute and change both bounds together, as in ∑k=1nak=∑j=0n−1aj+1\sum_{k=1}^{n} a_k = \sum_{j=0}^{n-1} a_{j+1}. There is no product rule, since ∑akbk\sum a_k b_k is generally not ∑ak⋅∑bk\sum a_k \cdot \sum b_k.

Lessons Sigma notation

Arithmetic and geometric sequences

FeatureArithmeticGeometric
Patternadd a common difference ddmultiply by a common ratio r≠0r \ne 0
Term nnan=a1+(n−1)da_n = a_1 + (n - 1)dan=a1rn−1a_n = a_1 r^{n-1}
Sum of the first nn termsSn=n(a1+an)2=n2[2a1+(n−1)d]S_n = \dfrac{n(a_1 + a_n)}{2} = \dfrac{n}{2}\big[2a_1 + (n - 1)d\big]Sn=a11−rn1−rS_n = a_1\dfrac{1 - r^n}{1 - r} for r≠1r \ne 1, and Sn=na1S_n = na_1 for r=1r = 1

The arithmetic sum is the number of terms times the average of the first and last, which you see by pairing the first term with the last. So 1+2+⋯+100=100⋅1012=50501 + 2 + \cdots + 100 = \frac{100 \cdot 101}{2} = 5050. The geometric sum comes from subtracting rSnrS_n from SnS_n, which cancels all but two terms. So 1+2+4+⋯+29=1−2101−2=10231 + 2 + 4 + \cdots + 2^9 = \frac{1 - 2^{10}}{1 - 2} = 1023.

Lessons Arithmetic sequences and series, Geometric sequences and series

Infinite geometric series

An infinite series converges when its partial sums approach a finite number, which is its sum. For a≠0a \ne 0,

∑n=1∞arn−1=a+ar+ar2+⋯=a1−rwhen ∣r∣<1,\begin{aligned} \sum_{n=1}^{\infty} ar^{n-1} &= a + ar + ar^2 + \cdots \\[0.8em] &= \frac{a}{1 - r} \qquad \text{when } \lvert r \rvert < 1, \end{aligned}

and the series diverges when ∣r∣≥1\lvert r \rvert \ge 1. With r=1r = 1 the partial sums nana grow without bound in size, and with r=−1r = -1 they alternate between aa and 00. Repeating decimals are geometric series, as in

0.36‾=36100+361002+⋯=36/1001−1/100=3699=411.\begin{aligned} 0.\overline{36} &= \frac{36}{100} + \frac{36}{100^2} + \cdots \\[0.8em] &= \frac{36/100}{1 - 1/100} = \frac{36}{99} = \frac{4}{11}. \end{aligned}

Lessons Infinite geometric series

Mathematical induction

To prove a statement P(n)P(n) for every integer n≥n0n \ge n_0, do two things.

  1. Base case. Show that P(n0)P(n_0) is true.
  2. Inductive step. Assume that P(k)P(k) is true for an arbitrary integer k≥n0k \ge n_0, the inductive hypothesis, and use it to prove P(k+1)P(k + 1).

Together these show that P(n)P(n) holds for every n≥n0n \ge n_0. Checking many values of nn is not a proof. For example, to show 1+3+⋯+(2n−1)=n21 + 3 + \cdots + (2n - 1) = n^2 for n≥1n \ge 1, the base case is 1=121 = 1^2. If the sum up to 2k−12k - 1 is k2k^2, adding the next odd number 2k+12k + 1 gives k2+2k+1=(k+1)2k^2 + 2k + 1 = (k + 1)^2.

Lessons Mathematical induction

The binomial theorem

For a nonnegative integer nn,

(a+b)n=∑k=0n(nk)an−kbk,(nk)=n!k! (n−k)!.\begin{gathered} (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k}b^k, \\[0.8em] \binom{n}{k} = \frac{n!}{k!\,(n - k)!}. \end{gathered}

The coefficient (nk)\binom{n}{k}, read nn choose kk, counts the ways to choose kk objects from nn. The coefficients form Pascal’s triangle, where each inner entry is the sum of the two above it,

(nk)=(n−1k−1)+(n−1k)for 1≤k≤n−1.\begin{gathered} \binom{n}{k} = \binom{n - 1}{k - 1} + \binom{n - 1}{k} \\[0.8em] \text{for } 1 \le k \le n - 1. \end{gathered}

The first five rows are

111121133114641\begin{array}{c} 1 \\ 1 \quad 1 \\ 1 \quad 2 \quad 1 \\ 1 \quad 3 \quad 3 \quad 1 \\ 1 \quad 4 \quad 6 \quad 4 \quad 1 \end{array}

For example (x+2)4=x4+8x3+24x2+32x+16(x + 2)^4 = x^4 + 8x^3 + 24x^2 + 32x + 16. The term containing bkb^k is term number k+1k + 1 of the expansion. For an exponent that is not a nonnegative integer the expansion never stops, and it becomes the binomial series of calculus.

Lessons The binomial theorem

The Bridge to Calculus

Calculus rests on one new idea, the limit. This unit meets it informally, through sequences that settle down, function values that approach a number, secant slopes that approach a tangent slope and rectangle sums that approach an area.

Limiting values

A sequence ana_n converges to LL when its terms get as close to LL as we like and stay that close for all large nn. Otherwise it diverges.

SequenceBehavior
1n\dfrac{1}{n}converges to 00
rnr^n with ∣r∣<1\lvert r \rvert < 1converges to 00
(1+1n)n\left(1 + \dfrac{1}{n}\right)^nconverges to ee
(−1)n(-1)^nbounded but divergent, it keeps jumping between −1-1 and 11
n2n^2divergent, it grows without bound

For a function we write lim⁡x→af(x)=L\lim_{x \to a} f(x) = L when f(x)f(x) gets as close to LL as we like for every x≠ax \ne a close enough to aa. The value f(a)f(a), and whether it exists at all, plays no part. The limit exists only when the approach from the left and the approach from the right lead to the same number. For example x2−1x−1\frac{x^2 - 1}{x - 1} is undefined at 11 but equals x+1x + 1 everywhere else, so its limit as x→1x \to 1 is 22.

Tables and graphs give evidence, not proof. A table can miss what happens between the inputs you chose, and a graphing window can hide a jump or fast oscillation, as with sin⁡1x\sin\frac{1}{x} near 00, which has no limit there.

Lessons Sequence behavior, Approaching a limit

Difference quotients

For h≠0h \ne 0, with aa and a+ha + h in the domain of ff, the difference quotient

f(a+h)−f(a)h\frac{f(a + h) - f(a)}{h}

is the slope of the secant line through (a,f(a))(a, f(a)) and (a+h,f(a+h))(a + h, f(a + h)). Simplify it algebraically before letting hh shrink. If it approaches a number as h→0h \to 0, that number is the instantaneous rate of change at aa, the slope of the tangent line. For f(x)=x2f(x) = x^2,

(a+h)2−a2h=2a+h⟶2a.\frac{(a + h)^2 - a^2}{h} = 2a + h \longrightarrow 2a.

At a corner there is no single tangent slope. For ∣x∣\lvert x \rvert at 00 the quotient is −1-1 for every h<0h < 0 and 11 for every h>0h > 0.

Lessons Difference quotients

Area with rectangles

To estimate the area under a nonnegative function ff on [a,b][a, b], split the interval into nn equal parts of width Δx=b−an\Delta x = \frac{b - a}{n} with endpoints xi=a+i Δxx_i = a + i\,\Delta x. The left and right sums are

Ln=∑i=0n−1f(xi) Δx,Rn=∑i=1nf(xi) Δx.\begin{gathered} L_n = \sum_{i=0}^{n-1} f(x_i)\,\Delta x, \\[0.8em] R_n = \sum_{i=1}^{n} f(x_i)\,\Delta x. \end{gathered}

If ff is increasing on [a,b][a, b], then Ln≤area≤RnL_n \le \text{area} \le R_n, and if ff is decreasing the inequalities reverse. For such an ff the gap between the two sums is ∣f(b)−f(a)∣ Δx\lvert f(b) - f(a) \rvert\,\Delta x, which shrinks as nn grows. For f(x)=x2f(x) = x^2 on [0,1][0, 1] with n=4n = 4, the sums are L4=0.21875L_4 = 0.21875 and R4=0.46875R_4 = 0.46875, on either side of the exact area 13\frac{1}{3}. Calculus finds that exact area as the limit of these sums.

Lessons Area with rectangles