Precalculus · Study tools

InteractiveBrown et al., §§1-2–1-6, 4-2, 5-2

What you may and may not do

Simplifying is not guessing what “looks right”. Every step has to be an instance of a rule that is true for all values of the letters. Most algebra mistakes come from applying a rule for products to a sum.

The idea

The rules you may use are few. Letters stand for real numbers, and any step must stay true whatever numbers they are.

The rules for sums and products (§1-3). Order and grouping don't matter:

a + b = b + a

,

a b = b a

,

(a + b) + c = a + (b + c)

,

(a b) c = a (b c)

. And the one rule that connects them, the distributive law:

a (b + c) = a b + a c .

It is the only rule that splits a sum apart. Everything else that “distributes” over a sum is a mistake.

Laws of exponents (§4-2, §5-2). For

a, b \neq 0

and integers

m, n

:

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

Each comes from counting factors:

a^{3} a^{2} = (a a a) (a a) = a^{5}

.

The traps. Powers, roots, reciprocals and logarithms behave well on products, and badly on sums:

true for productsfalse for sumswhat is true instead

(a b)^{2} = a^{2} b^{2}

(a + b)^{2} \neq a^{2} + b^{2}

(a + b)^{2} = a^{2} + 2 a b + b^{2}

\sqrt{a b} = \sqrt{a} \sqrt{b}

\sqrt{a + b} \neq \sqrt{a} + \sqrt{b}

nothing simpler

\frac{1}{a b} = \frac{1}{a} \cdot \frac{1}{b}

\frac{1}{a + b} \neq \frac{1}{a} + \frac{1}{b}

\frac{1}{a} + \frac{1}{b} = \frac{a + b}{a b}

\ln (a b) = \ln a + \ln b

\ln (a + b) \neq \ln a + \ln b

nothing simpler

How to check a step yourself. Plug in numbers. One pair of values where the two sides differ proves the step wrong: $(1 + 1)^{2} = 4$ but $1^{2} + 1^{2} = 2$ . Agreement at a few random values doesn't prove a step right, but a wrong step almost never survives it. The checker below does exactly this.

Explore: is this step legal?

Worked example

Simplify $\frac{(2 x^{3} y)^{2}}{4 x y^{- 1}}$ , assuming $x, y \neq 0$ .

Expand the numerator.

$(2 x^{3} y)^{2} = 2^{2} (x^{3})^{2} y^{2} = 4 x^{6} y^{2}$ , using $(a b)^{n} = a^{n} b^{n}$ and $(a^{m})^{n} = a^{m n}$ .
Divide, one letter at a time.

$\frac{4}{4} = 1$ , $\frac{x^{6}}{x} = x^{5}$ , $\frac{y^{2}}{y^{- 1}} = y^{2 - (- 1)} = y^{3}$ .
Answer, and check it with a number.

$x^{5} y^{3}$ . At $x = y = 1$ : the original is $\frac{4}{4} = 1$ and the answer is 1. At $x = 2, y = 1$ : $\frac{(16)^{2}}{8} = 32 = 2^{5}$ .

Check your understanding

$(x + 3)^{2}$ equals:

$x^{2} + 9$ Try $x = 1$ : $16 \neq 10$ . Squaring doesn't split a sum.
$x^{2} + 6 x + 9$

(x + 3)^{2} = (x + 3) (x + 3)

, and the distributive law gives

x^{2} + 3 x + 3 x + 9

. The middle term

6 x

is exactly what the trap forgets.

Which is true for all $x$ and $y$ ?

$2^{x + y} = 2^{x} \cdot 2^{y}$
$2^{x + y} = 2^{x} + 2^{y}$ At $x = y = 1$ : $4$ versus $4$ , fine, but at $x = y = 2$ : $16$ versus $8$ .
$2^{x y} = 2^{x} \cdot 2^{y}$ That multiplies exponents where it should add them.

Adding exponents means multiplying powers:

a^{m} a^{n} = a^{m + n}

. Note that one lucky agreement (

x = y = 1

) proves nothing; it takes only one disagreement to disprove.

$x^{- 2}$ means:

$\frac{1}{x^{2}}$
$- x^{2}$ A negative exponent means a reciprocal, not a negative number.

a^{- n} = 1 / a^{n}

is the only definition that keeps

a^{m} a^{n} = a^{m + n}

true:

x^{- 2} x^{2}

must be

x^{0} = 1

.

Practice

Find the errorA counterexample, then a fix

A student simplifies $\frac{x^{2} + 9}{x + 3} = x + 3$ . Show the step is wrong, and say what can actually be done.

Outline

It fails at $x = 1$ .
Why?
Left side $\frac{10}{4} = 2.5$ , right side $4$ .
The student treated $x^{2} + 9$ as $(x + 3)^{2}$ .
Why?
Cancelling $x + 3$ needs $x + 3$ to be a factor of the numerator, and $(x + 3)^{2} = x^{2} + 6 x + 9 \neq x^{2} + 9$ .
$x^{2} + 9$ has no real factors, so the fraction doesn't simplify.
Why?
$x^{2} + 9 \geq 9 > 0$ for every real $x$ , so it has no real roots, hence no factor $x - r$ .

When does squaring split?A small inequality

Show that $(a + b)^{2} \geq a^{2} + b^{2}$ when $a$ and $b$ have the same sign, and find when equality holds.

Outline

$(a + b)^{2} - (a^{2} + b^{2}) = 2 a b$ .
Why?
Expand: $(a + b)^{2} = a^{2} + 2 a b + b^{2}$ .
$2 a b \geq 0$ when $a, b$ have the same sign.
Why?
A product of two numbers of the same sign is nonnegative.
Equality holds exactly when $a = 0$ or $b = 0$ .
Why?
$2 a b = 0$ iff one factor is zero. So the trap $(a + b)^{2} = a^{2} + b^{2}$ is true only in this trivial case.

InteractiveBrown et al., §§6-1–6-5

Radicals

A square root splits over a product, $\sqrt{2 x} = \sqrt{2} \sqrt{x}$ , and never over a sum, $\sqrt{x + y} \neq \sqrt{x} + \sqrt{y}$ . A right triangle shows why at a glance.

The idea

Square root (§6-1). For

a \geq 0

,

\sqrt{a}

is the nonnegative number whose square is

a

. So

\sqrt{9} = 3

, not

\pm 3

: the equation

x^{2} = 9

has two solutions,

x = \pm \sqrt{9}

, but the symbol

\sqrt{9}

means just one number.

Product and quotient rules (§6-2). For

a, b \geq 0

(and

b > 0

in the quotient),

\sqrt{a b} = \sqrt{a} \sqrt{b}, \sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}} .

So

\sqrt{2 x} = \sqrt{2} \sqrt{x}

for

x \geq 0

, and

\sqrt{18} = \sqrt{9} \sqrt{2} = 3 \sqrt{2}

.

Show the computation

Both $\sqrt{a b}$ and $\sqrt{a} \sqrt{b}$ are nonnegative, and both square to $a b$ : $(\sqrt{a} \sqrt{b})^{2} = (\sqrt{a})^{2} (\sqrt{b})^{2} = a b$ . A nonnegative number is determined by its square, so they are equal.

Careful with letters. $\sqrt{x^{2}} = | x |$ , not $x$ : at $x = - 3$ , $\sqrt{(- 3)^{2}} = \sqrt{9} = 3$ . This matters whenever a variable could be negative.

Sums don't split. $\sqrt{9 + 16} = \sqrt{25} = 5$ , but $\sqrt{9} + \sqrt{16} = 7$ . In fact always $\sqrt{a + b} \leq \sqrt{a} + \sqrt{b}$ , with equality only if $a$ or $b$ is 0. The figure shows why: a right triangle with legs $\sqrt{a}$ and $\sqrt{b}$ has hypotenuse $\sqrt{a + b}$ (Pythagoras), and the straight side is shorter than the two legs together.

Rationalizing (§6-4). To clear a root from a denominator, multiply top and bottom by what removes it: $\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ , and for a sum use the conjugate, since $(a + \sqrt{b}) (a - \sqrt{b}) = a^{2} - b$ .

Equations with radicals (§6-5). Squaring both sides can create false solutions, because $u = v$ and $u = - v$ both square to $u^{2} = v^{2}$ . Always check answers in the original equation.

Explore

a b

Legs

\sqrt{a}

and

\sqrt{b}

; by Pythagoras the hypotenuse is

\sqrt{a + b}

.

\sqrt{a + b}

, the straight path, against

\sqrt{a} + \sqrt{b}

, the path along the legs.

Worked example

Solve $\sqrt{x + 3} = x - 3$ .

Square both sides and solve.

$x + 3 = x^{2} - 6 x + 9$ , so $x^{2} - 7 x + 6 = 0 = (x - 1) (x - 6)$ : $x = 1$ or $x = 6$ .
Check both in the original equation.

$x = 6$ : $\sqrt{9} = 3 = 6 - 3$ . ✓. $x = 1$ : $\sqrt{4} = 2$ but $1 - 3 = - 2$ . ✗.
Where did the false solution come from?

$x = 1$ solves $\sqrt{x + 3} = - (x - 3)$ , which squares to the same equation. A square root is never negative, so the right side must satisfy $x - 3 \geq 0$ anyway. The only solution is $x = 6$ .

Check your understanding

For every real $x$ , $\sqrt{x^{2}}$ equals:

$x$ Only when $x \geq 0$ . Try $x = - 5$ .
$| x |$
$\pm x$ $\sqrt{}$ means one number, the nonnegative one.

\sqrt{x^{2}}

is the nonnegative number whose square is

x^{2}

, and that is

| x |

.

$\sqrt{50}$ simplifies to:

$5 \sqrt{2}$
$25 \sqrt{2}$ $\sqrt{25} = 5$ , not 25.
$\sqrt{25} + \sqrt{25} = 10$ That splits a sum, $25 + 25$ ; the rule is for products, $25 \cdot 2$ .

50 = 25 \cdot 2

, so

\sqrt{50} = \sqrt{25} \sqrt{2} = 5 \sqrt{2}

. Look for the largest perfect-square factor.

Is $\sqrt{x^{2} + 16} = x + 4$ ?

Yes, by the product rule. $x^{2} + 16$ is a sum, not a product.
No: at $x = 3$ , the left is 5 and the right is 7.

The left side is the hypotenuse of a right triangle with legs

| x |

and 4; the right side is the sum of the legs. They agree only at

x = 0

.

Practice

RationalizeThe conjugate

Write $\frac{2}{\sqrt{5} - 1}$ with no root in the denominator.

Outline

Multiply top and bottom by the conjugate $\sqrt{5} + 1$ .
Why?
Multiplying by $\frac{\sqrt{5} + 1}{\sqrt{5} + 1} = 1$ doesn't change the value, and $(\sqrt{5} - 1) (\sqrt{5} + 1) = 5 - 1 = 4$ has no root.
The result is $\frac{\sqrt{5} + 1}{2}$ .
Why?
$\frac{2 (\sqrt{5} + 1)}{4} = \frac{\sqrt{5} + 1}{2} \approx 1.618$ , the golden ratio.

Prove the triangle pictureWithout the triangle

Prove $\sqrt{a + b} \leq \sqrt{a} + \sqrt{b}$ for $a, b \geq 0$ , with equality only if $a = 0$ or $b = 0$ .

Outline

Both sides are nonnegative, so it is enough to compare their squares.
Why?
For $u, v \geq 0$ , $u \leq v$ exactly when $u^{2} \leq v^{2}$ , because squaring is increasing on $[0, \infty)$ .
$(\sqrt{a} + \sqrt{b})^{2} = a + b + 2 \sqrt{a b} \geq a + b$ .
Why?
Expand, and note $\sqrt{a b} \geq 0$ .
Equality needs $\sqrt{a b} = 0$ .
Why?
That is $a b = 0$ : one of them is zero.

InteractiveBrown et al., §§5-4–5-7

Fractions and rational expressions

You may cancel a factor that the whole top and the whole bottom share. You may not cancel a term. This one rule, and the common denominator, are most of what calculus asks of fractions.

The idea

Cancelling (§5-4).

\frac{a c}{b c} = \frac{a}{b}

when

c \neq 0

: a common factor of the entire numerator and the entire denominator. So

\frac{x^{2} - 4}{x - 2} = \frac{(x - 2) (x + 2)}{x - 2} = x + 2 (x \neq 2),

but

\frac{x + 2}{x}

does not become

2

:

x

is a term of the numerator, not a factor of it.

The excluded value. $\frac{x^{2} - 4}{x - 2}$ is undefined at $x = 2$ , while $x + 2$ is defined there. The two expressions agree everywhere else, which is exactly the situation of a limit in calculus: $lim_{x \to 2} \frac{x^{2} - 4}{x - 2} = 4$ .

Adding (§5-6). Use a common denominator:

\frac{a}{b} + \frac{c}{d} = \frac{a d + b c}{b d}, so \frac{1}{x} + \frac{1}{y} = \frac{x + y}{x y} \neq \frac{1}{x + y} .

Complex fractions (§5-7). A fraction of fractions simplifies by multiplying its top and bottom by the least common denominator of all the small fractions.

Explore: is this step legal?

Worked example

Simplify $\frac{\frac{1}{x} - \frac{1}{2}}{x - 2}$ (for $x \neq 0, 2$ ). This exact expression appears when you compute the derivative of $1 / x$ at 2 from the definition.

Combine the top into one fraction.

$\frac{1}{x} - \frac{1}{2} = \frac{2 - x}{2 x}$ .
Divide by $x - 2$ .

$\frac{2 - x}{2 x} \cdot \frac{1}{x - 2} = \frac{- (x - 2)}{2 x (x - 2)}$ . Now $x - 2$ is a common factor.
Cancel, and read off the limit as $x \to 2$ .

$- \frac{1}{2 x}$ , which tends to $- \frac{1}{4}$ : the slope of $1 / x$ at $x = 2$ .

Check your understanding

$\frac{x + 5}{5}$ simplifies to:

$x$ 5 is a term on top, not a factor. Try $x = 5$ : $\frac{10}{5} = 2 \neq 5$ .
$\frac{x}{5} + 1$ , and nothing shorter.

Dividing a sum divides each term:

\frac{x + 5}{5} = \frac{x}{5} + \frac{5}{5}

. That is the distributive law, and it is the only way a denominator splits.

$\frac{1}{2} + \frac{1}{3}$ equals:

$\frac{1}{5}$ That adds the denominators. A half plus a third is more than a half.
$\frac{5}{6}$

\frac{3}{6} + \frac{2}{6} = \frac{5}{6}

. The same goes for letters:

\frac{1}{x} + \frac{1}{y} = \frac{x + y}{x y}

.

Are $\frac{x^{2} - 1}{x - 1}$ and $x + 1$ the same function?

Yes, since one simplifies to the other.They differ at one point.
Almost: they agree for every $x \neq 1$ , but the first is undefined at $x = 1$ .

Cancelling

x - 1

assumes

x - 1 \neq 0

. The limit as

x \to 1

is 2 even though the first expression has no value there: that is the whole idea of a limit.

Practice

A difference quotientThe step before every derivative

Simplify $\frac{(x + h)^{2} - x^{2}}{h}$ for $h \neq 0$ , and find its limit as $h \to 0$ .

Outline

The top is $2 x h + h^{2}$ .
Why?
$(x + h)^{2} = x^{2} + 2 x h + h^{2}$ ; the $x^{2}$ cancels.
$h$ is a factor of the top, so the quotient is $2 x + h$ .
Why?
$2 x h + h^{2} = h (2 x + h)$ , and $h \neq 0$ .
As $h \to 0$ it tends to $2 x$ : the derivative of $x^{2}$ .
Why?
$2 x + h$ is defined at $h = 0$ and continuous in $h$ .

Splitting a fractionPartial fractions in reverse

Show that $\frac{1}{x (x + 1)} = \frac{1}{x} - \frac{1}{x + 1}$ , and use it to add $\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{9 \cdot 10}$ .

Outline

Combine the right side over $x (x + 1)$ .
Why?
$\frac{1}{x} - \frac{1}{x + 1} = \frac{(x + 1) - x}{x (x + 1)} = \frac{1}{x (x + 1)}$ .
The sum telescopes to $1 - \frac{1}{10} = \frac{9}{10}$ .
Why?
$(1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{9} - \frac{1}{10})$ : every middle term cancels with the next.

InteractiveBrown et al., §§10-1–10-8

Exponential and logarithmic functions

A logarithm answers one question: to what power must the base be raised to get this number? Every law of logarithms is a law of exponents read backwards.

The idea

Rational exponents (§10-1). For

a > 0

,

a^{1 / n} = \sqrt[n]{a}

and

a^{m / n} = (\sqrt[n]{a})^{m}

. So

8^{2 / 3} = (\sqrt[3]{8})^{2} = 4

and

x^{1 / 2} = \sqrt{x}

. These definitions are the only ones that keep

(a^{m})^{n} = a^{m n}

true.

Inverse functions (§10-3). If

f

is one-to-one, its inverse

f^{- 1}

undoes it:

f^{- 1} (f (x)) = x

. Its graph is the graph of

f

reflected in the line

y = x

.

Logarithm (§10-4). For

b > 0

,

b \neq 1

, and

x > 0

:

\log_{b} x = y ⟺ b^{y} = x .

\log_{b}

is the inverse of

x \mapsto b^{x}

. So

\log_{2} 8 = 3

,

\log_{10} 0.01 = - 2

,

\log_{b} 1 = 0

,

\log_{b} b = 1

. The natural logarithm

\ln

is

\log_{e}

, with

e \approx 2.718

(§10-8).

Laws of logarithms (§10-5). For

x, y > 0

and any real

r

:

\log_{b} (x y) = \log_{b} x + \log_{b} y, \log_{b} \frac{x}{y} = \log_{b} x - \log_{b} y, \log_{b} (x^{r}) = r \log_{b} x, \log_{b} x = \frac{\ln x}{\ln b} .

Show the computation

Write $x = b^{u}$ and $y = b^{v}$ , so $u = \log_{b} x$ , $v = \log_{b} y$ . Then $x y = b^{u} b^{v} = b^{u + v}$ , so $\log_{b} (x y) = u + v$ . Likewise $x / y = b^{u - v}$ and $x^{r} = b^{r u}$ . For the change of base, take $\ln$ of $b^{u} = x$ : $u \ln b = \ln x$ .

And the trap: there is no law for $\log_{b} (x + y)$ . It is not $\log_{b} x + \log_{b} y$ (that is $\log_{b} (x y)$ ).

Solving equations (§10-6). If the unknown is in an exponent, take logarithms; if it is inside a logarithm, exponentiate.

Explore

Base b Point on y = bˣ: x

y = b^{x}

y = \log_{b} x

y = x

. The point

(x, b^{x})

and its mirror image

(b^{x}, x)

: the logarithm swaps input and output.

Worked example

Three standard tasks.

Expand $\ln (\frac{x^{3} \sqrt{y}}{(1 + x)^{2}})$ for $x, y > 0$ .

$\ln (x^{3}) + \ln (y^{1 / 2}) - \ln ((1 + x)^{2}) = 3 \ln x + \frac{1}{2} \ln y - 2 \ln (1 + x)$ . The $\ln (1 + x)$ stays as it is: it is the log of a sum.
Solve $3^{2 x - 1} = 7$ exactly.

Take $\ln$ : $(2 x - 1) \ln 3 = \ln 7$ , so $x = \frac{1}{2} (1 + \frac{\ln 7}{\ln 3}) \approx 1.39$ .
Solve $\ln (2 x - 1) = 3$ .

Exponentiate: $2 x - 1 = e^{3}$ , so $x = \frac{e^{3} + 1}{2} \approx 10.54$ . Check: $2 x - 1 = e^{3} > 0$ , so the log is defined.

Check your understanding

$\ln (x + y)$ equals:

$\ln x + \ln y$ That is $\ln (x y)$ .
$\ln x \cdot \ln y$ Try $x = y = 1$ : $\ln 2 \approx 0.69$ against $0$ .
Nothing simpler, in general.

Logarithms turn products into sums, not sums into anything nicer. If you meet

\ln (x + y)

, leave it, or factor the inside first:

\ln (x + x y) = \ln x + \ln (1 + y)

.

Condense $2 \ln (x + 1) - \frac{1}{2} \ln x + \ln 5$ into a single logarithm.

$\ln \frac{5 (x + 1)^{2}}{\sqrt{x}}$
$\ln (2 (x + 1) - \frac{1}{2} x + 5)$ Coefficients become exponents, and sums become products.

2 \ln (x + 1) = \ln (x + 1)^{2}

,

\frac{1}{2} \ln x = \ln \sqrt{x}

; then add the logs of the factors on top and subtract the log of the bottom.

$\log_{3} \frac{1}{9}$ equals:

$- 2$
$\frac{1}{3}$ Ask: 3 to what power is $\frac{1}{9}$ ?

3^{- 2} = \frac{1}{9}

. Logs of numbers between 0 and 1 are negative (for base

b > 1

).

Practice

Half-lifeGrowth and decay

A substance decays as $A (t) = A_{0} e^{- k t}$ . Its half-life is 5 years. Find $k$ , and how long until 10% remains.

Outline

$k = \frac{\ln 2}{5} \approx 0.139$ per year.
Why?
$e^{- 5 k} = \frac{1}{2}$ ; take $\ln$ : $- 5 k = - \ln 2$ .
10% remains after $\frac{\ln 10}{k} = 5 \log_{2} 10 \approx 16.6$ years.
Why?
$e^{- k t} = 0.1$ gives $t = \frac{\ln 10}{k} = \frac{5 \ln 10}{\ln 2}$ .

Derivatives that need log lawsA calculus warm-up

Using $\frac{d}{d x} \ln x = \frac{1}{x}$ and $\frac{d}{d x} e^{x} = e^{x}$ , find the derivatives of $\log_{3} x$ and $5^{2 x - 1}$ .

Outline

$\frac{d}{d x} \log_{3} x = \frac{1}{x \ln 3}$ .
Why?
Change of base: $\log_{3} x = \frac{\ln x}{\ln 3}$ , and $\ln 3$ is a constant.
$\frac{d}{d x} 5^{2 x - 1} = 2 \ln 5 \cdot 5^{2 x - 1}$ .
Why?
$5^{2 x - 1} = e^{(2 x - 1) \ln 5}$ ; by the chain rule the derivative is $e^{(2 x - 1) \ln 5} \cdot 2 \ln 5$ .

InteractiveBrown et al., §§12-1–12-4, 13-1–13-4

Radians and the unit circle

Sine and cosine are coordinates: walk a distance $θ$ around the circle of radius 1, and you arrive at the point $(\cos θ, \sin θ)$ . Every value, sign and symmetry follows from that picture.

The idea

Radians (§13-1). An angle of

θ

radians cuts off an arc of length

θ

on the unit circle. A full turn is the whole circumference,

2 π

, so

360^{\circ} = 2 π

,

180^{\circ} = π

, and

1^{\circ} = \frac{π}{180}

. Calculus uses radians because then

\frac{d}{d x} \sin x = \cos x

with no extra constant.

Circular functions (§13-2). The point at angle

θ

on the unit circle is

(\cos θ, \sin θ)

; also

\tan θ = \frac{\sin θ}{\cos θ}

. For an acute angle in a right triangle with hypotenuse 1, these are the familiar “opposite over hypotenuse” and “adjacent over hypotenuse” (§12-2).

Exact values (§12-4). Two triangles give everything: half a square (angles $45^{\circ}, 45^{\circ}$ , sides $1, 1, \sqrt{2}$ ) and half an equilateral triangle (angles $30^{\circ}, 60^{\circ}$ , sides $1, \sqrt{3}, 2$ ). Scaled to hypotenuse 1:

θ

0

\frac{π}{6}

\frac{π}{4}

\frac{π}{3}

\frac{π}{2}

\sin θ

0

\frac{1}{2}

\frac{\sqrt{2}}{2}

\frac{\sqrt{3}}{2}

1

\cos θ

1

\frac{\sqrt{3}}{2}

\frac{\sqrt{2}}{2}

\frac{1}{2}

0

Every other angle (§12-3). Find the reference angle (the acute angle to the $x$ -axis), take its value from the table, and attach the sign of the quadrant: $x$ (cosine) is negative on the left, $y$ (sine) is negative below.

Periodicity and symmetry (§13-3).

\sin (θ + 2 π) = \sin θ

,

\cos (θ + 2 π) = \cos θ

;

\sin (- θ) = - \sin θ

,

\cos (- θ) = \cos θ

;

\sin (π - θ) = \sin θ

,

\cos (π - θ) = - \cos θ

. Each is a reflection or a full turn of the circle.

Explore

Angle

Drag the point around the circle.

\cos θ

, its

x

-coordinate;

\sin θ

, its

y

-coordinate; the gold arc has length

θ

.

The same two numbers as

θ

goes once around:

\sin

and

\cos

.

Worked example

Find $\sin \frac{5 π}{6}$ , $\cos \frac{4 π}{3}$ and $\tan (- \frac{π}{4})$ exactly.

$\sin \frac{5 π}{6}$ .

$\frac{5 π}{6} = π - \frac{π}{6}$ : second quadrant, reference angle $\frac{π}{6}$ , and sine is positive above the axis. So $\sin \frac{5 π}{6} = \sin \frac{π}{6} = \frac{1}{2}$ .
$\cos \frac{4 π}{3}$ .

$\frac{4 π}{3} = π + \frac{π}{3}$ : third quadrant, reference angle $\frac{π}{3}$ , and cosine is negative on the left. So $- \cos \frac{π}{3} = - \frac{1}{2}$ .
$\tan (- \frac{π}{4})$ .

The point is $(\frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2})$ , so $\tan = \frac{- \sqrt{2} / 2}{\sqrt{2} / 2} = - 1$ .

Check your understanding

$135^{\circ}$ in radians is:

$\frac{3 π}{4}$
$\frac{3 π}{2}$ That is $270^{\circ}$ .

Multiply by

\frac{π}{180}

:

135 \cdot \frac{π}{180} = \frac{3 π}{4}

.

$\sin \frac{7 π}{6}$ equals:

$\frac{1}{2}$ Right size, wrong sign: $\frac{7 π}{6}$ is below the axis.
$- \frac{1}{2}$
$- \frac{\sqrt{3}}{2}$ That is the cosine.

\frac{7 π}{6} = π + \frac{π}{6}

: third quadrant, reference angle

\frac{π}{6}

, and sine is negative there.

Which is always true?

$\cos (- θ) = \cos θ$
$\sin (- θ) = \sin θ$ Reflecting in the $x$ -axis flips the $y$ -coordinate.

The angle

- θ

is the reflection of

θ

in the

x

-axis: same

x

-coordinate, opposite

y

-coordinate. Cosine is even, sine is odd.

Practice

Where

\frac{\sqrt{3}}{2}

comes fromHalf an equilateral triangle

Show that $\sin \frac{π}{3} = \frac{\sqrt{3}}{2}$ and $\cos \frac{π}{3} = \frac{1}{2}$ .

Outline

Cut an equilateral triangle of side 1 in half.
Why?
Its angles are all $\frac{π}{3}$ ; the altitude bisects the base, giving a right triangle with hypotenuse 1, short side $\frac{1}{2}$ , and angle $\frac{π}{3}$ opposite the long leg.
The long leg is $\frac{\sqrt{3}}{2}$ .
Why?
Pythagoras: $\sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$ .
Read off sine and cosine.
Why?
Opposite over hypotenuse $= \frac{\sqrt{3}}{2}$ ; adjacent over hypotenuse $= \frac{1}{2}$ .

A reference angleFourth quadrant

Find $\sin \frac{11 π}{6}$ and $\cos \frac{11 π}{6}$ .

Outline

$\frac{11 π}{6} = 2 π - \frac{π}{6}$ : fourth quadrant, reference angle $\frac{π}{6}$ .
Why?
It is $\frac{π}{6}$ short of a full turn, below the positive $x$ -axis.
$\sin = - \frac{1}{2}$ , $\cos = \frac{\sqrt{3}}{2}$ .
Why?
Below the axis $y < 0$ , right of it $x > 0$ ; the sizes come from $\frac{π}{6}$ .

InteractiveBrown et al., §§13-6–13-9

Trigonometric identities

Almost everything follows from two facts: the point $(\cos θ, \sin θ)$ is on the unit circle, and the addition formulas. The traps are the same as in algebra: $\sin$ of a sum is not the sum of the sines.

The idea

The Pythagorean identity (§13-6).

\sin^{2} θ + \cos^{2} θ = 1

, because

(\cos θ, \sin θ)

lies on

x^{2} + y^{2} = 1

. Dividing by

\cos^{2} θ

or

\sin^{2} θ

:

1 + \tan^{2} θ = \sec^{2} θ, 1 + \cot^{2} θ = \csc^{2} θ .

(Notation:

\sin^{2} θ

means

(\sin θ)^{2}

, not

\sin (θ^{2})

.)

Addition formulas (§13-7).

\sin (α \pm β) = \sin α \cos β \pm \cos α \sin β, \cos (α \pm β) = \cos α \cos β \mp \sin α \sin β .

Double and half angles (§13-8). Put

β = α

:

\sin 2 θ = 2 \sin θ \cos θ, \cos 2 θ = \cos^{2} θ - \sin^{2} θ = 1 - 2 \sin^{2} θ = 2 \cos^{2} θ - 1.

Solving the last two for the squares gives the forms calculus uses to integrate

\sin^{2}

and

\cos^{2}

:

\sin^{2} θ = \frac{1 - \cos 2 θ}{2}, \cos^{2} θ = \frac{1 + \cos 2 θ}{2} .

How to prove an identity. Work on one side only, usually the more complicated one, rewriting with the facts above until it becomes the other side. Writing everything in sines and cosines is a good default. To disprove one, a single angle is enough; the checker below tries many.

Explore: is this identity true?

Worked example

Prove $\frac{1 - \cos 2 x}{\sin 2 x} = \tan x$ (where both sides are defined).

Rewrite the top with a double-angle formula.

$1 - \cos 2 x = 1 - (1 - 2 \sin^{2} x) = 2 \sin^{2} x$ .
Rewrite the bottom.

$\sin 2 x = 2 \sin x \cos x$ .
Cancel.

$\frac{2 \sin^{2} x}{2 \sin x \cos x} = \frac{\sin x}{\cos x} = \tan x$ , cancelling the common factor $2 \sin x \neq 0$ .

Check your understanding

$\sin 2 x$ equals:

$2 \sin x$ At $x = \frac{π}{2}$ : $\sin π = 0$ but $2 \sin \frac{π}{2} = 2$ .
$2 \sin x \cos x$

The 2 is inside the function; it can't be pulled out. The addition formula with

α = β = x

gives the right answer.

$\sec^{2} θ - \tan^{2} θ$ equals:

1
0They differ by exactly 1: $\sec^{2} = 1 + \tan^{2}$ .

Divide

\sin^{2} + \cos^{2} = 1

by

\cos^{2}

. This is the identity behind the substitution

x = \tan θ

for integrals with

\sqrt{1 + x^{2}}

.

Is $\sin (α + β) = \sin α + \sin β$ ?

Yes, since sine is a function.Being a function says nothing about sums.
No: $α = β = \frac{π}{2}$ gives $0$ on the left and $2$ on the right.

The correct formula has four factors:

\sin α \cos β + \cos α \sin β

.

Practice

A product of squaresFor integration later

Write $\sin^{2} x \cos^{2} x$ using $\cos 4 x$ and no powers.

Outline

$\sin^{2} x \cos^{2} x = \frac{1}{4} \sin^{2} 2 x$ .
Why?
$\sin x \cos x = \frac{1}{2} \sin 2 x$ ; square it.
$\sin^{2} 2 x = \frac{1 - \cos 4 x}{2}$ .
Why?
The half-angle formula $\sin^{2} θ = \frac{1 - \cos 2 θ}{2}$ with $θ = 2 x$ .
So $\sin^{2} x \cos^{2} x = \frac{1 - \cos 4 x}{8}$ .
Why?
Multiply: $\frac{1}{4} \cdot \frac{1 - \cos 4 x}{2}$ . Check at $x = \frac{π}{4}$ : $\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$ and $\frac{1 - \cos π}{8} = \frac{2}{8}$ . ✓

The tangent of a sumFrom the addition formulas

Show $\tan (α + β) = \frac{\tan α + \tan β}{1 - \tan α \tan β}$ .

Outline

Write $\tan (α + β) = \frac{\sin (α + β)}{\cos (α + β)}$ and expand both.
Why?
$\frac{\sin α \cos β + \cos α \sin β}{\cos α \cos β - \sin α \sin β}$ .
Divide top and bottom by $\cos α \cos β$ .
Why?
Each $\frac{\sin}{\cos}$ becomes a tangent: the top becomes $\tan α + \tan β$ and the bottom $1 - \tan α \tan β$ .

InteractiveBrown et al., §§14-6–14-7

Inverse trigonometric functions

$\arcsin x$ is an angle: the one whose sine is $x$ . Because many angles share a sine, each inverse function picks one, from a fixed range. To simplify $\tan (\arccos u)$ and its relatives, draw the angle as a triangle.

The idea

The three inverse functions (§§14-6, 14-7). Each inverts the function on a range where it is one-to-one:

$\arcsin x$ (also written $\sin^{- 1} x$ ): for $x \in [- 1, 1]$ , the angle in $[- \frac{π}{2}, \frac{π}{2}]$ whose sine is $x$ .
$\arccos x$ : for $x \in [- 1, 1]$ , the angle in $[0, π]$ whose cosine is $x$ .
$\arctan x$ : for every real $x$ , the angle in $(- \frac{π}{2}, \frac{π}{2})$ whose tangent is $x$ .

Note

\sin^{- 1} x

is not

\frac{1}{\sin x}

(that is

\csc x

).

Undoing works one way. $\sin (\arcsin x) = x$ for every $x \in [- 1, 1]$ . But $\arcsin (\sin θ) = θ$ only when $θ$ is in the range $[- \frac{π}{2}, \frac{π}{2}]$ : $\arcsin (\sin \frac{3 π}{4}) = \arcsin \frac{\sqrt{2}}{2} = \frac{π}{4}$ .

The triangle method. To simplify

f (\arccos u)

, call the angle

θ = \arccos u

, so

\cos θ = u = \frac{u}{1}

. Draw a right triangle with adjacent side

u

and hypotenuse 1; Pythagoras gives the opposite side

\sqrt{1 - u^{2}}

. Then read off any function of

θ

:

\sin (\arccos u) = \sqrt{1 - u^{2}}, \tan (\arccos u) = \frac{\sqrt{1 - u^{2}}}{u}, \cos (\arcsin u) = \sqrt{1 - u^{2}}, \sin (\arctan u) = \frac{u}{\sqrt{1 + u^{2}}} .

The signs are right for every allowed

u

, because each range was chosen to make the root nonnegative where it appears.

These are exactly the simplifications needed after a trigonometric substitution in an integral.

Explore

Outer function

Inner function

u

The angle

θ

from the inner function, drawn as a right triangle. The side the inner function fixes is in gold; the side found by Pythagoras is in blue.

Worked example

Simplify $\tan (\arccos \frac{x + 1}{2})$ and $\sin (\arctan (x - 4))$ .

For the first, set up the triangle.

$θ = \arccos \frac{x + 1}{2}$ , so $\cos θ = \frac{x + 1}{2}$ : adjacent $x + 1$ , hypotenuse 2. Opposite: $\sqrt{4 - (x + 1)^{2}} = \sqrt{3 - 2 x - x^{2}}$ .
Read off the tangent.

$\tan θ = \frac{\sqrt{3 - 2 x - x^{2}}}{x + 1}$ , for $- 3 \leq x \leq 1$ , $x \neq - 1$ .
For the second, $θ = \arctan (x - 4)$ .

$\tan θ = \frac{x - 4}{1}$ : opposite $x - 4$ , adjacent 1, hypotenuse $\sqrt{1 + (x - 4)^{2}}$ . So $\sin θ = \frac{x - 4}{\sqrt{1 + (x - 4)^{2}}}$ , for every real $x$ .

Check your understanding

$\arcsin (\sin \frac{3 π}{4})$ equals:

$\frac{3 π}{4}$ $\frac{3 π}{4}$ is outside arcsin's range $[- \frac{π}{2}, \frac{π}{2}]$ .
$\frac{π}{4}$

\sin \frac{3 π}{4} = \frac{\sqrt{2}}{2}

, and the angle in

[- \frac{π}{2}, \frac{π}{2}]

with that sine is

\frac{π}{4}

.

$\cos (\arcsin u)$ for $u \in [- 1, 1]$ equals:

$\sqrt{1 - u^{2}}$
$\pm \sqrt{1 - u^{2}}$ $\arcsin u$ is in $[- \frac{π}{2}, \frac{π}{2}]$ , where cosine is never negative.

The range of arcsin was chosen in the right half of the circle, so the cosine is

\geq 0

and only the positive root appears.

$\sin^{- 1} x$ means:

$\arcsin x$ , an angle.
$\frac{1}{\sin x}$ That is $\csc x$ , written $(\sin x)^{- 1}$ .

An unfortunate notation:

\sin^{2} x

means

(\sin x)^{2}

, but

\sin^{- 1} x

means the inverse function. Writing

\arcsin

avoids the confusion.

Practice

Cosine of an arctangentThe triangle method

Show $\cos (\arctan x) = \frac{1}{\sqrt{1 + x^{2}}}$ for every real $x$ .

Outline

Triangle: opposite $x$ , adjacent 1, hypotenuse $\sqrt{1 + x^{2}}$ .
Why?
$\tan θ = \frac{x}{1}$ with $θ = \arctan x$ ; Pythagoras gives the hypotenuse.
So $\cos θ = \frac{1}{\sqrt{1 + x^{2}}}$ , positive.
Why?
Adjacent over hypotenuse; and $θ \in (- \frac{π}{2}, \frac{π}{2})$ , where cosine is positive, so no sign is lost.

Two inverse functions at onceAn addition formula

Show $\arcsin x + \arccos x = \frac{π}{2}$ for $x \in [- 1, 1]$ .

Outline

Let $θ = \arcsin x$ . Then $\frac{π}{2} - θ \in [0, π]$ .
Why?
$θ \in [- \frac{π}{2}, \frac{π}{2}]$ .
$\cos (\frac{π}{2} - θ) = \sin θ = x$ .
Why?
The addition formula: $\cos (\frac{π}{2} - θ) = \cos \frac{π}{2} \cos θ + \sin \frac{π}{2} \sin θ = \sin θ$ .
So $\frac{π}{2} - θ = \arccos x$ .
Why?
It is the angle in $[0, π]$ whose cosine is $x$ , which is the definition of $\arccos x$ .

InteractiveBrown et al., §§11-1–11-6

Sequences and their limits

A sequence has limit $L$ if its terms eventually stay as close to $L$ as anyone could ask. Making “as close as anyone could ask” precise is the first step from algebra into calculus.

The idea

Sequences (§§11-1–11-3). A sequence is an infinite list

a_{1}, a_{2}, a_{3}, \dots

, usually given by a formula for

a_{n}

. Arithmetic: add the same

d

each time,

a_{n} = a_{1} + (n - 1) d

. Geometric: multiply by the same

r

,

a_{n} = a_{1} r^{n - 1}

.

The idea of a limit. $a_{n} = \frac{1}{n}$ gets closer and closer to 0. But “closer and closer” isn't enough: $1 + \frac{1}{n}$ also gets closer and closer to 0 at every step, and its limit is 1. What we mean is that the terms get and stay within any tolerance of $L$ .

Limit of a sequence.

lim_{n \to \infty} a_{n} = L

means: for every tolerance

ε > 0

, there is a point

N

such that every term after it is within

ε

of

L

:

| a_{n} - L | < ε for all n > N .

Picture a band of half-width

ε

around

L

. However narrow the band, from some point on the whole tail of the sequence lies inside it. A smaller

ε

usually needs a larger

N

.

Working out limits.

Basic limits: $\frac{1}{n} \to 0$ ; $r^{n} \to 0$ if $| r | < 1$ ; $r^{n}$ grows without bound if $r > 1$ , and has no limit if $r \leq - 1$ .
Limits respect sums, products and quotients (if the bottom's limit isn't 0).
Dominant terms: divide top and bottom by the highest power of $n$ in the denominator. $\frac{3 n^{2} + 1}{n^{2} - 5} = \frac{3 + 1 / n^{2}}{1 - 5 / n^{2}} \to 3$ .
Squeeze: if $b_{n} \leq a_{n} \leq c_{n}$ and $b_{n}, c_{n} \to L$ , then $a_{n} \to L$ . For example $- \frac{1}{n} \leq \frac{(- 1)^{n}}{n} \leq \frac{1}{n}$ , so $\frac{(- 1)^{n}}{n} \to 0$ .

Infinite geometric series (§11-6). The partial sums

S_{n} = a + a r + \dots + a r^{n - 1} = a \frac{1 - r^{n}}{1 - r}

form a sequence, and if

| r | < 1

then

r^{n} \to 0

, so

a + a r + a r^{2} + \dots = lim_{n \to \infty} S_{n} = \frac{a}{1 - r} .

For example

0.999 \dots = \frac{9}{10} + \frac{9}{100} + \dots = \frac{9 / 10}{1 - 1 / 10} = 1

.

The same definition, written for functions, is the $ε$ – $δ$ definition of continuity in Real Analysis, and its sequences module takes this one further.

Explore

Sequence

Tolerance ε

The terms

a_{n}

(dots), the band

L \pm ε

(gold), and the point

N

after which every term stays inside (dashed). Terms outside the band are red.

Worked example

Show that $a_{n} = \frac{2 n + 1}{n + 3}$ has limit 2, and find $N$ for $ε = 0.1$ .

Guess the limit with dominant terms.

$\frac{2 n + 1}{n + 3} = \frac{2 + 1 / n}{1 + 3 / n} \to \frac{2}{1} = 2$ .
Measure the distance to 2 exactly.

$a_{n} - 2 = \frac{2 n + 1 - 2 (n + 3)}{n + 3} = \frac{- 5}{n + 3}$ , so $| a_{n} - 2 | = \frac{5}{n + 3}$ .
When is that less than $ε$ ?

$\frac{5}{n + 3} < ε ⟺ n > \frac{5}{ε} - 3$ . For $ε = 0.1$ : $n > 47$ , so $N = 47$ works. For any $ε$ , $N = \frac{5}{ε}$ works: that proves the limit.

Check your understanding

A sequence has its odd-numbered terms all positive and its even-numbered terms all negative. Must it diverge?

Yes: it keeps jumping across zero.The jumps can shrink.
No: $\frac{(- 1)^{n + 1}}{n}$ alternates in sign and has limit 0.

Alternating signs only force divergence if the terms stay away from 0. If they shrink to 0, the sequence converges to 0, as the squeeze argument shows.

$| a_{n} | \to 1$ . Does $a_{n}$ have a limit?

Yes, 1 or −1.It could switch between them forever.
Not necessarily: $(- 1)^{n}$ has $| a_{n} | = 1$ but no limit.

For the limit 0 it is different:

a_{n} \to 0

exactly when

| a_{n} | \to 0

, because

| a_{n} - 0 | = | | a_{n} | - 0 |

. The special role of 0 is the point.

$0.999 \dots$ (nines forever) equals:

1
Slightly less than 1.There is no positive gap: any gap would be bigger than $10^{- n}$ for some $n$ , but the difference after $n$ nines is exactly $10^{- n}$ .

It is the infinite geometric series with

a = 0.9

,

r = 0.1

, whose sum is

\frac{0.9}{0.9} = 1

: the limit of

0.9, 0.99, 0.999, \dots

.

Practice

Squeeze to infinityA lower bound is enough

Show that $a_{n} = \frac{n^{2} (2 + \sin^{2} n)}{n + 100}$ grows without bound.

Outline

$a_{n} \geq \frac{2 n^{2}}{n + 100}$ .
Why?
$\sin^{2} n \geq 0$ , so $2 + \sin^{2} n \geq 2$ .
$\frac{2 n^{2}}{n + 100} \geq n$ once $n \geq 100$ .
Why?
$2 n^{2} \geq n (n + 100) ⟺ n^{2} \geq 100 n ⟺ n \geq 100$ .
So $a_{n} \geq n$ for $n \geq 100$ , and $a_{n} \to \infty$ .
Why?
A sequence that stays above one that grows without bound does too: for any $M$ , $a_{n} > M$ once $n > max (M, 100)$ .

A repeating decimalGeometric series

Write $0. \overset{―}{27} = 0.272727 \dots$ as a fraction.

Outline

It is a geometric series with $a = \frac{27}{100}$ , $r = \frac{1}{100}$ .
Why?
$0.27 + 0.0027 + 0.000027 + \dots$ : each block is the previous one divided by 100.
Its sum is $\frac{27 / 100}{99 / 100} = \frac{27}{99} = \frac{3}{11}$ .
Why?
$\frac{a}{1 - r}$ , valid since $| r | < 1$ . Check: $3 \div 11 = 0.2727 \dots$