0.1 Proofs, identities, and toolkit functions

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Appendix

Important proofs and derivations

Product Rule

$\log_{a} x y = \log_{a} x + \log_{a} y$

Proof:

Let $m = \log_{a} x$ and $n = \log_{a} y .$

Write in exponent form.

$x = a^{m}$ and $y = a^{n} .$

Multiply.

$x y = a^{m} a^{n} = a^{m + n}$

$\begin{matrix} a^{m + n} & = & x y \\ \log_{a} (x y) & = & m + n \\ = & \log_{a} x + \log_{b} y \end{matrix}$

Change of Base Rule

$\begin{array}{l} \log_{a} b = \frac{\log_{c} b}{\log_{c} a} \\ \log_{a} b = \frac{1}{\log_{b} a} \end{array}$

where $x$ and $y$ are positive, and $a > 0, a \neq 1.$

Proof:

Let $x = \log_{a} b .$

Write in exponent form.

$a^{x} = b$

Take the $\log_{c}$ of both sides.

$\begin{matrix} \log_{c} a^{x} & = & \log_{c} b \\ x \log_{c} a & = & \log_{c} b \\ x & = & \frac{\log_{c} b}{\log_{c} a} \\ \log_{a} b & = & \frac{\log_{c} b}{\log_{a} b} \end{matrix}$

When $c = b,$

$\log_{a} b = \frac{\log_{b} b}{\log_{b} a} = \frac{1}{\log_{b} a}$

Heron’s Formula

$A = \sqrt{s (s - a) (s - b) (s - c)}$

where $s = \frac{a + b + c}{2}$

Proof:

Let $a,$ $b,$ and $c$ be the sides of a triangle, and $h$ be the height.

A triangle with sides labeled: a, b and c. A line runs through the center of the triangle, bisecting the top angle; this line is labeled: h.

So $s = \frac{a + b + c}{2}$ .

We can further name the parts of the base in each triangle established by the height such that $p + q = c .$

A triangle with sides labeled: a, b, and c. A line runs through the center of the triangle bisecting the angle at the top; this line is labeled: h. The two new line segments on the base of the triangle are labeled: p and q.

Using the Pythagorean Theorem, $h^{2} + p^{2} = a^{2}$ and $h^{2} + q^{2} = b^{2} .$

Since $q = c - p,$ then $q^{2} = {(c - p)}^{2} .$ Expanding, we find that $q^{2} = c^{2} - 2 c p + p^{2} .$

We can then add $h^{2}$ to each side of the equation to get $h^{2} + q^{2} = h^{2} + c^{2} - 2 c p + p^{2} .$

Substitute this result into the equation $h^{2} + q^{2} = b^{2}$ yields $b^{2} = h^{2} + c^{2} - 2 c p + p^{2} .$

Then replacing $h^{2} + p^{2}$ with $a^{2}$ gives $b^{2} = a^{2} - 2 c p + c^{2} .$

Solve for $p$ to get

$p = \frac{a^{2} + b^{2} - c^{2}}{2 c}$

Since $h^{2} = a^{2} - p^{2},$ we get an expression in terms of $a,$ $b,$ and $c .$

$\begin{matrix} h^{2} & = & a^{2} - p^{2} \\ = & (a + p) (a - p) \\ = & [a + \frac{(a^{2} + c^{2} - b^{2})}{2 c}] [a - \frac{(a^{2} + c^{2} - b^{2})}{2 c}] \\ = & \frac{(2 a c + a^{2} + c^{2} - b^{2}) (2 a c - a^{2} - c^{2} + b^{2})}{4 c^{2}} \\ = & \frac{({(a + c)}^{2} - b^{2}) (b^{2} - {(a - c)}^{2})}{4 c^{2}} \\ = & \frac{(a + b + c) (a + c - b) (b + a - c) (b - a + c)}{4 c^{2}} \\ = & \frac{(a + b + c) (- a + b + c) (a - b + c) (a + b - c)}{4 c^{2}} \\ = & \frac{2 s \cdot (2 s - a) \cdot (2 s - b) (2 s - c)}{4 c^{2}} \end{matrix}$

Therefore,

$\begin{matrix} h^{2} & = & \frac{4 s (s - a) (s - b) (s - c)}{c^{2}} \\ h & = & \frac{2 \sqrt{s (s - a) (s - b) (s - c)}}{c} \end{matrix}$

And since $A = \frac{1}{2} c h,$ then

$\begin{matrix} A & = & \frac{1}{2} c \frac{2 \sqrt{s (s - a) (s - b) (s - c)}}{c} \\ = & \sqrt{s (s - a) (s - b) (s - c)} \end{matrix}$

Properties of the Dot Product

$u \cdot v = v \cdot u$

Proof:

$\begin{matrix} u \cdot v & = ⟨ u_{1}, u_{2}, ... u_{n} ⟩ \cdot ⟨ v_{1}, v_{2}, ... v_{n} ⟩ \\ = u_{1} v_{1} + u_{2} v_{2} + ... + u_{n} v_{n} \\ = v_{1} u_{1} + v_{2} u_{2} + ... + v_{n} v_{n} \\ = ⟨ v_{1}, v_{2}, ... v_{n} ⟩ \cdot ⟨ u_{1}, u_{2}, ... u_{n} ⟩ \\ = v \cdot u \end{matrix}$

$u \cdot (v + w) = u \cdot v + u \cdot w$

Proof:

$\begin{matrix} u \cdot (v + w) & = ⟨ u_{1}, u_{2}, ... u_{n} ⟩ \cdot (⟨ v_{1}, v_{2}, ... v_{n} ⟩ + ⟨ w_{1}, w_{2}, ... w_{n} ⟩) \\ = ⟨ u_{1}, u_{2}, ... u_{n} ⟩ \cdot ⟨ v_{1} + w_{1}, v_{2} + w_{2}, ... v_{n} + w_{n} ⟩ \\ = ⟨ u_{1} (v_{1} + w_{1}), u_{2} (v_{2} + w_{2}), ... u_{n} (v_{n} + w_{n}) ⟩ \\ = ⟨ u_{1} v_{1} + u_{1} w_{1}, u_{2} v_{2} + u_{2} w_{2}, ... u_{n} v_{n} + u_{n} w_{n} ⟩ \\ = ⟨ u_{1} v_{1}, u_{2} v_{2}, ..., u_{n} v_{n} ⟩ + ⟨ u_{1} w_{1}, u_{2} w_{2}, ..., u_{n} w_{n} ⟩ \\ = ⟨ u_{1}, u_{2}, ... u_{n} ⟩ \cdot ⟨ v_{1}, v_{2}, ... v_{n} ⟩ + ⟨ u_{1}, u_{2}, ... u_{n} ⟩ \cdot ⟨ w_{1}, w_{2}, ... w_{n} ⟩ \\ = u \cdot v + u \cdot w \end{matrix}$

$u \cdot u = {| u |}^{2}$

Proof:

$\begin{matrix} u \cdot u & = ⟨ u_{1}, u_{2}, ... u_{n} ⟩ \cdot ⟨ u_{1}, u_{2}, ... u_{n} ⟩ \\ = u_{1} u_{1} + u_{2} u_{2} + ... + u_{n} u_{n} \\ = u_{1}^{2} + u_{2}^{2} + ... + u_{n}^{2} \\ = | ⟨ u_{1}, u_{2}, ... u_{n} ⟩ |^{2} \\ = v \cdot u \end{matrix}$

Standard Form of the Ellipse centered at the Origin

$1 = \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}$

Derivation

An ellipse consists of all the points for which the sum of distances from two foci is constant:

$\sqrt{{(x - (- c))}^{2} + {(y - 0)}^{2}} + \sqrt{{(x - c)}^{2} + {(y - 0)}^{2}} = constant$

An ellipse centered at the origin on an x, y-coordinate plane. Points C1 and C2 are plotted at the points (0, b) and (0, -b) respectively; these points appear on the ellipse. Points V1 and V2 are plotted at the points (-a, 0) and (a, 0) respectively; these points appear on the ellipse. Points F1 and F2 are plotted at the points (-c, 0) and (c, 0) respectively; these points appear on the x-axis, but not the ellipse. The point (x, y) appears on the ellipse in the first quadrant. Dotted lines extend from F1 and F2 to the point (x, y).

Consider a vertex.

An ellipse centered at the origin. The points C1 and C2 are plotted at the points (0, b) and (0, -b) respectively; these points are on the ellipse. The points V1 and V2 are plotted at the points (-a, 0) and (a, 0) respectively; these points are on the ellipse. The points F1 and F2 are plotted at the points (-c, 0) and (c, 0) respectively; these points are on the x-axis and not on the ellipse. A line extends from the point F1 to a point (x, y) which is at the point (a, 0). A line extends from the point F2 to the point (x, y) as well.

Then, $\sqrt{{(x - (- c))}^{2} + {(y - 0)}^{2}} + \sqrt{{(x - c)}^{2} + {(y - 0)}^{2}} = 2 a$

Consider a covertex.

Then $b^{2} + c^{2} = a^{2} .$

$\begin{matrix} \sqrt{{(x - (- c))}^{2} + {(y - 0)}^{2}} + \sqrt{{(x - c)}^{2} + {(y - 0)}^{2}} & = & 2 a \\ \sqrt{{(x + c)}^{2} + y^{2}} & = & 2 a - \sqrt{{(x - c)}^{2} + y^{2}} \\ {(x + c)}^{2} + y^{2} & = & {(2 a - \sqrt{{(x - c)}^{2} + y^{2}})}^{2} \\ x^{2} + 2 c x + c^{2} + y^{2} & = & 4 a^{2} - 4 a \sqrt{{(x - c)}^{2} + y^{2}} + {(x - c)}^{2} + y^{2} \\ x^{2} + 2 c x + c^{2} + y^{2} & = & 4 a^{2} - 4 a \sqrt{{(x - c)}^{2} + y^{2}} + x^{2} - 2 c x + y^{2} \\ 2 c x & = & 4 a^{2} - 4 a \sqrt{{(x - c)}^{2} + y^{2}} - 2 c x \\ 4 c x - 4 a^{2} & = & 4 a \sqrt{{(x - c)}^{2} + y^{2}} \\ - \frac{1}{4 a} (4 c x - 4 a^{2}) & = & \sqrt{{(x - c)}^{2} + y^{2}} \\ a - \frac{c}{a} x & = & \sqrt{{(x - c)}^{2} + y^{2}} \\ a^{2} - 2 x c + \frac{c^{2}}{a^{2}} x^{2} & = & {(x - c)}^{2} + y^{2} \\ a^{2} - 2 x c + \frac{c^{2}}{a^{2}} x^{2} & = & x^{2} - 2 x c + c^{2} + y^{2} \\ a^{2} + \frac{c^{2}}{a^{2}} x^{2} & = & x^{2} + c^{2} + y^{2} \\ a^{2} + \frac{c^{2}}{a^{2}} x^{2} & = & x^{2} + c^{2} + y^{2} \\ a^{2} - c^{2} & = & x^{2} - \frac{c^{2}}{a^{2}} x^{2} + y^{2} \\ a^{2} - c^{2} & = & x^{2} (1 - \frac{c^{2}}{a^{2}}) + y^{2} \end{matrix}$

Let $1 = \frac{a^{2}}{a^{2}} .$

\begin{matrix} a^{2} - c^{2} & = & x^{2} (\frac{a^{2} - c^{2}}{a^{2}}) + y^{2} \\ 1 & = & \frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2} - c^{2}} \end{matrix}

Because $b^{2} + c^{2} = a^{2},$ then $b^{2} = a^{2} - c^{2} .$

\begin{matrix} 1 & = & \frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2} - c^{2}} \\ 1 & = & \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} \end{matrix}

Standard Form of the Hyperbola

$1 = \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}}$

Derivation

A hyperbola is the set of all points in a plane such that the absolute value of the difference of the distances between two fixed points is constant.

Side-by-side graphs of hyperbole. In Diagram 1: The foci F’ and F are labeled and can be found a little in front of the opening of the hyperbola. A point P at (x,y) on the right curve is labeled. A line extends from the F’ focus to the point P labeled: D1. A line extends from the F focus to the point P labeled: D2. In Diagram 2: The foci F’ and F are labeled and can be found a little in front of the opening of the hyperbola. A point V is labeled at the vertex of the right hyperbola. A line extends from the F’ focus to the point V labeled: D1. A line extends from the F focus to the point V labeled: D2.

Diagram 1: The difference of the distances from Point P to the foci is constant:

$\sqrt{{(x - (- c))}^{2} + {(y - 0)}^{2}} - \sqrt{{(x - c)}^{2} + {(y - 0)}^{2}} = constant$

Diagram 2: When the point is a vertex, the difference is $2 a .$

$\sqrt{{(x - (- c))}^{2} + {(y - 0)}^{2}} - \sqrt{{(x - c)}^{2} + {(y - 0)}^{2}} = 2 a$

$\begin{matrix} \sqrt{{(x - (- c))}^{2} + {(y - 0)}^{2}} - \sqrt{{(x - c)}^{2} + {(y - 0)}^{2}} & = & 2 a \\ \sqrt{{(x + c)}^{2} + y^{2}} - \sqrt{{(x - c)}^{2} + y^{2}} & = & 2 a \\ \sqrt{{(x + c)}^{2} + y^{2}} & = & 2 a + \sqrt{{(x - c)}^{2} + y^{2}} \\ {(x + c)}^{2} + y^{2} & = & (2 a + \sqrt{{(x - c)}^{2} + y^{2}}) \\ x^{2} + 2 c x + c^{2} + y^{2} & = & 4 a^{2} + 4 a \sqrt{{(x - c)}^{2}} + y^{2} \\ x^{2} + 2 c x + c^{2} + y^{2} & = & 4 a^{2} + 4 a \sqrt{{(x - c)}^{2} + y^{2}} + x^{2} - 2 c x + y^{2} \\ 2 c x & = & 4 a^{2} + 4 a \sqrt{{(x - c)}^{2} + y^{2}} - 2 c x \\ 4 c x - 4 a^{2} & = & 4 a \sqrt{{(x - c)}^{2} + y^{2}} \\ c x - a^{2} & = & a \sqrt{{(x - c)}^{2} + y^{2}} \\ {(c x - a^{2})}^{2} & = & a^{2} ({(x - c)}^{2} + y^{2}) \\ c^{2} x^{2} - 2 a^{2} c^{2} x^{2} + a^{4} & = & a^{2} x^{2} - 2 a^{2} c^{2} x^{2} + a^{2} c^{2} + a^{2} y^{2} \\ c^{2} x^{2} + a^{4} & = & a^{2} x^{2} + a^{2} c^{2} + a^{2} y^{2} \\ a^{4} - a^{2} c^{2} & = & a^{2} x^{2} - c^{2} x^{2} + a^{2} y^{2} \\ a^{2} (a^{2} - c^{2}) & = & (a^{2} - c^{2}) x^{2} + a^{2} y^{2} \\ a^{2} (a^{2} - c^{2}) & = & (c^{2} - a^{2}) x^{2} - a^{2} y^{2} \end{matrix}$

Define $b$ as a positive number such that $b^{2} = c^{2} - a^{2} .$

\begin{matrix} a^{2} b^{2} & = & b^{2} x^{2} - a^{2} y^{2} \\ \frac{a^{2} b^{2}}{a^{2} b^{2}} & = & \frac{b^{2} x^{2}}{a^{2} b^{2}} - \frac{a^{2} y^{2}}{a^{2} b^{2}} \\ 1 & = & \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} \end{matrix}

Trigonometric identities

Pythagorean Identity	$\begin{array}{l} \cos^{2} t + \sin^{2} t = 1 \\ 1 + \tan^{2} t = \sec^{2} t \\ 1 + \cot^{2} t = \csc^{2} t \end{array}$
Even-Odd Identities	$\begin{array}{l} \cos (- t) = c o s t \\ \sec (- t) = \sec t \\ \sin (- t) = - \sin t \\ \tan (- t) = - \tan t \\ \csc (- t) = - \csc t \\ \cot (- t) = - \cot t \end{array}$
Cofunction Identities	$\begin{array}{l} \cos t = \sin (\frac{π}{2} - t) \\ \sin t = \cos (\frac{π}{2} - t) \\ \tan t = \cot (\frac{π}{2} - t) \\ \cot t = \tan (\frac{π}{2} - t) \\ \sec t = \csc (\frac{π}{2} - t) \\ \csc t = \sec (\frac{π}{2} - t) \end{array}$
Fundamental Identities	$\begin{array}{l} \tan t = \frac{\sin t}{\cos t} \\ \sec t = \frac{1}{\cos t} \\ \csc t = \frac{1}{\sin t} \\ c o t t = \frac{1}{tan t} = \frac{cos t}{sin t} \end{array}$
Sum and Difference Identities	$\begin{array}{l} \cos (α + β) = \cos α \cos β - \sin α \sin β \\ \cos (α - β) = \cos α \cos β + \sin α \sin β \\ \sin (α + β) = \sin α \cos β + \cos α \sin β \\ \sin (α - β) = \sin α \cos β - \cos α \sin β \\ \tan (α + β) = \frac{\tan α + \tan β}{1 - \tan α \tan β} \\ \tan (α - β) = \frac{\tan α - \tan β}{1 + \tan α \tan β} \end{array}$
Double-Angle Formulas	$\begin{array}{l} \sin (2 θ) = 2 \sin θ \cos θ \\ \cos (2 θ) = \cos^{2} θ - \sin^{2} θ \\ \cos (2 θ) = 1 - 2 \sin^{2} θ \\ \cos (2 θ) = 2 \cos^{2} θ - 1 \\ \tan (2 θ) = \frac{2 \tan θ}{1 - \tan^{2} θ} \end{array}$
Half-Angle Formulas	$\begin{array}{l} \sin \frac{α}{2} = \pm \sqrt{\frac{1 - \cos α}{2}} \\ \cos \frac{α}{2} = \pm \sqrt{\frac{1 + \cos α}{2}} \\ \tan \frac{α}{2} = \pm \sqrt{\frac{1 - \cos α}{1 + \cos α}} \\ \tan \frac{α}{2} = \frac{\sin α}{1 + \cos α} \\ \tan \frac{α}{2} = \frac{1 - \cos α}{\sin α} \end{array}$
Reduction Formulas	$\begin{array}{l} \sin^{2} θ = \frac{1 - \cos (2 θ)}{2} \\ \cos^{2} θ = \frac{1 + \cos (2 θ)}{2} \\ \tan^{2} θ = \frac{1 - \cos (2 θ)}{1 + \cos (2 θ)} \end{array}$
Product-to-Sum Formulas	$\begin{array}{l} \cos α \cos β = \frac{1}{2} [\cos (α - β) + \cos (α + β)] \\ \sin α \cos β = \frac{1}{2} [\sin (α + β) + \sin (α - β)] \\ \sin α \sin β = \frac{1}{2} [\cos (α - β) - \cos (α + β)] \\ \cos α \sin β = \frac{1}{2} [\sin (α + β) - \sin (α - β)] \end{array}$
Sum-to-Product Formulas	$\begin{array}{l} \sin α + \sin β = 2 \sin (\frac{α + β}{2}) \cos (\frac{α - β}{2}) \\ \sin α - \sin β = 2 \sin (\frac{α - β}{2}) \cos (\frac{α + β}{2}) \\ \cos α - \cos β = - 2 \sin (\frac{α + β}{2}) \sin (\frac{α - β}{2}) \\ \cos α + \cos β = 2 \cos (\frac{α + β}{2}) \cos (\frac{α - β}{2}) \end{array}$
Law of Sines	$\begin{array}{l} \frac{\sin α}{a} = \frac{\sin β}{b} = \frac{\sin γ}{c} \\ \frac{a}{\sin α} = \frac{b}{\sin β} = \frac{c}{\sin γ} \end{array}$
Law of Cosines	$\begin{matrix} a^{2} = b^{2} + c^{2} - 2 b c \cos α \\ b^{2} = a^{2} + c^{2} - 2 a c \cos β \\ c^{2} = a^{2} + b^{2} - 2 a b cos γ \end{matrix}$

Toolkit functions

Three graphs side-by-side. From left to right, graph of the identify function, square function, and square root function. All three graphs extend from -4 to 4 on each axis.

Three graphs side-by-side. From left to right, graph of the cubic function, cube root function, and reciprocal function. All three graphs extend from -4 to 4 on each axis.

Three graphs side-by-side. From left to right, graph of the absolute value function, exponential function, and natural logarithm function. All three graphs extend from -4 to 4 on each axis.

Trigonometric functions

Unit Circle

Angle	$0$	$\frac{π}{6}, or 30 °$	$\frac{π}{4}, or 45 °$	$\frac{π}{3}, or 60 °$	$\frac{π}{2}, or 90 °$
Cosine	1	$\frac{\sqrt{3}}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{1}{2}$	0
Sine	0	$\frac{1}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{3}}{2}$	1
Tangent	0	$\frac{\sqrt{3}}{3}$	1	$\sqrt{3}$	Undefined
Secant	1	$\frac{2 \sqrt{3}}{3}$	$\sqrt{2}$	2	Undefined
Cosecant	Undefined	2	$\sqrt{2}$	$\frac{2 \sqrt{3}}{3}$	1
Cotangent	Undefined	$\sqrt{3}$	1	$\frac{\sqrt{3}}{3}$	0

Questions & Answers

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