付録

Review of Pre-Calculus微積分の前提となる数学の復習

A = area,

V = Volume, and

S = lateral surface area

A = 面積

、

V = 体積,

、

S = 側面積

。

OpenStax, Rice University; Calculus Volume 1. · CC BY-NC-SA 4.0

OpenStax, Rice University; Calculus Volume 1. · CC BY-NC-SA 4.0

\begin{matrix} x^{m} x^{n} & = & x^{m + n} & \frac{x^{m}}{x^{n}} & = & x^{m - n} & {(x^{m})}^{n} & = & x^{m n} \\ x^{− n} & = & \frac{1}{x^{n}} & {(x y)}^{n} & = & x^{n} y^{n} & {(\frac{x}{y})}^{n} & = & \frac{x^{n}}{y^{n}} \\ x^{1 / n} & = & \sqrt[n]{x} & \sqrt[n]{x y} & = & \sqrt[n]{x} \sqrt[n]{y} & \sqrt[n]{\frac{x}{y}} & = & \frac{\sqrt[n]{x}}{\sqrt[n]{y}} \\ x^{m / n} & = & \sqrt[n]{x^{m}} = {(\sqrt[n]{x})}^{m} \end{matrix}

\begin{matrix} x^{m} x^{n} & = & x^{m + n} & \frac{x^{m}}{x^{n}} & = & x^{m - n} & {(x^{m})}^{n} & = & x^{m n} \\ x^{− n} & = & \frac{1}{x^{n}} & {(x y)}^{n} & = & x^{n} y^{n} & {(\frac{x}{y})}^{n} & = & \frac{x^{n}}{y^{n}} \\ x^{1 / n} & = & \sqrt[n]{x} & \sqrt[n]{x y} & = & \sqrt[n]{x} \sqrt[n]{y} & \sqrt[n]{\frac{x}{y}} & = & \frac{\sqrt[n]{x}}{\sqrt[n]{y}} \\ x^{m / n} & = & \sqrt[n]{x^{m}} = {(\sqrt[n]{x})}^{m} \end{matrix}

\begin{matrix} x^{2} - y^{2} & = & (x + y) (x - y) \\ x^{3} + y^{3} & = & (x + y) (x^{2} - x y + y^{2}) \\ x^{3} - y^{3} & = & (x - y) (x^{2} + x y + y^{2}) \end{matrix}

\begin{matrix} x^{2} - y^{2} & = & (x + y) (x - y) \\ x^{3} + y^{3} & = & (x + y) (x^{2} - x y + y^{2}) \\ x^{3} - y^{3} & = & (x - y) (x^{2} + x y + y^{2}) \end{matrix}

a x^{2} + b x + c = 0,

then

x = \frac{− b \pm \sqrt{b^{2} - 4 c a}}{2 a} .

英語のヒント

a x^{2} + b x + c = 0

ならば、

x = \frac{− b \pm \sqrt{b^{2} - 4 c a}}{2 a}

。

{(a + b)}^{n} = a^{n} + (\begin{array}{l} n \\ 1 \end{array}) a^{n - 1} b + (\begin{array}{l} n \\ 2 \end{array}) a^{n - 2} b^{2} + \dots + (\begin{matrix} n \\ n - 1 \end{matrix}) a b^{n - 1} + b^{n},

{(a + b)}^{n} = a^{n} + (\begin{array}{l} n \\ 1 \end{array}) a^{n - 1} b + (\begin{array}{l} n \\ 2 \end{array}) a^{n - 2} b^{2} + \dots + (\begin{matrix} n \\ n - 1 \end{matrix}) a b^{n - 1} + b^{n}

where

(\begin{array}{l} n \\ k \end{array}) = \frac{n (n - 1) (n - 2) \dots (n - k + 1)}{k (k - 1) (k - 2) \dots 3 \cdot 2 \cdot 1} = \frac{n!}{k! (n - k)!}

英語のヒント

ここで、

(\begin{array}{l} n \\ k \end{array}) = \frac{n (n - 1) (n - 2) \dots (n - k + 1)}{k (k - 1) (k - 2) \dots 3 \cdot 2 \cdot 1} = \frac{n!}{k! (n - k)!}

。

\begin{matrix} sin θ = \frac{opp}{hyp} & csc θ = \frac{hyp}{opp} \\ cos θ = \frac{adj}{hyp} & sec θ = \frac{hyp}{adj} \\ tan θ = \frac{opp}{adj} & cot θ = \frac{adj}{opp} \end{matrix}

\begin{matrix} sin θ = \frac{対辺}{斜辺} & csc θ = \frac{斜辺}{対辺} \\ cos θ = \frac{隣辺}{斜辺} & sec θ = \frac{斜辺}{隣辺} \\ tan θ = \frac{対辺}{隣辺} & cot θ = \frac{隣辺}{対辺} \end{matrix}

OpenStax, Rice University; Calculus Volume 1. · CC BY-NC-SA 4.0

\begin{matrix} {sin}^{2} θ + {cos}^{2} θ & = & 1 & sin (− θ) & = & − sin θ \\ 1 + {tan}^{2} θ & = & {sec}^{2} θ & cos (− θ) & = & cos θ \\ 1 + {cot}^{2} θ & = & {csc}^{2} θ & tan (− θ) & = & − tan θ \\ sin (\frac{π}{2} - θ) & = & cos θ & sin (θ + 2 π) & = & sin θ \\ cos (\frac{π}{2} - θ) & = & sin θ & cos (θ + 2 π) & = & cos θ \\ tan (\frac{π}{2} - θ) & = & cot θ & tan (θ + π) & = & tan θ \end{matrix}

\begin{matrix} {sin}^{2} θ + {cos}^{2} θ & = & 1 & sin (− θ) & = & − sin θ \\ 1 + {tan}^{2} θ & = & {sec}^{2} θ & cos (− θ) & = & cos θ \\ 1 + {cot}^{2} θ & = & {csc}^{2} θ & tan (− θ) & = & − tan θ \\ sin (\frac{π}{2} - θ) & = & cos θ & sin (θ + 2 π) & = & sin θ \\ cos (\frac{π}{2} - θ) & = & sin θ & cos (θ + 2 π) & = & cos θ \\ tan (\frac{π}{2} - θ) & = & cot θ & tan (θ + π) & = & tan θ \end{matrix}

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

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

OpenStax, Rice University; Calculus Volume 1. · CC BY-NC-SA 4.0

\begin{matrix} a^{2} & = & b^{2} + c^{2} - 2 b c cos A \\ b^{2} & = & a^{2} + c^{2} - 2 a c cos B \\ c^{2} & = & a^{2} + b^{2} - 2 a b cos C \end{matrix}

\begin{matrix} a^{2} & = & b^{2} + c^{2} - 2 b c cos A \\ b^{2} & = & a^{2} + c^{2} - 2 a c cos B \\ c^{2} & = & a^{2} + b^{2} - 2 a b cos C \end{matrix}

\begin{matrix} sin (x + y) & = & sin x cos y + cos x sin y \\ sin (x - y) & = & sin x cos y - cos x sin y \\ cos (x + y) & = & cos x cos y - sin x sin y \\ cos (x - y) & = & cos x cos y + sin x sin y \\ tan (x + y) & = & \frac{tan x + tan y}{1 - tan x tan y} \\ tan (x - y) & = & \frac{tan x - tan y}{1 + tan x tan y} \end{matrix}

\begin{matrix} sin (x + y) & = & sin x cos y + cos x sin y \\ sin (x - y) & = & sin x cos y - cos x sin y \\ cos (x + y) & = & cos x cos y - sin x sin y \\ cos (x - y) & = & cos x cos y + sin x sin y \\ tan (x + y) & = & \frac{tan x + tan y}{1 - tan x tan y} \\ tan (x - y) & = & \frac{tan x - tan y}{1 + tan x tan y} \end{matrix}

\begin{matrix} sin 2 x & = & 2 sin x cos x \\ cos 2 x & = & {cos}^{2} x - {sin}^{2} x = 2 {cos}^{2} x - 1 = 1 - 2 {sin}^{2} x \\ tan 2 x & = & \frac{2 tan x}{1 - {tan}^{2} x} \end{matrix}

\begin{matrix} sin 2 x & = & 2 sin x cos x \\ cos 2 x & = & {cos}^{2} x - {sin}^{2} x = 2 {cos}^{2} x - 1 = 1 - 2 {sin}^{2} x \\ tan 2 x & = & \frac{2 tan x}{1 - {tan}^{2} x} \end{matrix}

\begin{matrix} {sin}^{2} x & = & \frac{1 - cos 2 x}{2} \\ {cos}^{2} x & = & \frac{1 + cos 2 x}{2} \end{matrix}

\begin{matrix} {sin}^{2} x & = & \frac{1 - cos 2 x}{2} \\ {cos}^{2} x & = & \frac{1 + cos 2 x}{2} \end{matrix}