数学公式まとめ

三角関数

加法定理

$\sin{\left(\alpha \pm \beta\right)} = \sin{\alpha}\cos{\beta} \pm \cos{\alpha}\sin{\beta}$ $\cos{\left(\alpha \pm \beta\right)} = \cos{\alpha}\cos{\beta} \mp \sin{\alpha}\sin{\beta}$ $\tan{\left(\alpha \pm \beta\right)} =\frac{\tan{\alpha} \pm \tan{\beta}}{1 \mp \tan{\alpha}\tan{\beta}}$

$\sin{\left( -\theta \right)} =-\sin{ \theta }$ $\cos{\left( -\theta \right)} = \cos{ \theta }$ $\tan{\left( – \theta\right)} =- \tan{ \theta }$ $\sin{\left( \theta \pm \frac{\pi}{2} \right)} = \pm \cos{ \theta }$ $\cos{\left( \theta \pm \frac{\pi}{2} \right)} = \mp \sin{ \theta }$ $\tan{\left( \theta \pm \frac{\pi}{2} \right)} =- \frac{1}{\tan{\theta}}$ $\sin{\left( \theta \pm \pi \right)} =- \sin{ \theta }$ $\cos{\left( \theta \pm \pi \right)} =- \cos{ \theta }$ $\tan{\left( \theta \pm \pi \right)} = \tan{\theta}$

倍角の公式

$\sin{2\alpha} = 2 \sin{\alpha}\cos{\alpha}$ $\cos{2\alpha} = \cos^2{\alpha} – \sin^2{\alpha} = 1- 2 \sin^{\alpha} = 2\cos^2{\alpha} -1$ $\tan{2\alpha} = \frac{2 \tan{\alpha}}{1-\tan^2{\theta}}$

半角の公式

$\sin^2{\left( \frac{\alpha}{2} \right)} = \frac{1-\cos{\alpha}}{2}$ $\cos^2{\left( \frac{\alpha}{2} \right)} = \frac{1+\cos{\alpha}}{2}$ $\tan^2{\left( \frac{\alpha}{2} \right)} = \frac{1-\cos{\alpha}}{1+\cos{\alpha}}$

和積の公式

$\sin{A}+\sin{B} = 2\sin{\left(\frac{A+B}2{}\right)}\cos{\left(\frac{A-B}2{}\right)}$ $\sin{A}-\sin{B} = 2\cos{\left(\frac{A+B}2{}\right)}\sin{\left(\frac{A-B}2{}\right)}$ $\cos{A}+\cos{B} = 2\cos{\left(\frac{A+B}2{}\right)}\cos{\left(\frac{A-B}2{}\right)}$ $\cos{A}+\cos{B} = 2\sin{\left(\frac{A+B}2{}\right)}\sin{\left(\frac{A-B}2{}\right)}$

三角関数の合成

$a \sin{\theta} + b \cos{\theta} = \sqrt{a^2 + b^2}\sin{\left( \theta + \alpha\right)}$

ただし, $$\alpha$$は次式を満たす. $\sin{\alpha} = \frac{b}{\sqrt{a^2 + b^2}} \, \quad \cos{\alpha} = \frac{a}{\sqrt{a^2 + b^2}}$

指数関数・対数関数

指数関数に登場する定義

$a^0 = 1$ $a^{-n} = \frac{1}{a^n}$

指数関数の公式

$a^r a^s = a^{r+s}$ $\left( a^r \right)^s = a^{rs}$ $\left( ab \right)^r = a^r b^r$

対数関数に登場する定義

$a^p = M \iff p = \log_{a}{M} \iff \log_{a}{a^p} = p$ $\log_{a}{a} = 1 , \quad \log_{a}{1}=0 , \quad \log_{a}{\frac{1}{a}} =-1$

対数関数の公式

$\log_{a}{MN} = \log_{a}{M} + \log_{a}{N}$ $\log_{a}{\frac{M}{N}} = \log_{a}{M} – \log_{a}{N}$ $\log_{a}{M^{k}} = k \log_{a}{M}$ $\log_{a}{b} = \frac{ \log_{c}{b} }{ \log_{c}{a} }$ $\log_{a}{b} = \frac{1}{\log_{b}{a}}$ $\left( \log_{a}{b} \right) \left( \log_{b}{a} \right) = 1$

ベクトル

ベクトルの大きさ

ベクトル $$\overrightarrow{a} = \left( a_1 , a_2\right)$$の大きさは次式で表される. $\left| \overrightarrow{a} \right| = \sqrt{a_1^2 + a_2^2}$ $$O \left( 0, 0 \right)$$, $$A \left( a_1, a_2 \right)$$, $$B \left( b_1 , b_2\right)$$の時, $\begin{gathered} \overrightarrow{AB} = \overrightarrow{OB} – \overrightarrow{OA} = \left( b_1-a_1, b_2-a_2\right) \\ \left| \overrightarrow{AB} \right| = \sqrt{ \left( b_1-a_1 \right)^2 + \left( b_2-a_2 \right)^2 } \end{gathered}$

内積

$$\overrightarrow{a} = \left( a_1 , a_2\right) \neq \overrightarrow{0}$$, $$\overrightarrow{b} = \left( b_1 , b_2\right) \neq \overrightarrow{0}$$, ベクトル $$\overrightarrow{a}$$$$\overrightarrow{a}$$のなす角が $$\theta$$の時, 内積を次式で定義する. $\begin{gathered} \overrightarrow{a} \cdot \overrightarrow{b} = \left| \overrightarrow{a} \right| \left| \overrightarrow{b} \right| \cos{\theta} \\ \overrightarrow{a} \cdot \overrightarrow{b} = a_1 b_1 + a_2 b_2 \end{gathered}$

ベクトルの平行条件

$\overrightarrow{a} \parallel \overrightarrow{b} \quad \iff \quad \overrightarrow{b} = k \overrightarrow{a} \quad \iff \quad a_1 b_2 – a_2 b_1 = 0$

ベクトルの垂直条件

$\overrightarrow{a} \perp \overrightarrow{b} \iff \overrightarrow{a} \cdot \overrightarrow{b} = 0 \iff a_1 b_1 + a_2 b_2 = 0$

微分・積分

ネイピア数

\begin{align} e^x &= \lim_{n \to \infty} \left( 1+ \frac{x}{n} \right) \\ &= 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \end{align} $e = 2.71828\cdots$

微分の公式

$$k$$, $$l$$を定数とする. $\frac{d }{dx}f(x) = f'(x)$ $\frac{d}{dx}\left( k f(x) + l g(x) \right) = \left( k f'(x) + l g'(x) \right)$ $\frac{d}{dx}\left( f(x) g(x) \right)' = f'(x)g(x) + f(x) g'(x)$ $\frac{d}{dx}\left\{\frac{f(x) }{ g(x) }\right\}' = \frac{f'(x)g(x) – f(x) g'(x)}{\left\{ g^{2}(x)\right\}}$ $\frac{d}{dx}\left\{ f(g(x)) \right\}' = f'(g(x)) g'(x)$ $\frac{d }{dx} x^{n} = nx^{n-1}$ $\frac{d }{dx} \sin{x} = \cos{x}$ $\frac{d }{dx} \cos{x} =-\sin{x}$ $\frac{d }{dx} \tan{x} = \frac{1}{\cos^2{x}}$ $\frac{d }{dx} \sin{Ax} = A\cos{x}$ $\frac{d }{dx} \cos{Ax} =-A\sin{x}$ $\frac{d }{dx} e^{x} = e^{x}$ $\frac{d }{dx} e^{Ax} = Ae^{Ax}$ $\frac{d }{dx} a^{x} = a^x \log_e a$ $\frac{d }{dx} \log_e{\left| x \right|} = \frac{1}{x}$ $\frac{d }{dx} \log_a{\left| x \right|} = \frac{1}{x\log_{e}{a}}$

近似式

$f(a + h) = f(a) + f'(a) h$ $\left( 1+ x \right)^n = a + nx$

不定積分

$$F'(x)=f(x)$$の時, 不定積分を次式で定義する. $\int f(x) \ dx = F(x) + C$ $$C$$は積分定数である.

積分の公式

$\int k f(x)\ dx = k \int f(x)\ dx$ $\int \left\{ k f(x) + l g(x) \right\} \ dx = k \int f(x) \ dx + l \int g(x) \ dx$ $\int x^{n} \ dx = \frac{1}{n+1}x^{n+1} + C$ $\int \sin(\theta) \ d\theta =-\cos{\theta} + C$ $\int \cos(\theta) \ d\theta = \sin{\theta} + C$ $\int \sin(A\theta) \ d\theta =-\frac{1}{A} \cos{\theta} + C$ $\int \cos(A\theta) \ d\theta = \frac{1}{A} \sin{\theta} + C$ $\int e^x \ dx = e^x + C$ $\int e^{Ax} \ dx = \frac{1}{A}e^{Ax} + C$ $\int a^{x} \ dx = \frac{a^{x}}{\log_{e}{a}} + C$ $\int \frac{1}{x} \ dx = \log_{e}{\left| x \right|} + C$

置換積分

$\int f(x) \ dx = \int f(g(t))g'(t) \ dt$ $\int f(g(x))g'(x) \ dx = \int f(t)$ $\int \left\{ f(x)\right\}^{\alpha} g'(x) \ dx = \frac{ \left\{g(x)\right\}^{\alpha+1}}{\alpha+1} + C$ $\int \frac{g'(x)}{g(x)} \ dx = \log_{e}{\left\| g(x) \right\|} + C$

部分積分

$\int f(x) g'(x) \ dx = f(x)g(x) – \int f'(x)g(x) \ dx$

1. ちなみに, 過去の東大入試で加法定理の証明をする問題が出ました.