数式サンプル
$2$次方程式
2次方程式($a\neq0$) \[ ax^2+bx+c=0 \] の解の公式は \[ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \] または \[ x=\frac{-b\pm\sqrt{D}}{2a} \] である。ここで、判別式(Discriminant) \[ D=b^2-4ac \] は \[ \left\{\begin{array}{llc} >0 & 解xは2個の異なる実数 & \frac{-b+\sqrt{D}}{2a}、\frac{-b-\sqrt{D}}{2a}\\ =0 & 解xは1個の実数(重解) & -\frac{b}{2a}\\ <0 & 解xは2個の異なる複素数 & 実数解を持たない \end{array}\right. \] である。解を$\alpha$、$\beta$とするとき、 \[ \alpha+\beta=-\frac{b}{a} \] \[ \alpha\beta=\frac{c}{a} \] \[ a\left(x-\alpha\right)\left(x-\beta\right)=0 \] の関係がある。解の$1$次の係数が$2$の倍数のとき \[ ax^2+2b'x+c=0 \] の解の公式は \[ x=\frac{-b'\pm\sqrt{b'^2-ac}}{a} \] または、 \[ x=\frac{-b'\pm\sqrt{D}}{a} \] であり、判別式は \[ D'=b'^2-ac \] となる。 \[ \left\{\begin{array}{ll} >0 解xは2個の異なる実数 \\ =0 解xは1個の実数(重解) \\ <0 解xは2個の異なる複素数 \end{array}\right. \]
展開と因数分解
\[ acx^2+\left(ad+bc\right)x+bd=\left(ax+b\right)\left(cx+d\right) \] \[ x^2+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right) \] \[ cx^2+2acx+ca^2=c\left(x^2+2ax+a^2\right)=c\left(x+a\right)^2 \] \[ x^2+2xy+y^2=\left(x+y\right)^2 \] \[ x^2-2xy+y^2=\left(x-y\right)^2 \] \[ x^2-a^2=\left(x+a\right)\left(x-a\right) \] \[ x^2+y^2+z^2-2xy-2yz-2zx=\left(x+y+z\right)^2 \] \[ x^3+3x^2y+3xy^2+y^3=\left(x+y\right)^3 \] \[ x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3 \] \[ x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right) \] \[ x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right) \] \[ x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right) \] \[ x^4+x^2+1=\left(x^2+x+1\right)\left(x^2-x+1\right) \]相加平均と相乗平均
$x>0$、$y>0$のとき \[ \frac{x+y}{2}\geq\sqrt{xy} \] $x>0$、$y>0$、$z>0$のとき \[ \frac{x+y+z}{3}\geq\sqrt[3]{xyz} \]シュワルツの不等式
\[ \left(a^2+b^2\right)\left(x^2+y^2\right)\geq\left(ax+by\right)^2 \] 等号は$\frac{x}{a}=\frac{y}{b}$のとき。\[ \left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\geq\left(ax+by+cz\right)^2 \] 等号は$\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$のとき。
三角不等式
\[ \left|x\right|-\left|y\right|\leq\left|x+y\right|\leq\left|x\right|+\left|y\right| \]三角関数
\[ \sin\left(\theta\pm\phi\right)=\sin\theta\cos\phi\pm\cos\theta\sin\phi \] \[ \cos\left(\theta\pm\phi\right)=\cos\theta\cos\phi\mp\sin\theta\sin\phi \] \[ \tan\left(\theta\pm\phi\right)=\frac{\tan\theta\pm\tan\phi}{1\mp\tan\theta\tan\phi} \] \[ \sin\left(2\theta\right)=2\sin\theta\cos\theta \] \[ \cos\left(2\theta\right)=\cos^2\theta-\sin^2\theta \] \[ \cos\left(2\theta\right)=2\cos^2\theta-1 \] \[ \cos\left(2\theta\right)=1-2\sin^2\theta \] \[ \tan\left(2\theta\right)=\frac{2\tan\theta}{1-\tan^2\theta} \] \[ \sin\left(3\theta\right)=3\sin\theta-4\sin^3\theta \] \[ \cos\left(3\theta\right)=4\cos^3\theta-3\cos\theta \] \[ \sin\theta\cos\phi=\frac{1}{2}\left\{\sin\left(\theta+\phi\right)+\sin\left(\theta-\phi\right)\right\} \] \[ \cos\theta\cos\phi=\frac{1}{2}\left\{\cos\left(\theta+\phi\right)+\cos\left(\theta-\phi\right)\right\} \] \[ \sin\theta\sin\phi=\frac{1}{2}\left\{\cos\left(\theta+\phi\right)-\cos\left(\theta-\phi\right)\right\} \] \[ \sin\theta+\sin\phi=2\sin\frac{\theta+\phi}{2}\cos\frac{\theta-\phi}{2} \] \[ \sin\theta-\sin\phi=2\cos\frac{\theta+\phi}{2}\sin\frac{\theta-\phi}{2} \] \[ \cos\theta+\cos\phi=2\cos\frac{\theta+\phi}{2}\cos\frac{\theta-\phi}{2} \] \[ \cos\theta-\cos\phi=-2\sin\frac{\theta+\phi}{2}\sin\frac{\theta-\phi}{2} \] \[ \sin^2\theta=\frac{1-\cos\left(2\theta\right)}{2} \] \[ \cos^2\theta=\frac{1+\cos\left(2\theta\right)}{2} \] \[ \sin\theta\cos\theta=\frac{\sin\left(2\theta\right)}{2} \] \[ \sin^2\theta+\cos^2\theta=1 \] \[ \tan\theta=\frac{\sin\theta}{\cos\theta} \] \[ \tan^2\theta+1=\frac{1}{\cos^2\theta} \] $ \tan\alpha=\frac{b}{a} $とするとき、 \[ a\sin\theta+b\cos\theta=\sqrt{a^2+b^2}\sin\left(\theta+\alpha\right) \] $ \tan\alpha=\frac{a}{b} $とするとき、 \[ a\sin\theta+b\cos\theta=\sqrt{a^2+b^2}\cos\left(\theta-\alpha\right) \]\[ \sin \] | \[ \cos \] | \[ \tan \] |
---|---|---|
\[ \sin\left(-\theta\right)=-\sin\theta \] | \[ \cos\left(-\theta\right)=\cos\theta \] | \[ \tan\left(-\theta\right)=-\tan\theta \] |
\[ \sin\left(\frac{\pi}{2}-\theta\right)=\cos\theta \] | \[ \cos\left(\frac{\pi}{2}-\theta\right)=\sin\theta \] | \[ \tan\left(\frac{\pi}{2}-\theta\right)=\frac{1}{\tan\theta} \] |
\[ \sin\left(\frac{\pi}{2}+\theta\right)=\cos\theta \] | \[ \cos\left(\frac{\pi}{2}+\theta\right)=-\sin\theta \] | \[ \tan\left(\frac{\pi}{2}+\theta\right)=-\frac{1}{\tan\theta} \] |
\[ \sin\left(\pi-\theta\right)=\sin\theta \] | \[ \cos\left(\pi-\theta\right)=-\cos\theta \] | \[ \tan\left(\pi-\theta\right)=-\tan\theta \] |
\[ \sin\left(\pi+\theta\right)=-\sin\theta \] | \[ \cos\left(\pi+\theta\right)=-\cos\theta \] | \[ \tan\left(\pi+\theta\right)=\tan\theta \] |
\[ \mathrm{rad} \] | \[ \sin \] | \[ \cos \] |
---|---|---|
\[ 0 \] | \[ \sin0=0 \] | \[ \cos0=1 \] |
\[ \frac{\pi}{6} \] | \[ \sin\frac{\pi}{6}=\frac{1}{2} \] | \[ \cos\frac{\pi}{6}=\frac{\sqrt{3}}{2} \] |
\[ \frac{\pi}{4} \] | \[ \sin\frac{\pi}{4}=\frac{\sqrt{2}}{2} \] | \[ \cos\frac{\pi}{4}=\frac{\sqrt{2}}{2} \] |
\[ \frac{\pi}{3} \] | \[ \sin\frac{\pi}{3}=\frac{\sqrt{3}}{2} \] | \[ \cos\frac{\pi}{3}=\frac{1}{2} \] |
\[ \frac{\pi}{2} \] | \[ \sin\frac{\pi}{2}=1 \] | \[ \cos\frac{\pi}{2}=0 \] |
\[ \frac{2\pi}{3} \] | \[ \sin\frac{2\pi}{3}=\frac{\sqrt{3}}{2} \] | \[ \cos\frac{2\pi}{3}=-\ \frac{1}{2} \] |
\[ \frac{3\pi}{4} \] | \[ \sin\frac{3\pi}{4}=\frac{\sqrt{2}}{2} \] | \[ \cos\frac{3\pi}{4}=-\ \frac{\sqrt{2}}{2} \] |
\[ \frac{5\pi}{6} \] | \[ \sin\frac{5\pi}{6}=\frac{1}{2} \] | \[ \cos\frac{5\pi}{6}=-\ \frac{\sqrt{3}}{2} \] |
\[ \pi \] | \[ \sin\pi=0 \] | \[ \cos\pi=1 \] |
\[ \frac{7\pi}{6} \] | \[ \sin\frac{7\pi}{6}=-\ \frac{1}{2} \] | \[ \cos\frac{7\pi}{6}=-\ \frac{\sqrt{3}}{2} \] |
\[ \frac{5\pi}{4} \] | \[ \sin\frac{5\pi}{4}=-\ \frac{\sqrt{2}}{2} \] | \[ \cos\frac{5\pi}{4}=-\ \frac{\sqrt{2}}{2} \] |
\[ \frac{4\pi}{3} \] | \[ \sin\frac{4\pi}{3}=-\ \frac{\sqrt{3}}{2} \] | \[ \cos\frac{4\pi}{3}=-\ \frac{1}{2} \] |
\[ \frac{3\pi}{2} \] | \[ \sin\frac{3\pi}{2}=-1 \] | \[ \cos\frac{3\pi}{2}=0 \] |
\[ \frac{5\pi}{3} \] | \[ \sin\frac{5\pi}{3}=-\ \frac{\sqrt{3}}{2} \] | \[ \cos\frac{5\pi}{3}=\frac{1}{2} \] |
\[ \frac{7\pi}{4} \] | \[ \sin\frac{7\pi}{4}=-\ \frac{\sqrt{2}}{2} \] | \[ \cos\frac{7\pi}{4}=\frac{\sqrt{2}}{2} \] |
\[ \frac{11\pi}{6} \] | \[ \sin\frac{11\pi}{6}=-\ \frac{1}{2} \] | \[ \cos\frac{11\pi}{6}=\frac{\sqrt{3}}{2} \] |
\[ 2\pi \] | \[ \sin2\pi=0 \] | \[ \cos2\pi=1 \] |
\[ \theta+2\pi \] | \[ \sin\left(\theta+2\pi\right)=\sin\theta \] | \[ \cos\left(\theta+2\pi\right)=\cos\theta \] |
\[ \theta+2n\pi \]$n$:整数 | \[ \sin\left(\theta+2n\pi\right)=\sin\theta \] | \[ \cos\left(\theta+2n\pi\right)=\cos\theta \] |
対数関数
\[ \log_xy=a \] \[ y=x^a \] \[ \log_xx=1 \] \[ \log_x1=0 \] \[ x^{\log_xy}=y \] \[ \log_x x^a=a \] \[ \log_a x^b=b\log_ax \] \[ \log_a\frac{1}{x^b}=-b\log_ax \] \[ \log_a\left(xy\right)=\log_ax+\log_ay \] \[ \log_a\frac{x}{y}=\log_ax-\log_ay \] \[ \log_ax^{y}=y\log_ax \] \[ \log_ab=\frac{\log_cb}{\log_ca} \] \[ \log_ax\left\{\begin{array}{lcl} <0 & & x<1\\ =0 & & x=1\\ >0 & & x>1 \end{array}\right. \]冪乗、累乗、power
\[ \left(ab\right)^{x}=a^{x}b^{x} \] \[ \left(\frac{a}{b}\right)^{x}=\frac{a^{x}}{b^{x}} \] \[ \left(a^x\right)^y=a^{xy} \] \[ \frac{a^x}{a^y}=a^{x-y} \] \[ a^xa^y=a^{x+y} \] \[ a^{\frac{1}{n}}=\sqrt[n]{a} \] \[ a^{-\frac{p}{n}}=\frac{1}{\sqrt[n]{a^p}} \] \[ a^{\frac{p}{n}}=\sqrt[n]{a^p} \] \[ a^{-n}=\frac{1}{a^n} \] \[ a^{-1}=\frac{1}{a} \] \[ a^0=1 \] \[ e^1=e=2.718\cdots \] \[ \sqrt[n]{a}\sqrt[n]{b}=\sqrt[n]{ab} \] \[ \frac{\sqrt[n]{a}}{\sqrt[n]{b}}=\sqrt[n]{\frac{a}{b}} \] \[ \left(\sqrt[n]{a}\right)^m=\sqrt[n]{a^{m}} \] \[ \sqrt[m]{\sqrt[n]{a}}=\sqrt[mn]{a} \] \[ \sqrt[n]{a^{m}}=\sqrt[np]{a^{mp}} \]平均変化率
\[ \frac{f\left(a+h\right)-f\left(a\right)}{h} \]微分係数
\[ f'\left(a\right)=\lim_{h\rightarrow0}\frac{f\left(a+h\right)-f\left(a\right)}{h} \]導関数
\[ f'\left(x\right)=\lim_{h\rightarrow0}\frac{f\left(x+h\right)-f\left(a\right)}{h} \]微分
\[ \frac{{\rm d}}{{\rm d}x}f\left(x\right)=f'\left(x\right) \] $c$が定数のとき \[ \left(c\right)'=0 \] \[ \left(x^n\right)'=nx^{n-1} \] \[ \left(\sin x\right)'=\cos x \] \[ \left(\cos x\right)'=-\sin x \] \[ \left(\tan x\right)'=\frac{1}{\cos^2x} \] \[ \left(\sin^{-1}x\right)'=\frac{1}{\sqrt{1-x^2}} \] \[ \left(\cos^{-1}x\right)'=-\frac{1}{\sqrt{1-x^2}} \] \[ \left(\tan^{-1}x\right)'=\frac{1}{1+x^2} \] \[ \left(\log_ex\right)'=\frac{1}{x} \] \[ \left(e^x\right)'=e^x \] \[ \left(a^x\right)'=a^x\log_ea \] \[ \left(\log_ex\right)'=\frac{1}{x} \] \[ \left(f+g\right)'=f'+g' \] \[ \left(cf\right)'=cf' \] \[ \left(fg\right)'=f'g+fg' \] \[ \left(\frac{1}{f}\right)'=-\frac{f'}{f^2} \] \[ \left(\frac{f}{g}\right)'=-\frac{f'g-fg'}{g^2} \] \[ \left(f\left(g\left(x\right)\right)\right)'=\frac{{\rm d}f}{{\rm d}g}\frac{{\rm d}g}{{\rm d}x} \] \[ \frac{\mathrm{d}f}{\mathrm{d}x}=\frac{\frac{{\rm d}f}{{\rm d}t}}{\frac{{\rm d}x}{{\rm d}t}} \] \[ \frac{\mathrm{d}f}{\mathrm{d}x}=\frac{1}{\frac{\mathrm{d}x}{\mathrm{d}f}} \] \[ \left(ax^2+bx+c\right)'=2ax+b\\ \] \[ \left(x^n\sin x\right)'=nx^{n-1}\sin x+x^n\cos x\\ \] \[ \left(\log_e\sin x\right)'=\frac{\cos x}{\sin x}=\frac{1}{\tan x}\\ \] \[ \left\{\begin{array}{l}f\left(g\right)=\log_eg\\g\left(x\right)=\sin x\end{array}\right.\\ \] \begin{eqnarray*} & &\left\{\begin{array}{l}\frac{{\rm d}f}{{\rm d}g}=\frac{1}{g}=\frac{1}{\sin x}\\ \frac{{\rm d}g}{{\rm d}x}=\cos x\end{array}\right.\\ &=&\left(f\left(g\left(x\right)\right)\right)'=\frac{{\rm d}f}{{\rm d}g}\frac{{\rm d}g}{{\rm d}x}=\frac{1}{\sin x}\cos x \end{eqnarray*} \[ \left(\frac{1}{\left(\sqrt{x^2+2x+3}\right)^3}\right)'=-\ \frac{3}{2}\frac{2x+2}{\sqrt{x^2+2x+3}^5}=-\ \frac{3\left(x+1\right)}{\sqrt{x^2+2x+3}^5}\\ \] \[ \left\{\begin{array}{l}f\left(g\right)=g^{-3/2}\\g\left(x\right)=x^2+2x+3\end{array}\right.\\ \] \[ \left\{\begin{array}{l}\frac{{\rm d}f}{{\rm d}g}=-\ \frac{3}{2}g^{-5/2}=-\ \frac{3}{2}\frac{1}{\left(x^2+2x+3\right)^{5/2}}\\ \frac{{\rm d}f}{{\rm d}g}=2x+2=2\left(x+1\right)\end{array}\right.\\ =\left(f\left(g\left(x\right)\right)\right)'=\frac{{\rm d}f}{{\rm d}g}\frac{{\rm d}g}{{\rm d}x}=-\ \frac{3}{2}\frac{1}{\left(x^2+2x+3\right)^{5/2}}2\left(x+1\right)=-\ \frac{3\left(x+1\right)}{\left(x^2+2x+3\right)^{5/2}}\\ \]接戦の方程式
\[ y=f'\left(a\right)\left(x-a\right)+f\left(a\right) \]平均値の定理
\[ \frac{f\left(b\right)-f\left(b\right)}{b-a}=f'\left(c\right) \] を満たす$c\left(a < c < b\right)$が存在する。積分
不定積分 \[ \int\ f\left(x\right){\rm d}x \] 例) \[ \int x{\rm d}x=\frac{x^2}{2}+C \] 定積分 \[ \int_a^bf\left(x\right){\rm d}x \] 例) \[ \int_0^1x{\rm d}x=\left[\frac{x^2}{2}\right]_0^1=\frac{1^2}{2}-\frac{0^2}{2}=\frac{1}{2} \]\[ \int x^n{\rm d}x=\frac{x^{n+1}}{n+1}+C\ ,\ n\neq-1 \] \[ \int x^{-1}{\rm d}x=\log_e\left|x\right|+C \] \[ \int\sin x{\rm d}x=-\cos x+C \] \[ \int\cos x{\rm d}x=\sin x+C \] \[ \int\tan x{\rm d}x=-\log_e\left|\cos x\right|+C \] \[ \int\frac{1}{\cos^2x}{\rm d}x=\tan x+C \] \[ \int\frac{1}{\sin^2x}{\rm d}x=-\frac{1}{\tan x}+C \] \[ \int e^x{\rm d}x=e^x+C \] \[ \int a^x{\rm d}x=\frac{a^x}{\log_ea}+C \] \[ \int \log_ex{\rm d}x=x\left(\log_ex-1\right)+C \] \[ \int \frac{1}{x^2+a^2}{\rm d}x=\frac{1}{a}\tan^{-1}\frac{x}{a}+C \] \[ \int\frac{1}{\sqrt{a^2-x^2}}{\rm d}x=\sin^{-1}\frac{x}{a}+C \] \[ \int\frac{1}{\sqrt{x^2+a}}{\rm d}x=\log_e\left|x+\sqrt{x^2+a}\right|+C \] \[ \int\left(f+g\right){\rm d}x=\int f{\rm d}x+\int g{\rm d}x \] \[ \int\frac{f'}{f}{\rm d}x=\log_e\left|f\right|+C \] $\sin x=t$とすると、 \[ \int f\left(\sin x\right)\cos x{\rm d}x=\int f\left(t\right){\rm d}t \] $\cos x=t$とすると、 \[ \int f\left(\cos x\right)\sin x{\rm d}x=-\int f\left(t\right){\rm d}t \] $\tan x=t$とすると、 \[ \int f\left(\tan x\right)\frac{1}{\cos^2x}{\rm d}x=\int f\left(t\right){\rm d}t \] $\tan x=t$とすると、 \[ \int f\left(\sin x,\cos x\right){\rm d}x=\int f\left(\frac{2t}{1+t^2},\frac{1-t^2}{1+t^2}\right)\frac{2}{1+t^2}{\rm d}t \] \[ \int_a^bf{\rm d}x=\int_a^cf{\rm d}x+\int_c^bf{\rm d}x \] \[ \int_a^bf{\rm d}x=-\int_b^af{\rm d}x \] $c$が定数のとき、 \[ \int cf\left(x\right){\rm d}x=c\int f\left(x\right){\rm d}x \] 部分積分 \[ \int f'g{\rm d}x=fg-\int fg'{\rm d}x \] \[ \int_a^b f'g{\rm d}x=\left[fg\right]_a^b-\int_a^b fg'{\rm d}x \] 合成関数 \[ \int \frac{{\rm d}f}{{\rm d}g}\frac{{\rm d}g}{{\rm d}x}{\rm d}x=f\left(g\left(x\right)\right)+C \] \[ \frac{{\rm d}}{{\rm d}x}\int_a^xf\left(t\right){\rm d}t=f\left(x\right) \] \[ \frac{{\rm d}}{{\rm d}x}\int_x^{x+a}f\left(t\right){\rm d}t=f\left(x+a\right)-f\left(x\right) \] $f$が偶関数のとき、 \[ \int_{-a}^{a}fx{\rm d}x=2\int_{0}^{a}fx{\rm d}x \] $f$が奇関数のとき、 \[ \int_{-a}^{a}fx{\rm d}x=0 \] \[ \int_{0}^{\pi}\sin\left(\pi-x\right){\rm d}x=\int_{0}^{\pi}\sin x{\rm d}x \] \[ \int_{\alpha}^{\beta}\left(x-\alpha\right)\left(x-\beta\right){\rm d}x=-\frac{1}{6}\left(\beta-\alpha\right)^3 \] \[ \frac{{\rm d}}{{\rm d}x}\int x{\rm d}x=\frac{{\rm d}}{{\rm d}x}\left(\frac{x^2}{2}+C\right)=x \] \[ \int\frac{{\rm d}}{{\rm d}x}f\left(x\right){\rm d}x=f\left(x\right)+C \] \[ \int\left(\frac{{\rm d}}{{\rm d}x}x\right){\rm d}x=\int1{\rm d}x=x+C \] $f$と$g$に挟まれた面積は \[ S=\int_a^b\left|f-g\right|{\rm d}x \] 位置$x$での断面積がと$S$のとき、体積は \[ V=\int_a^bS{\rm d}x \] $x$軸回りに回転させたとき、体積は \[ V=\pi\int_a^bf^2{\rm d}x \] $x=x\left(t\right)$、$y=y\left(t\right)$の曲線の長さ \[ s=\int_a^b\sqrt{x'^2+y'^2}{\rm d}t \] $y=y\left(x\right)$の曲線の長さ \[ s=\int_a^b\sqrt{1+y'^2}{\rm d}x \] 区分求積法 \[ \lim_{n\rightarrow\infty}\sum_{k=0}^{n-1}f\left(k_x\right)\varDelta x=\int_{a}^{b}f\left(x\right)\mathrm{d}x \] \begin{eqnarray*} & &\int x\cos x^2{\rm d}x=\frac{1}{2}\int\left(2x\right)\left(\cos x^2\right){\rm d}x=\frac{1}{2}\int\frac{{\rm d}g}{{\rm d}x}\frac{{\rm d}f}{{\rm d}g}{\rm d}x=\frac{1}{2}\sin x^2+C\\ & &\downarrow 予想する \\ & &\left\{\begin{array}{l}f=\sin g\\g=x^2\end{array}\right. \left\{\begin{array}{l}\frac{{\rm d}f}{{\rm d}x}=\cos g\\ \frac{{\rm d}g}{{\rm d}x}=2x\end{array}\right. \end{eqnarray*}
剰余の定理
$ F\left(x\right) $ を$x-\alpha$で割ったとき、余りを$r$とする。 \[ F\left(x\right)=\left(x-\alpha\right)Q\left(x\right)+r \] そのとき、 \[ F\left(\alpha\right)=r \]因数定理
\[ \frac{P\left(x\right)}{x-a}=Q\left(x\right)\hspace{1cm}(余り=0) \] のとき、 \[ P\left(a\right)=0 \]絶対値
\[ \left|x\right|=\left\{\begin{array}{rcl} x & & x\geq0 \\ -x & & x\leq0 \end{array}\right. \] \[ \left|x\right|\geq0 \]有理化
\[ \frac{1}{\sqrt{x}+\sqrt{y}}=\frac{\sqrt{x}-\sqrt{y}}{x-y} \] 証明 \[ \frac{1}{\sqrt{x}+\sqrt{y}}=\frac{1}{\sqrt{x}+\sqrt{y}}\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{x}-\sqrt{y}}{x-y} \]正弦定理
\[ \frac{a}{\sin A}=2R \] \[ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R \]余弦定理
\[ a^2=b^2+c^2-2bc\cos A \] \begin{eqnarray*} a^2&=&b^2+c^2-2bc\cos A\\ b^2&=&c^2+a^2-2ca\cos B\\ c^2&=&a^2+b^2-2ab\cos C \end{eqnarray*}三角形の面積
\[ S=\frac{1}{2}ab\sin C \] \[ S=\frac{1}{2}ar+\frac{1}{2}br+\frac{1}{2}cr \] \[ S=\frac{\left(a+b+c\right)r}{2} \] \begin{eqnarray*} S=\frac{1}{2}ab\sin C=\frac{1}{2}bc\sin A=\frac{1}{2}ca\sin B\\ =\frac{1}{2}ar+\frac{1}{2}br+\frac{1}{2}cr=\frac{\left(a+b+c\right)r}{2} \end{eqnarray*} \[ S=\frac{1}{2}\sqrt{\left|\overrightarrow{AB}\right|^2\left|\overrightarrow{AC}\right|^2-\left(\overrightarrow{AB}\cdotp\overrightarrow{AC}\right)^2} \]中線定理
点$M$が辺$BC$の中点のとき \[ \overline{AB}^2+\overline{AC}^2=2\left(\overline{AM}^2+\overline{BM}^2\right) \]内積
\[ \overrightarrow{a}\cdotp\overrightarrow{b}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos\theta \] \[ \overrightarrow{a}\cdotp\overrightarrow{b}=a_1b_1+a_2b_2 \] $\theta=90^\circ$のとき、 \[ \overrightarrow{a}\cdotp\overrightarrow{b}=0 \] \[ \cos\theta=\frac{\overrightarrow{a}\cdotp\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|} \] \[ \cos\theta=\frac{a_1b_1+a_2b_2}{\sqrt{a_1^2+a_2^2}\sqrt{b_1^2+b_2^2}} \] \[ \overrightarrow{a}\cdotp\overrightarrow{b}=\overrightarrow{b}\cdotp\overrightarrow{a} \] \[ \left(c\overrightarrow{a}\right)\cdotp\overrightarrow{b}=\overrightarrow{b}\cdotp\left(c\overrightarrow{a}\right)=\left(c\overrightarrow{b}\cdotp\overrightarrow{a}\right) \] \[ \left(\overrightarrow{a}+\overrightarrow{b}\right)\cdotp\overrightarrow{c}=\overrightarrow{a}\cdotp\overrightarrow{c}+\overrightarrow{b}\cdotp\overrightarrow{c} \]内分点
\[ \overrightarrow{a}=\left(\begin{array}{c} x_1 \\ y_1 \end{array}\right)\ ,\ \overrightarrow{b}=\left(\begin{array}{c} x_2 \\ y_2 \end{array}\right)\\ \] $ \overrightarrow{a} $ 、 $ \overrightarrow{b} $ を$m:n$に内分する点 \[ \frac{m\overrightarrow{b}+n\overrightarrow{a}}{m+n}=\left(\begin{array}{c} \frac{mx_2+nx_1}{m+n} \\ \frac{my_2+ny_1}{m+n}\end{array}\right) \]外分点
\[ \overrightarrow{a}\ ,\ \overrightarrow{b} \] $ \overrightarrow{a} $ 、 $ \overrightarrow{b} $ を$m:n$に外分する点 \[ \frac{m\overrightarrow{b}-n\overrightarrow{a}}{m-n}=\left(\begin{array}{c} \frac{mx_2-nx_1}{m-n} \\ \frac{my_2-ny_1}{m-n}\end{array}\right) \]重心
\[ \frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{3} \]内心
\[ \overrightarrow{i}=\frac{a\overrightarrow{a}+b\overrightarrow{b}+c\overrightarrow{c}}{a+b+c} \]平面の方程式
点$\left(x_0,y_0,z_0\right)$を通り、法線ベクトルが$\overrightarrow{n}=\left(a,b,c\right)$の平面の方程式は \[ a\left(x-x_0\right)+b\left(y-y_0\right)+c\left(z-z_0\right)=0 \] 一般形は \[ ax+by+cz+d=0 \]直線の方程式
点$\left(x_0,y_0,z_0\right)$を通り、法線ベクトルが$\overrightarrow{d}=\left(\ell,m,n\right)$の平面の方程式は \[ \frac{x-x_0}{\ell}=\frac{y-y_0}{m}=\frac{z-z_0}{n} \] 媒介変数で表すと \[ x=x_0+\ell t \] \[ y=y_0+mt \] \[ z=z_0+nt \]点と直線の距離
点$\left(x_0,y_0\right)$と直線$ax+by+c=0$との距離(点から直線へ下ろした垂線の長さ)は \[ \frac{\left|ax_0+by_0+c\right|}{\sqrt{a^2+b^2}} \]2直線の間の角度
直線$m_1x+n_1$、$m_2x+n_2$の間の角度は $m_1m_2\neq-1$のとき \[ \tan\theta=\pm\frac{m_2-m_1}{1+m_2m_1} \] $m_1m_2=-1$のとき \[ \theta=90^\circ \]円
原点$\left(0,0\right)$を中心とする半径$1$の円 \[ x^2+y^2=1 \] 点$\left(a,b\right)$を中心とする半径$1$の円 \[ \left(x-a\right)^2+\left(y-b\right)^2=1 \] 原点$\left(0,0\right)$を中心とする半径$r$の円 \[ x^2+y^2=r^2 \] \[ \left\{\begin{array}{c} x=r\cos\theta \\ y=r\sin\theta \end{array}\right. \] 点$\left(a,b\right)$を中心とする半径$r$の円 \[ \left(x-a\right)^2+\left(y-b\right)^2=r^2 \] \[ \left\{\begin{array}{c} x=a+r\cos\theta \\ y=b+r\sin\theta \end{array}\right. \] 原点と点$\left(a,b\right)$を両端とする線分が直径の円 \[ \left(x-\frac{a}{2}\right)^2+\left(y-\frac{b}{2}\right)^2=\frac{a^2+b^2}{4} \]楕円
\[ \bar{F_1P}+\bar{F_2P}=定数 \] を満たす点$P$の集合。$F_1$、$F_2$は焦点。楕円の方程式(標準系)は \[ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \] ($a>b>0$)
焦点の座標 \[ \left(\pm\sqrt{a^{2},b^{2}},0\right) \] 長軸の長さ \[ 2a \] 短軸の長さ \[ 2b \]