初等函数

双曲正弦:

$$shx = \frac{{{e^x} - {e^{ - x}}}}{2}$$

双曲余弦:

$$chx = \frac{{{e^x} + {e^{ - x}}}}{2}$$

双曲正切:

$$thx = \frac{{shx}}{{chx}} = \frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}$$

反双曲：

$$arshx = \ln (x + \sqrt {{x^2} + 1} )$$
$$archx = \pm \ln (x + \sqrt {{x^2} - 1} )$$
$$arthx = \frac{1}{2}\ln \frac{{1 + x}}{{1 - x}}$$

重要极限

$$ \begin{gathered} \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1 \\\\ \mathop {\lim }\limits_{x \to \infty } {(1 + \frac{1}{x})^x} = e = 2.718281828459045... \end{gathered} $$

导数公式

$$(C)' = 0 \qquad ({x^n})' = n{x^{n - 1}} \qquad (\sin x)' = \cos x \qquad (\cos x)' = - \sin x$$
$$(tgx)' = {\sec ^2}x \qquad (ctgx)' = - {\csc ^2}x \qquad
(\sec x)' = \sec xtgx \qquad (\csc x)' = - \csc xctgx$$
$$({a^x})' = {a^x}\ln a \qquad ({e^x})' = {e^x} \qquad
({\log _a}x)' = \frac{1}{{x\ln a}} \qquad
(\ln x)' = \frac{1}{x}$$
$$(\arcsin x)' = \frac{1}{{\sqrt {1 - {x^2}} }} \qquad (\arccos x)' = - \frac{1}{{\sqrt {1 - {x^2}} }}$$
$$(arctgx)' = \frac{1}{{1 + {x^2}}} \qquad
(arcctgx)' = - \frac{1}{{1 + {x^2}}}$$

积分公式

$$ \int {tgxdx = - \ln \left| {\cos x} \right| + C} \qquad \int {ctgxdx = \ln \left| {\sin x} \right| + C} $$

$$ \int {\sec xdx = \ln \left| {\sec x + tgx} \right| + C} \qquad \int {\csc xdx = \ln \left| {\csc x - ctgx} \right| + C} $$

$$ \int {\frac{{dx}}{{{a^2} + {x^2}}} = \frac{1}{a}arctg\frac{x}{a} + } C \qquad \int {\frac{{dx}}{{{x^2} - {a^2}}} = \frac{1}{{2a}}\ln \left| {\frac{{x - a}}{{x + a}}} \right| + C} $$

$$ \int {\frac{{dx}}{{{a^2} - {x^2}}} = \frac{1}{{2a}}\ln \frac{{a + x}}{{a - x}} + C} \qquad \int {\frac{{dx}}{{\sqrt {{a^2} - {x^2}} }}} = \arcsin \frac{x}{a} + C $$

$$ \int {\frac{{dx}}{{\sqrt {{x^2} \pm {a^2}} }} = \ln (x + \sqrt {{x^2} \pm {a^2}} ) + C} \qquad \int {{a^x}dx = \frac{{{a^x}}}{{\ln a}}} + C $$

$$ \int {\frac{{dx}}{{{{\cos }^2}x}} = \int {{{\sec }^2}xdx = tgx + C} } \qquad \int {\frac{{dx}}{{{{\sin }^2}x}} = \int {{{\csc }^2}xdx = - ctgx + C} } $$

$$ \int {\sec x \cdot tgx} dx = \sec x + C \qquad \int {\csc x \cdot ctgxdx = - \csc x + C} $$

$$ \int {chxdx = shx + C} \qquad \int {shxdx = chx + C} $$

$$ {I_n} = \int\limits_0^{\frac{\pi }{2}} {{{\sin }^n}xdx = } \int\limits_0^{\frac{\pi }{2}} {{{\cos }^n}xdx} = \frac{{n - 1}}{n}{I_{n - 2}} $$

$$ \int {\sqrt {{x^2} + {a^2}} dx = \frac{x}{2}\sqrt {{x^2} + {a^2}} + \frac{{{a^2}}}{2}\ln (x + \sqrt {{x^2} + {a^2}} ) + C} $$

$$ \int {\sqrt {{x^2} - {a^2}} dx = \frac{x}{2}\sqrt {{x^2} - {a^2}} - \frac{{{a^2}}}{2}\ln \left| {x + \sqrt {{x^2} - {a^2}} } \right|} + C $$

$$ \int {\sqrt {{a^2} - {x^2}} dx = \frac{x}{2}\sqrt {{a^2} - {x^2}} + \frac{{{a^2}}}{2}\arcsin \frac{x}{a} + C} $$

泰勒公式

通用形式

$$ f(x) = f({x_0})(x - {x_0}) + \frac{{f''({x_0})}}{{2!}}{(x - {x_0})^2} + \cdots + \frac{{{f^{(n)}}({x_0})}}{{n!}}{(x - {x_0})^n} + \cdots $$

余项：

$$ {R_n} = \frac{{{f^{(n + 1)}}(\xi )}}{{(n + 1)!}}{(x - {x_0})^{n + 1}} $$

$f(x)$ 可以展开成泰勒级数的充要条件是

$$\mathop {\lim }\limits_{n \to \infty } {R_n} = 0$$

当${x_0} = 0$时, 该式即为麦克劳林公式：

$$f(x) = f(0) + f'(0)x + \frac{{f''(0)}}{{2!}}{x^2} + \cdots + \frac{{{f^{(n)}}(0)}}{{n!}}{x^n} + \cdots $$

常用几类

$$ {(1 + x)^m} = 1 + mx + \frac{{m(m - 1)}}{{2!}}{x^2} + \cdots + \frac{{m(m - 1) \cdots (m - n + 1)}}{{n!}}{x^n} + \cdots ( - 1 \lt x \lt 1) $$

$$ \sin x = x - \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} - \cdots + {( - 1)^{n - 1}}\frac{{{x^{2n - 1}}}}{{(2n - 1)!}} + \cdots ( - \infty \lt x \lt + \infty ) $$

高斯公式

$$ \iiint \limits_{\Omega} (\frac{{\partial P}}{{\partial x}} + \frac{{\partial Q}}{{\partial y}} + \frac{{\partial R}}{{\partial z}})dv = \mathop{{\int\\!\\!\\!\\!\int}\mkern-21mu \bigcirc} \limits_{\sum}{Pdydz + Qdzdx + Rdxdy}= \mathop{{\int\\!\\!\\!\\!\int}\mkern-21mu \bigcirc}\limits_{\sum} {(P\cos \alpha + Q\cos \beta + R\cos \gamma )ds} $$

斯托克斯公式

$$ \iint \limits_{\sum} (\frac{{\partial R}}{{\partial y}} - \frac{{\partial Q}}{{\partial z}})dydz + (\frac{{\partial P}}{{\partial z}} - \frac{{\partial R}}{{\partial x}})dzdx + (\frac{{\partial Q}}{{\partial x}} - \frac{{\partial P}}{{\partial y}})dxdy = \oint\limits_{\Gamma} {Pdx + Qdy + Rdz} $$

上式左端可写为

$$ \iint \limits_{\sum} { \begin{vmatrix} {dydz} &\quad {dzdx} &\quad {dxdy} \\\\ {\frac{\partial }{{\partial x}}} &\quad {\frac{\partial }{{\partial y}}} &\quad {\frac{\partial }{{\partial z}}} \\\\ P &\quad Q &\quad R \end{vmatrix} } = \iint\limits_{\sum} \begin{vmatrix} {\cos \alpha }&{\cos \beta }&{\cos \gamma } \\\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}} \\\\ P&Q&R \end{vmatrix} $$

欧拉公式

$$ {e^{ix}} = \cos x + i\sin x $$

其中

$$ \cos x = \frac{{{e^{ix}} + {e^{ - ix}}}}{2} \qquad \sin x = \frac{{{e^{ix}} - {e^{ - ix}}}}{2} $$

三角级数

$$ f(t) = A_{0} + \sum\limits_{n = 1}^\infty A_{n}\sin (n\omega t + \phi_{n}) = \frac{a_{0}}{2} + \sum\limits_{n = 1}^\infty {(a_{n}\cos nx + b_{n}\sin nx)} $$

其中

$$ {a_0} = a{A_0}, \quad {a_n} = {A_n}\sin {\phi _n}, \quad {b_n} = {A_n}\cos {\phi _n} \quad \omega t = x $$

正交性：

$$1, \quad \sin x, \quad \cos x, \quad \sin 2x, \quad \cos 2x, \quad \cdots, \quad \sin nx, \quad \cos nx$$

任意两个不同项的乘积在$[ - \pi ,\pi ]$

三角公式

（1）、诱导公式

（2）、三角函数的有理式

$$ \sin x = \frac{{2u}}{{1 + {u^2}}} \qquad \cos x = \frac{{1 - {u^2}}}{{1 + {u^2}}} \qquad u = tg\frac{x}{2} \qquad dx = \frac{{2du}}{{1 + {u^2}}} $$

（3）、和差角公式

$$ \sin (\alpha \pm \beta ) = \sin \alpha \cos \beta \pm \cos \alpha \sin \beta $$

$$ \cos (\alpha \pm \beta ) = \cos \alpha \cos \beta \mp \sin \alpha \sin \beta $$

$$ tg(\alpha \pm \beta ) = \frac{{tg\alpha \pm tg\beta }}{{1 \mp tg\alpha \cdot tg\beta } }$$ $$ctg(\alpha \pm \beta) = \frac{ctg \alpha \cdot ctg\beta \mp 1}{ctg\beta \pm ctg\alpha }$$ （4）、和差化积公式 $$\sin \alpha + \sin \beta = 2\sin \frac{{\alpha + \beta }}{2}\cos \frac{{\alpha - \beta }}{2}$$ $$\sin \alpha - \sin \beta = 2\cos \frac{{\alpha + \beta }}{2}\sin \frac{{\alpha - \beta }}{2}$$ $$\cos \alpha + \cos \beta = 2\cos \frac{{\alpha + \beta }}{2}\cos \frac{{\alpha - \beta }}{2}$$ $$\cos \alpha - \cos \beta = 2\sin \frac{{\alpha + \beta }}{2}\sin \frac{{\alpha - \beta }}{2}$$ （5）、积化和差公式 $$\sin \alpha \sin \beta = - \frac{1}{2} [ \cos (\alpha + \beta ) - \cos (\alpha - \beta ) ]$$ $$\cos \alpha \cos \beta = \frac{1}{2} [ \cos (\alpha + \beta ) + \cos (\alpha - \beta ) ]$$ $$\sin \alpha \cos \beta = \frac{1}{2} [ \sin (\alpha + \beta ) + \sin (\alpha - \beta ) ]$$ $$\cos \alpha \sin \beta = \frac{1}{2} [ \sin (\alpha + \beta ) - \sin (\alpha - \beta ) ]$$ （6）、倍角公式 $$\sin 2\alpha = 2\sin \alpha \cos \alpha$$ $$\cos 2\alpha = 2{\cos ^2}\alpha - 1 = 1 - 2{\sin ^2}\alpha = {\cos ^2}\alpha - {\sin ^2}\alpha$$ $$ctg2\alpha = \frac{{ct{g^2}\alpha - 1}}{{2ctg\alpha }}$$ $$tg2\alpha = \frac{{2tg\alpha }}{{1 - t{g^2}\alpha }}$$ $$tg3\alpha = \frac{{3tg\alpha - t{g^3}\alpha }}{{1 - 3t{g^2}\alpha }}$$ $$\sin 3\alpha = 3\sin \alpha - 4{\sin ^3}\alpha$$ $$\cos 3\alpha = 4{\cos ^3}\alpha - 3\cos \alpha$$ （7）、半角公式 $$\sin \frac{\alpha }{2} = \pm \sqrt {\frac{{1 - \cos \alpha }}{2}}$$ $$\cos \frac{\alpha }{2} = \pm \sqrt {\frac{{1 + \cos \alpha }}{2}}$$ $$tg\frac{\alpha }{2} = \pm \sqrt {\frac{{1 - \cos \alpha }}{{1 + \cos \alpha }}} = \frac{{1 - \cos \alpha }}{{\sin \alpha }} = \frac{{\sin \alpha }}{{1 + \cos \alpha }}$$ $$ctg\frac{\alpha }{2} = \pm \sqrt {\frac{{1 + \cos \alpha }}{{1 - \cos \alpha }}} = \frac{{1 + \cos \alpha }}{{\sin \alpha }} = \frac{{\sin \alpha }}{{1 - \cos \alpha }}$$ （8）、正弦定理余弦定理 $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \qquad c^2 = a^2 + b^2 - 2ab \cos C$$ （9）、反三角函数性质 $$\arcsin x = \frac{\pi }{2} - \arccos x \qquad arctgx = \frac{\pi }{2} - arcctgx$$