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高等数学基本公式(三角函数、导数公式、 积分公式)
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一、三角公式
(1)、三角函数的有理式
$$\sin x = \frac{{2u}}{{1 + {u^2}}}, \cos x = \frac{{1 - {u^2}}}{{1 + {u^2}}}, u = tg\frac{x}{2}, dx = \frac{{2du}}{{1 + {u^2}}}$$
(2)、和差角公式
$$\begin{array}{c}\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta,\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta\\tg(\alpha\pm\beta)=\frac{tg\alpha\pm tg\beta}{1\mp tg\alpha⋅tg\beta},ctg(\alpha\pm\beta)=\frac{ctg\alpha⋅ctg\beta\mp1}{ctg\beta\pm ctg\alpha}\end{array}$$
(3)、和差化积公式
$$\begin{array}{c}\sin\alpha+\sin\beta=2\sin\frac{\alpha+\beta}2\cos\frac{\alpha-\beta}2,\sin\alpha-\sin\beta=2\cos\frac{\alpha+\beta}2\sin\frac{\alpha-\beta}2\\\cos\alpha+\cos\beta=2\cos\frac{\alpha+\beta}2\cos\frac{\alpha-\beta}2,\cos\alpha-\cos\beta=2\sin\frac{\alpha+\beta}2\sin\frac{\alpha-\beta}2\end{array}$$
(4)、积化和差公式
$$\begin{array}{c}\sin\alpha\sin\beta=-\frac12\lbrack\cos(\alpha+\beta)-\cos(\alpha-\beta)\rbrack,\cos\alpha\cos\beta=\frac12\lbrack\cos(\alpha+\beta)+\cos(\alpha-\beta)\rbrack\\\sin\alpha\cos\beta=\frac12\lbrack\sin(\alpha+\beta)+\sin(\alpha-\beta)\rbrack,\cos\alpha\sin\beta=\frac12\lbrack\sin(\alpha+\beta)-\sin(\alpha-\beta)\rbrack\end{array}$$
(5)、倍角公式
$$\begin{array}{c}\sin2\alpha=2\sin\alpha\cos\alpha,\cos2\alpha=2\cos^2\alpha-1=1-2\sin^2\alpha=\cos^2\alpha-\sin^2\alpha\\ctg2\alpha=\frac{ctg^2\alpha-1}{2ctg\alpha},tg2\alpha=\frac{2tg\alpha}{1-tg^2\alpha},tg3\alpha=\frac{3tg\alpha-tg^3\alpha}{1-3tg^2\alpha}\\\sin3\alpha=3\sin\alpha-4\sin^3\alpha,\cos3\alpha=4\cos^3\alpha-3\cos\alpha\end{array}$$
(6)、半角公式
$$\begin{array}{c}\sin\frac\alpha2=\pm\sqrt{\frac{1-\cos\alpha}2},\cos\frac\alpha2=\pm\sqrt{\frac{1+\cos\alpha}2}\\tg\frac\alpha2=\pm\sqrt{\frac{1-\cos\alpha}{1+\cos\alpha}}=\frac{1-\cos\alpha}{\sin\alpha}=\frac{\sin\alpha}{1+\cos\alpha}\\ctg\frac\alpha2=\pm\sqrt{\frac{1+\cos\alpha}{1-\cos\alpha}}=\frac{1+\cos\alpha}{\sin\alpha}=\frac{\sin\alpha}{1-\cos\alpha}\end{array}$$
(7)、正弦定理/余弦定理
$$\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}} = 2R,{c^2} = {a^2} + {b^2} - 2ab\cos C$$
(8)、反三角函数性质
$$\arcsin x = \frac{\pi }{2} - \arccos x,arctgx = \frac{\pi }{2} - arcctgx$$
(9)、诱导公式
| 函数/角 | sin | cos | tg | ctg |
| -α | -sinα | cosα | -tgα | -ctgα |
| 90°-α | cosα | sinα | ctgα | tgα |
| 90°+α | cosα | -sinα | -ctgα | -tgα |
| 180°-α | sinα | -cosα | -tgα | -ctgα |
| 180°+α | -sinα | -cosα | tgα | ctgα |
| 270°-α | -cosα | -sinα | ctgα | tgα |
| 270°+α | -cosα | sinα | -ctgα | -tgα |
| 360°-α | -sinα | cosα | -tgα | -ctgα |
| 360°+α | sinα | cosα | tgα | ctgα |
二、初等函数
$$\begin{array}{c} 双曲正弦:shx=\frac{e^x-e^{-x}}2,双曲余弦:chx=\frac{e^x+e^{-x}}2,双曲正切:thx=\frac{shx}{chx}=\frac{e^x-e^{-x}}{e^x+e^{-x}}\\arshx=\ln(x+\sqrt{x^2+1}),archx=\pm\ln(x+\sqrt{x^2-1}),arthx=\frac12\ln\frac{1+x}{1-x}\end{array}$$
三、 重要极限
$$\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1$$
$$\mathop {\lim }\limits_{x \to \infty } {(1 + \frac{1}{x})^x} = e = 2.718281828459045...$$
四、导数公式
$$(C)' = 0, ({x^n})' = n{x^{n - 1}}, (\sin x)' = \cos x, (\cos x)' = - \sin x$$
$$(tgx)' = {\sec ^2}x, (ctgx)' = - {\csc ^2}x $$
$$(\sec x)' = \sec xtgx, (\csc x)' = - \csc xctgx $$
$$({a^x})' = {a^x}\ln a, ({e^x})' = {e^x}$$
$$({\log _a}x)' = \frac{1}{{x\ln a}}, (\ln x)' = \frac{1}{x}$$
$$(\arcsin x)' = \frac{1}{{\sqrt {1 - {x^2}} }}, (\arccos x)' = - \frac{1}{{\sqrt {1 - {x^2}} }}$$
$$(arctgx)' = \frac{1}{{1 + {x^2}}}, (arcctgx)' = - \frac{1}{{1 + {x^2}}}$$
五、积分公式
$$\begin{array}{c}\int tgxdx=-\ln{\vert\cos x\vert}+C,\int ctgxdx=\ln{\vert\sin x\vert}+C\\\int\sec xdx=\ln{\vert\sec x+tgx\vert}+C,\int\csc xdx=\ln{\vert\csc x-ctgx\vert}+C\\\int\frac{dx}{a^2+x^2}=\frac1aarctg\frac xa+C,\int\frac{dx}{x^2-a^2}=\frac1{2a}\ln{\vert\frac{x-a}{x+a}\vert}+C\\\int\frac{dx}{a^2-x^2}=\frac1{2a}\ln\frac{a+x}{a-x}+C,\int\frac{dx}{\sqrt{a^2-x^2}}=\arcsin\frac xa+C\\\int\frac{dx}{\cos^2x}=\int\sec^2xdx=tgx+C,\int\frac{dx}{\sin^2x}=\int\csc^2xdx=-ctgx+C\\\int\sec x⋅tgxdx=\sec x+C,\int\csc x⋅ctgxdx=-\csc x+C\\\int a^xdx=\frac{a^x}{\ln a}+C,\int shxdx=chx+C\\\int chxdx=shx+C,\int{\frac{dx}{\sqrt{x^2\pm a^2}}=\ln(x+\sqrt{x^2\pm a^2})+C}\\I_n=\int\limits_0^\frac\pi2{\sin^nxdx=}\int\limits_0^\frac\pi2\cos^nxdx=\frac{n-1}nI_{n-2}\\\int{\sqrt{x^2+a^2}dx=\frac x2\sqrt{x^2+a^2}+\frac{a^2}2\ln(x+\sqrt{x^2+a^2})+C}\\\int\sqrt{x^2-a^2}dx=\frac x2\sqrt{x^2-a^2}-\frac{a^2}2\ln{\vert x+\sqrt{x^2-a^2}\vert}+C\\\int\sqrt{a^2-x^2}dx=\frac x2\sqrt{a^2-x^2}+\frac{a^2}2\arcsin\frac xa+C\end{array}$$
六、泰勒公式
6.1、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:
$$f(x) = f(0) + f'(0)x + \frac{{f''(0)}}{{2!}}{x^2} + \cdots + \frac{{{f^{(n)}}(0)}}{{n!}}{x^n} + \cdots $$
6.2、常用几类
$${(1 + x)^m} = 1 + mx + \frac{{m(m - 1)}}{{2!}}{x^2} + \cdots + \frac{{m(m - 1) \cdots (m - n + 1)}}{{n!}}{x^n} + \cdots ( - 1 < x < 1)$$
$$\sin x = x - \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} - \cdots + {( - 1)^{n - 1}}\frac{{{x^{2n - 1}}}}{{(2n - 1)!}} + \cdots ( - \infty < x < + \infty )$$
七、高斯公式
$$\iiint\limits_{\mathrm\Omega}{(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z})dv=\underset\sum{\int{}\int\hspace{-1.167em}\bigcirc}Pdydz+Qdzdx+Rdxdy}=\underset\sum{\int{}\int\hspace{-1.167em}\bigcirc}{(P\cos\alpha+Q\cos\beta+R\cos\gamma)ds}$$
八、斯托克斯公式
$$\iint\limits_\sum{(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z})dydz+(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x})dzdx+(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dxdy=\oint\limits_{\mathrm\Gamma}Pdx+Qdy+Rdz}$$
上式左端可写为
$$\iint\limits_\sum{\vert\begin{array}{ccc}dydz&dzdx&dxdy\\\frac\partial{\partial x}&\frac\partial{\partial y}&\frac\partial{\partial z}\\P&Q&R\end{array}\vert}=\iint\limits_\sum{\vert\begin{array}{ccc}\cos\alpha&\cos\beta&\cos\gamma\\\frac\partial{\partial x}&\frac\partial{\partial y}&\frac\partial{\partial z}\\P&Q&R\end{array}\vert}$$
九、欧拉公式
$$e^{ix}=\cos x+i\sin x,其中{\{\begin{array}{c}\cos x=\frac{e^{ix}+e^{-ix}}2\\\sin x=\frac{e^{ix}-e^{-ix}}2\end{array}}$$
十、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$${a_0} = a{A_0},{a_n} = {A_n}\sin {\phi _n},{b_n} = {A_n}\cos {\phi _n},\omega t = x$$
正交性:$$1,\sin x,\cos x,\sin 2x,\cos 2x,\cdots,\sin nx,\cos nx,任意两个不同项的乘积在[ - \pi ,\pi ]$$