Table of Contents
Trigonometry
Identities
Pythagorean Identities
cos 2 θ + sin 2 θ = 1 tan 2 θ + 1 = sec 2 θ cot 2 θ + 1 = csc 2 θ \begin{gathered}
\cos^2{\theta} + \sin^2{\theta} = 1 \\
\tan^2{\theta} + 1 = \sec^2{\theta} \\
\cot^2{\theta} + 1 = \csc^2{\theta}
\end{gathered} cos 2 θ + sin 2 θ = 1 tan 2 θ + 1 = sec 2 θ cot 2 θ + 1 = csc 2 θ Angle-Sum Identities
sin ( α + β ) = sin α cos β + sin β cos α cos ( α + β ) = cos α cos β − sin α sin β \begin{aligned}
\sin\p{\alpha + \beta} &= \sin{\alpha}\cos{\beta} + \sin{\beta}\cos{\alpha} \\
\cos\p{\alpha + \beta} &= \cos{\alpha}\cos{\beta} - \sin{\alpha}\sin{\beta}
\end{aligned} sin ( α + β ) cos ( α + β ) = sin α cos β + sin β cos α = cos α cos β − sin α sin β From these, you can derive the double-angle identities by plugging in α = β = θ \alpha = \beta = \theta α = β = θ
sin 2 θ = 2 sin θ cos θ cos 2 θ = cos 2 θ − sin 2 θ = 2 cos 2 θ − 1 = 1 − 2 sin 2 θ . \begin{aligned}
\sin{2\theta} &= 2\sin{\theta}\cos{\theta} \\
\cos{2\theta}
&= \cos^2{\theta} - \sin^2{\theta} \\
&= 2\cos^2{\theta} - 1 \\
&= 1 - 2\sin^2{\theta}.
\end{aligned} sin 2 θ cos 2 θ = 2 sin θ cos θ = cos 2 θ − sin 2 θ = 2 cos 2 θ − 1 = 1 − 2 sin 2 θ . If you solve for cos 2 θ \cos^2{\theta} cos 2 θ sin 2 θ \sin^2{\theta} sin 2 θ
cos 2 θ = 1 + cos 2 θ 2 sin 2 θ = 1 − cos 2 θ 2 . \begin{aligned}
\cos^2{\theta} &= \frac{1 + \cos{2\theta}}{2} \\
\sin^2{\theta} &= \frac{1 - \cos{2\theta}}{2}.
\end{aligned} cos 2 θ sin 2 θ = 2 1 + cos 2 θ = 2 1 − cos 2 θ . You can also derive the sum-to-product identities from these, but I'm not sure if you'll need them.
Integrals
Note that a lot of the following derivatives and integrals look really similar. You really only need to remember the ones for sin x \sin{x} sin x tan x \tan{x} tan x sec x \sec{x} sec x cos x \cos{x} cos x cot x \cot{x} cot x csc x \csc{x} csc x
∫ tan x d x = − ln ∣ sec x ∣ + C ∫ cot x d x = − ln ∣ csc x ∣ + C ∫ sec x d x = − ln ∣ sec x + tan x ∣ + C ∫ csc x d x = − ln ∣ csc x + cot x ∣ + C \begin{aligned}
\int \tan{x} \,\diff{x} &= \phantom{-}\ln{\abs{\sec{x}}} + C \\
\int \cot{x} \,\diff{x} &= -\ln{\abs{\csc{x}}} + C \\
\int \sec{x} \,\diff{x} &= \phantom{-}\ln{\abs{\sec{x} + \tan{x}}} + C \\
\int \csc{x} \,\diff{x} &= -\ln{\abs{\csc{x} + \cot{x}}} + C
\end{aligned} ∫ tan x d x ∫ cot x d x ∫ sec x d x ∫ csc x d x = − ln ∣ sec x ∣ + C = − ln ∣ csc x ∣ + C = − ln ∣ sec x + tan x ∣ + C = − ln ∣ csc x + cot x ∣ + C Inverse Functions
Derivatives
d d x arcsin x = − 1 1 − x 2 d d x arctan x = − 1 1 + x 2 d d x arcsec x = − 1 ∣ x ∣ x 2 − 1 d d x arccos x = − 1 1 − x 2 d d x arccot x = − 1 1 + x 2 d d x arccsc x = − 1 ∣ x ∣ x 2 − 1 \begin{aligned}
\deriv{}{x} \arcsin{x} &= \phantom{-}\frac{1}{\sqrt{1 - x^2}} \\
\deriv{}{x} \arctan{x} &= \phantom{-}\frac{1}{1 + x^2} \\
\deriv{}{x} \arcsec{x} &= \phantom{-}\frac{1}{\abs{x}\sqrt{x^2 - 1}} \\
\deriv{}{x} \arccos{x} &= -\frac{1}{\sqrt{1 - x^2}} \\
\deriv{}{x} \arccot{x} &= -\frac{1}{1 + x^2} \\
\deriv{}{x} \arccsc{x} &= -\frac{1}{\abs{x}\sqrt{x^2 - 1}} \\
\end{aligned} d x d arcsin x d x d arctan x d x d arcsec x d x d arccos x d x d arccot x d x d arccsc x = − 1 − x 2 1 = − 1 + x 2 1 = − ∣ x ∣ x 2 − 1 1 = − 1 − x 2 1 = − 1 + x 2 1 = − ∣ x ∣ x 2 − 1 1 Integrals
Any time you have a formula for a derivative, you automatically get a formula for an integral by going backwards:
∫ 1 1 − x 2 d x = arcsin x + C ∫ 1 1 + x 2 d x = arctan x + C ∫ 1 ∣ x ∣ x 2 − 1 d x = arcsec x + C \begin{aligned}
\int \frac{1}{\sqrt{1 - x^2}} \,\diff{x} &= \arcsin{x} + C \\
\int \frac{1}{1 + x^2} \,\diff{x} &= \arctan{x} + C \\
\int \frac{1}{\abs{x}\sqrt{x^2 - 1}} \,\diff{x} &= \arcsec{x} + C
\end{aligned} ∫ 1 − x 2 1 d x ∫ 1 + x 2 1 d x ∫ ∣ x ∣ x 2 − 1 1 d x = arcsin x + C = arctan x + C = arcsec x + C Series
Convergence Tests
Theorem (divergence test)
Let { a n } \set{a_n} { a n } lim n → ∞ a n \displaystyle \lim_{n\to\infty} a_n n → ∞ lim a n 0 0 0 ∑ n = 0 ∞ a n \displaystyle \sum_{n=0}^\infty a_n n = 0 ∑ ∞ a n
Theorem (direct comparison test)
Suppose 0 ≤ a n ≤ b n 0 \leq a_n \leq b_n 0 ≤ a n ≤ b n n ≥ M n \geq M n ≥ M M > 0 M > 0 M > 0
If ∑ n = 0 ∞ a n \displaystyle \sum_{n=0}^\infty a_n n = 0 ∑ ∞ a n ∑ n = 0 ∞ b n \displaystyle \sum_{n=0}^\infty b_n n = 0 ∑ ∞ b n
If ∑ n = 0 ∞ b n \displaystyle \sum_{n=0}^\infty b_n n = 0 ∑ ∞ b n ∑ n = 0 ∞ a n \displaystyle \sum_{n=0}^\infty a_n n = 0 ∑ ∞ a n
Theorem (limit comparison test)
Theorem (ratio test)
Let L = lim n → ∞ ∣ a n + 1 a n ∣ \displaystyle L = \lim_{n\to\infty} \abs{\frac{a_{n+1}}{a_n}} L = n → ∞ lim ∣ ∣ a n a n + 1 ∣ ∣
If L < 1 L < 1 L < 1 ∑ n = 1 ∞ a n \displaystyle \sum_{n=1}^\infty a_n n = 1 ∑ ∞ a n
If L > 1 L > 1 L > 1 ∑ n = 1 ∞ a n \displaystyle \sum_{n=1}^\infty a_n n = 1 ∑ ∞ a n
If L = 1 L = 1 L = 1
Theorem (root test)
Let L = lim n → ∞ ∣ a n ∣ 1 / n \displaystyle L = \lim_{n\to\infty} \abs{a_n}^{1/n} L = n → ∞ lim ∣ a n ∣ 1/ n
If L < 1 L < 1 L < 1 ∑ n = 1 ∞ a n \displaystyle \sum_{n=1}^\infty a_n n = 1 ∑ ∞ a n
If L > 1 L > 1 L > 1 ∑ n = 1 ∞ a n \displaystyle \sum_{n=1}^\infty a_n n = 1 ∑ ∞ a n
If L = 1 L = 1 L = 1
Theorem (alternating series test)
Let { a n } \set{a_n} { a n } a n = ( − 1 ) n b n a_n = \p{-1}^n b_n a n = ( − 1 ) n b n b n ≥ 0 b_n \geq 0 b n ≥ 0 lim n → ∞ b n = 0 \displaystyle\lim_{n\to\infty} b_n = 0 n → ∞ lim b n = 0 ∑ n = 0 ∞ a n \displaystyle \sum_{n=0}^\infty a_n n = 0 ∑ ∞ a n
Common Power Series Expansions
f ( x ) = T ( x ) 1 1 − x = ∑ n = 0 ∞ x n = 1 + x + x 2 + x 3 + ⋯ for x ∈ ( − 1 , 1 ) e x = ∑ n = 0 ∞ x n n ! = 1 + x + x 2 2 ! + x 3 3 ! + ⋯ for x ∈ R sin x = ∑ n = 0 ∞ ( − 1 ) n x 2 n + 1 ( 2 n + 1 ) ! = x − x 3 3 ! + x 5 5 ! − x 7 7 ! + ⋯ for x ∈ R cos x = ∑ n = 0 ∞ ( − 1 ) n x 2 n ( 2 n ) ! = 1 − x 2 2 ! + x 4 4 ! − x 6 6 ! + ⋯ for x ∈ R \begin{array}{rcllll}
\hline \\[-2ex]
f\p{x}
& =
& T\p{x} \\[1ex] \hline \\[-2ex]
\displaystyle\frac{1}{1 - x}
& =
& \displaystyle\sum_{n=0}^\infty x^n
& =
& \displaystyle 1 + x + x^2 + x^3 + \cdots
& \text{for } x \in \p{-1, 1} \\[3ex]
e^x & =
& \displaystyle\sum_{n=0}^\infty \frac{x^n}{n!}
& =
& \displaystyle 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots
& \text{for } x \in \R \\[3ex]
\sin{x}
& =
& \displaystyle\sum_{n=0}^\infty \p{-1}^n\frac{x^{2n+1}}{\p{2n+1}!}
& =
& \displaystyle x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots
& \text{for } x \in \R \\[3ex]
\cos{x}
& =
& \displaystyle\sum_{n=0}^\infty \p{-1}^n\frac{x^{2n}}{\p{2n}!}
& =
& \displaystyle 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots
& \text{for } x \in \R \\[3ex]\hline
\end{array} f ( x ) 1 − x 1 e x sin x cos x = = = = = T ( x ) n = 0 ∑ ∞ x n n = 0 ∑ ∞ n ! x n n = 0 ∑ ∞ ( − 1 ) n ( 2 n + 1 ) ! x 2 n + 1 n = 0 ∑ ∞ ( − 1 ) n ( 2 n ) ! x 2 n = = = = 1 + x + x 2 + x 3 + ⋯ 1 + x + 2 ! x 2 + 3 ! x 3 + ⋯ x − 3 ! x 3 + 5 ! x 5 − 7 ! x 7 + ⋯ 1 − 2 ! x 2 + 4 ! x 4 − 6 ! x 6 + ⋯ for x ∈ ( − 1 , 1 ) for x ∈ R for x ∈ R for x ∈ R Finding Power Series Expansions
Usually, you will want to start from one of the common power series expansions above and use the following operations to get new ones:
Multiplying by C x k Cx^k C x k
Plugging in C x k Cx^k C x k
Differentiating
Integrating
Important: If you differentiate or integrate a power series, the endpoints of the interval of convergence stays the same, but you have to check whether the new series converges or diverges at the endpoints.
Taylor Polynomials
The n n n f ( x ) f\p{x} f ( x ) a a a
T n f ( x ) = ∑ k = 0 n f ( k ) ( a ) k ! ( x − a ) k . T_nf\p{x} = \sum_{k=0}^n \frac{f^{\p{k}\p{a}}}{k!} \p{x - a}^k. T n f ( x ) = k = 0 ∑ n k ! f ( k ) ( a ) ( x − a ) k . Error Bound
Theorem (error bound)
Let f ( x ) f\p{x} f ( x ) n + 1 n+1 n + 1 T n f ( x ) T_nf\p{x} T n f ( x ) n n n f f f a a a
∣ T n f ( x ) − f ( x ) ∣ ≤ K ( n + 1 ) ! ∣ x − a ∣ n + 1 , \abs{T_nf\p{x} - f\p{x}} \leq \frac{K}{\p{n+1}!} \abs{x - a}^{n+1}, ∣ T n f ( x ) − f ( x ) ∣ ≤ ( n + 1 ) ! K ∣ x − a ∣ n + 1 , where K K K ∣ f ( n + 1 ) ∣ \abs{f^{\p{n+1}}} ∣ ∣ f ( n + 1 ) ∣ ∣ x x x a a a [ a , x ] \br{a, x} [ a , x ] [ x , a ] \br{x, a} [ x , a ] a ≤ x a \leq x a ≤ x x ≤ a x \leq a x ≤ a