Maths Formula Compendium

Trigonometry

Name$0\degree$$30\degree (\frac{\pi}{6})$$45\degree (\frac{\pi}{4})$$60\degree (\frac{\pi}{3})$$90\degree (\frac{\pi}{2})$$120\degree (\frac{2\pi}{3})$$135\degree (\frac{3\pi}{4})$$150\degree (\frac{5\pi}{6})$$180\degree (\pi)$
$\sin$$0$$\frac{1}{2}$$\frac{1}{\sqrt{2}}$$\frac{\sqrt{3}}{2}$$1$$\frac{\sqrt{3}}{2}$$\frac{1}{\sqrt{2}}$$\frac{1}{2}$$0$
$\cos$$1$$\frac{\sqrt{3}}{2}$$\frac{1}{\sqrt{2}}$$\frac{1}{2}$$0$$-\frac{1}{2}$$-\frac{1}{\sqrt{2}}$$-\frac{\sqrt{3}}{2}$$-1$
$\tan$$0$$\frac{1}{\sqrt{3}}$$1$$\sqrt{3}$$\infin$$-\sqrt{3}$$-1$$-\frac{1}{\sqrt{3}}$$0$
$\cot$$\infin$$\sqrt{3}$$1$$\frac{1}{\sqrt{3}}$$0$$-\frac{1}{\sqrt{3}}$$-1$$\sqrt{3}$$-\infin$
$\sec$$1$$\frac{2}{\sqrt{3}}$$\sqrt{2}$$ 2 $$\infin$$-2$$-\sqrt{2}$$-\frac{2}{\sqrt{3}}$$-1$
$\cosec$$\infin$$2$$\sqrt{2}$$\frac{2}{\sqrt{3}}$$1$$\frac{2}{\sqrt{3}}$$\sqrt{2}$$2$$\infin$

Inverse Trigonometry

Domain (value) ($x$)Range (Angle) ($\theta$)
$sin^{-1}x$$[-1,1]$$[-\pi/2, \pi/2]$
$\cos^{-1}x$$[-1. 1]$$[0, \pi]$
$\tan^{-1}x$$R$$(-\pi/2, \pi/2)$
$\cot^{-1}x$$R$$(0, \pi)$
$\sec^{-1}x$$R- (-1, 1)$$[0, \pi-{\pi/2}$
$\cosec^{-1}x$$R-(-1, 1)$$[-\pi/2, \pi/2] - {0}$
Property 1Property 2
$\sin^{-1}(\sin\theta) = \theta$$\sin(\sin^{-1}x) = x$
$\cos^{-1}(\cos\theta) = \theta$$\cos(\cos^{-1}x) = x$
$\tan^{-1}(\tan\theta) = \theta$$\tan(\tan^{-1}x) = x$
$\cosec^{-1}(\cosec\theta) = \theta$$\cosec(\cosec^{-1}x) = x$
$\sec^{-1}(\sec\theta) = \theta$$\sec(\sec^{-1}x) = x$
$\cot^{-1}(\cot\theta) = \theta$$\cot(\cot^{-1}x) = x$
Property 3
$sin^{-1}(-x) = -\sin^{-1}{x}$$cos^{-1}(-x) \pi -\cos^{-1}{x}$
$tan^{-1}(-x) = -\tan^{-1}{x}$$cot^{-1}(-x) \pi -\cot^{-1}{x}$
$cosec^{-1}(-x) = -\cosec^{-1}{x}$$sec^{-1}(-x) \pi -\sec^{-1}{x}$
Property 4
$sin^{-1}(\frac{1}{x}) = \cosec^{-1}x$
$cos^{-1}(\frac{1}{x}) = \sec^{-1}x$
$tan^{-1}(\frac{1}{x}) = \cot^{-1}x$
Property 5
$sin^{-1}x + \cos ^{-1}x = \frac{\pi}{2}$
$tan^{-1}x + \cot ^{-1}x = \frac{\pi}{2}$
$sec^{-1}x + \cosec ^{-1}x = \frac{\pi}{2}$
Property 6
$\sin^{-1} + \sin^{-1}y = \sin^{-1} \big(x\sqrt{1-y^2} + y\sqrt{1-x^2}\big)$
$\sin^{-1} - \sin^{-1}y = \sin^{-1} \big(x\sqrt{1-y^2} - y\sqrt{1-x^2}\big)$
Property 7
$\cos^{-1} + \cos^{-1}y = \cos^{-1} \big(xy - \sqrt{1-x^2} \sqrt{1-y^2}\big)$
$\cos^{-1} - \cos^{-1}y = \cos^{-1} \big(xy + \sqrt{1-x^2} \sqrt{1-y^2}\big)$
Property 8
$\tan^{-1} - \tan^{-1}{y} = \tan^{-1}\Big(\frac{x+y}{1-xy}\Big)$
$\tan^{-1} + \tan^{-1}{y} = \tan^{-1}\Big(\frac{x-y}{1+xy}\Big)$

Property 9

$2\sin^{-1}x = 2 \sin^{-1}(\theta/2)$

$3\sin^{-1}x = \sin^{-1}(3x-4x^2)$

$2\cos^{-1}x = \cos^{-1}(2x^2-1)$

$3\cos^{-1}x = \cos^{-1}(4x^3 -3x)$

$$ 2 \tan^{-1}x = \tan^{-1}\Big(\frac{2x}{1-x^2}\Big) \\ 3\tan^{-1}x = \tan^{-1}\Big(\frac{3x-x^2}{1-3x^2}\Big) $$

Property 10

$$ 2\tan^{-1}x = \begin{cases}\sin^{-1}\Big(\frac{2x}{1+x^2}\Big)\\cos^{-1}\Big(\frac{1-x^2}{1+x^2}\Big)\end{cases} $$

$1-\cos\theta = 2 \sin^2(\theta/2)$

$1+ \cos\theta = 2,\cos^2(\theta/2)$

$\sin\theta = 2\sin(\theta/2)\cos(\theta/2)$

$2\sin^{-1}x + \sin^{-1}(-x) = \cos^{1}x$

$$ \begin{split}\cos2\theta &= \cos^2\theta - \sin^2\theta \\& = 2\cos^2\theta -1 \\ &= 1 - 2\sin^2\theta \end{split} $$

ExpressionSubstitutionSubstitution
$a^2 + x^2$$x = a\tan\theta$$x = a\cot\theta$
$a^2-x^2$$x=a\sin\theta$$x=a\cos\theta$
$x^2-a^2$$x=a\sec\theta$$x=a\cosec\theta$
$\frac{a-x}{a+x}$$x=a\cos2\theta$
$\frac{a^2+x^2}{a^2-x^2}$$x=a^2\cos2\theta$

Line Equations

  1. $ax + by + c=0$

    $m = -a/b$

  2. One Point form of line

    $$ y - y_1 = m (x-x_1) $$

  3. Two point form of line

    $$ y-y_1 = \frac{y_2-y_1}{x_2-x_1}{x-x_1} $$

  4. Intercept Form of line

    $$ \frac{x}{a} + \frac{y}{b} = 1 $$

  1. Normal form of line

    $$ x\cos \theta + y\sin\theta = P $$

  1. Point Slope form

    $$ y = mx+ c \\ \text{Where $m$ is slope of line defined as } m = \frac{y_2 - y_1}{x_2 - x_1} $$

Distance of a point from a line

$$ \text{Dist}_{PA}= \Bigg|{\frac{ax_1 + by_1 + c}{\sqrt{a^2+b^2}}}\Bigg| $$

Distance between two lines

$y = mx + c_1 \qquad y= mx+c_2$

$$ \text{d} = \Bigg| \frac{c_1 - c_2}{\sqrt{1+m^2}}\Bigg| = \Bigg|\frac{c_1 - c_2}{\sqrt{a^2+b^2}}\Bigg| $$

Angle between two lines

Where $m_1$ and $m_2$ are slopes of two lines.

$$ \tan\theta = \frac{m_2 - m_1}{1+m_2m_1} $$

  • If line $l_1$ and $l_2$ are orthogonal to each other, then. $m_1m_2 = -1$

  • Collinearity of points

    Slope of $AB$ = Slope of $AC$

Shapes CSA(Curved Surface Area), T(Total)SA, and volume

Frustum

$\text{CSA} = \pi l (r_1+ r_2)$

$\text{TSA} = \pi r_1^2 + \pi r_2^2 + \pi l(r_1+r_2)$

$\text{Volume} = \frac{1}{3}\pi h (r_1^2 + r_2^2 + r_1r_2)$

where $l= \sqrt{h^2+(r_1-r_2)^2}$

Adjoint and Inverse

$A(\text{adj} A) = |A| I_n = (\text{adj}) A$
$A^{-1} = \frac{1}{|A|} (\text{adj} A)$$|\text{adj} A| = |A|^{n-1}$
$(A^\top)^{-1} = (A^{-1})^\top$$\text{adj} \space(adj A) = |A|^{n-2} A$
$|A \enspace adj A| = |A|^{n}$$|\text{adj} \space(adj A)| = |A|^{(n-1)^{2}}$
$|A^\top| = |A| $
$AA^{-1} = I_n$
$(A^{-1})^{-1} = A$
$(AB)^{-1} = B^{-1}A^{-1}$
$\text{adj}\space AB = (\text{adj} B)(\text{adj} A)$
$|AB| = |A| |B|$
$\text{adj}A^\top = (\text{adj} A)\top$
$|KA| = K^n |A|$
$AA^{-1}=I$
$A^{-1}I = A^{-1}$

Finding Log

  1. Given we need to find $\log$ of $\log 15.27$

    • Move the decimal after 1st digit and introduce power of 10.

    $$ \log \textcolor{#f56c42}1.\textcolor{#f56c42}5\textcolor{#ad42f5}2\textcolor{#f542bc}7 \times 10^{\textcolor{#42f5f2}1} $$

    When the decimal is moved in left/right: $$ (-)\medspace \overrightarrow{\text{introduce negative powers}} \qquad \overleftarrow{\text{introduce positive powers}} \medspace(+) $$

  2. Look for $\textcolor{#f56c42}{15}$th row and column with label $\textcolor{#ad42f5}2$. which is $\bold{\textcolor{#07fc03}{1818}}$.

  3. Add Mean difference from column $\textcolor{#f542bc}7$ in the corresponding row. which is $\textcolor{#07fc03}{20}$

    $$ 1818 + 20 = \bold{\textcolor{#07fc03}{1838}} $$

  4. Write the exponent, insert decimal and write the value calculated in Step 3.

    $$ \textcolor{#42f5f2}1.\textcolor{#07fc03}{1838} $$

  5. So $\log 15.27 = \bold{\textcolor{#07fc03}{1.1838}}$

Finding AntiLog

  1. Given we need to find Antilog of $15.5932$ $$ \log k = 15.5932 \\ k = Antilog (\medspace \textcolor{#42f5f2}{15}\textcolor{#f56c42}{.59}\textcolor{#ad42f5}3\textcolor{#f542bc}2\medspace) $$

  2. Look for $0\textcolor{#f56c42}{.59}$th row and column with label $\textcolor{#ad42f5}3$. Which is $\bold{\textcolor{#07fc03}{3917}}$.

  3. Add Mean difference from column $\textcolor{#f542bc}2$ to previous result. Which is $\textcolor{#07fc03}2$

    $$ 3917 + 2 = \bold{\textcolor{#07fc03}{3919}} $$

  4. Add $1$ to characteristic $\textcolor{#42f5f2}{15} = \textcolor{#07fc03}{16}$ and add insert the decimal from left calculated in step 2. That means we need to add decimal after $\textcolor{#07fc03}{16}$th position

    $$ 3919\enspace0000\enspace0000\enspace000 .\\ \implies k = \bold{\textcolor{#07fc03}{3.919\times10^{15}}} $$

Quickly write the exponential form

Since, we want to put decimal just after 1st digit. We need to move decimal from 16th position to right after 1st digit; which will introduce +ve powers. ($16 -1 = \textcolor{#07fc03}{15}$). Since we moved 15 positions left.

$3.919 \times 10^{\textcolor{#07fc03}{15}}$