$ x^2 \pm bx + c $ = $ \left(x \pm {b \over 2} \right)^2 - \left(b \over 2\right)^2 + c $

Make sure the coefficient of $x^2$ is equals to $1$ before applying the formula (see part ii and iii below).

$$ a(x - h)^2 + k $$

$ \text{If } a > 0,$ $ \text{ expression has minimum value of } k \text{ when } x = h $

$ \text{If } a < 0,$ $ \text{ expression has maximum value of } k \text{ when } x = h $

(a) Solve $(x - a)(x - b) > 0$, given that $ b > a $.

Solutions

$$ x < a \text{ or } x > b $$

(b) Solve $(x - a)(x - b) < 0,$ given that $ b > a $:

Solutions

$$ a < x < b $$

1. If equation has two real and distinct roots, then $b^2 - 4ac$ $ > 0 $

2. If equation has two real and equal roots (or one real root), then $b^2 - 4ac$ $ = 0 $

3. If equation has real roots (either one or two real roots), then $b^2 - 4ac$ $ \ge 0 $

4. If equation has no real roots, then $b^2 - 4ac$ $ < 0 $

$\text{Condition 1: } $ $ a > 0 $

$\text{Condition 2: } b^2 - 4ac $ $ < 0 $

$\text{Condition 1: } $ $ a < 0 $

$\text{Condition 2: } b^2 - 4ac $ $ < 0 $

1. If line meets curve at two points, then $b^2 - 4ac$ $ > 0 $

2. If line meets curve once or is tangent to the curve, then $b^2 - 4ac$ $ = 0 $

3. If line meets curve once or twice, then $b^2 - 4ac$ $ \ge 0 $

4. If line does not meet the curve, then $b^2 - 4ac$ $ < 0 $

$ \sqrt{a} \times \sqrt{a} = $ $ \phantom{.} a $

$ \sqrt{a} \times \sqrt{b} = $ $ \sqrt{a \times b} $

$ { \sqrt{a} \over \sqrt{b} }= $ $ \sqrt{a \over b} $

$ {2 \over 3 \sqrt{5}} \times $ $ {\sqrt{5} \over \sqrt{5}} = {2\sqrt{5} \over 3(5)} = {2\sqrt{5} \over 15} $

$ {1 \over 4 + \sqrt{5}} \times $ $ {4 - \sqrt{5} \over 4 - \sqrt{5}} = {4 - \sqrt{5} \over (4)^2 - (\sqrt{5})^2} = {4 - \sqrt{5} \over 11} $

$ \text{Polynomial} = $ $ \text{Divisor} \times \text{Quotient} + \text{Remainder} $

If the polynomial $f(x)$ has a linear factor $x + a$, then $f(-a)$ = $ \phantom{.} 0 $

If $x + a$ is not a factor of the polynomial $f(x)$, then when $f(x)$ is divided by $x + a$, remainder = $ f(-a) $

$ a^3 + b^3 = $ $ (a + b)(a^2 - ab + b^2) $

$ a^3 - b^3 = $ $ (a - b)(a^2 + ab + b^2) $

For an algebraic fraction in the form ${f(x) \over g(x)}$,

• If the degree of $f(x)$ < the degree of $g(x)$, then it is a proper fraction
• If the degree of $f(x)$ ≥ the degree of $g(x)$, then it is an improper fraction

Linear factors: $ {mx + n \over (ax + b)(cx - d)} = $ ${A \over ax + b} + {B \over cx - d}$

Repeated linear factors: $ {mx + n \over (ax + b)(cx - d)^2} = $ ${A \over ax + b} + {B \over cx - d} + {C \over (cx - d)^2}$

Quadratic factor that cannot be factorised: $ {mx + n \over (ax + b)(x^2 + d)} = $ ${A \over ax + b} + {Bx + C \over x^2 + d}$

1. Perform long division and express result in the form $ {\text{Polynomial} \over \text{Divisor}} = $ $ \text{Quotient} + { \text{Remainder} \over \text{Divisor} } $
2. Break down the proper fraction into partial fractions

\begin{align} (a+b)^n & =a^n+\binom{n}{1}a^{n-1}b+\binom{n}{2}a^{n-2}b^2+...+\binom{n}{r}a^{n-r}b^r+ \ldots + b^n \\ \\ \text{where } n \text{ is a positive integer and }&\binom{n}{r}={n!\over r!(n-r)!}={n(n-1) \ldots (n-r+1)\over r!} \end{align}

${n \choose 1}$ = $ \phantom{.} n $

${n \choose 2}$ = $ {n(n - 1) \over 2!} $

${n \choose 3}$ = $ {n(n - 1)(n - 2) \over 3!} $

$\text{General term, } T_{r + 1} = $ $ {n \choose r} a^{n - r} b^r $

$ \text{In the expansion of } (a + b)^n, \text{ there are } $$ n + 1 $ $ \text{ terms} $

$ a^m \times a^n = (a^m)(a^n) = \phantom{.} $ $ a^{m + n} $

$ a^m \div a^n = {a^m \over a^n} = \phantom{.} $ $ a^{m - n} $

$ a^m \times b^m = (a^m)(b^m) = \phantom{.} $ $ (ab)^m $

$ a^m \div b^m = {a^m \over b^m} = \phantom{.} $ $ \left(a \over b \right)^m $

$ (a^m)^n = \phantom{.} $ $ a^{mn} $

$ a^0 = \phantom{.} $ $ 1 $

$ a^{-n} = \phantom{.} $ $ {1 \over a^n} $

$ {1 \over a^{-n}} = \phantom{.} $ $ a^{-(-n)} = a^n $

$ \left(a \over b\right)^{-n} = \phantom{.} $ $ \left(b \over a\right)^n $

$ a^{1 \over n} = \phantom{.} $ $ \sqrt[n]{a} $

$ a^{1 \over 2} = \phantom{.} $ $ \sqrt{a} $

$ a^{m \over n} = \phantom{.} $ $ \sqrt[n]{a^m} $

(a) Solve $3^x = -3$

Solutions

(b) Solve $3^x = 81$

Solutions

(c) Solve $3^x = 4$

Solutions

(a) Graph of $y = a^x$, for $a > 1$

Shape

(b) Graph of $y = a^x$, for $0 < a < 1$

Shape

$ y = \log_a x \phantom{.} \Longleftrightarrow \phantom{.} x = $ $ \phantom{.} a^y $

1. $a$ is a positive real number and not equals to 1
2. $x$ is a positive real number

$ \log_a 1 = $ $ \phantom{.} 0 $

$ \log_a a = $ $ \phantom{.} 1 $

$ \lg x = $ $ \log_{10} x $ and $\lg 10 = $ $ \log_{10} 10 = 1 $

$ \ln x = $ $ \log_{e} x $ and $\ln e = $ $ \log_e e = 1 $

Product law: $ \log_a xy = $ $ \log_a x + \log_a y $

Quotient law: $ \log_a {x \over y} = $ $ \log_a x - \log_a y $

Power law: $ \log_a x^r = $ $ r \log_a x $

$ \log_a b = $ $ {\log_c b \over \log_c a} $

(a) Graph of $y = \log_a x, a > 1$

Shape

(b) Graph of $y = \log_a x, 0 < a < 1$

Shape

$ \text{Length of } AB $ = $ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $

$ \text{Gradient of } AB $ = $ {y_2 - y_1 \over x_2 - x_1} $

$ \text{Midpoint of } AB $ = $ \left( {x_1 + x_2 \over 2}, {y_1 + y_2 \over 2} \right) $

$ \text{Equation: } $$ y = mx + c $

$ m \text{ represents the } $$ \text{gradient of the straight line} $

$ c \text{ represents the } $$ y\text{-intercept of the straight line} $

$ \text{Equation of horizontal line passing though } (a, b) \text{:} $$ \phantom{.} y = b $

$ \text{Gradient of line} $ = $ \phantom{0} 0 \phantom{0} $

$ \text{Equation of vertical line passing though } (a, b) \text{:} $$ \phantom{.} x = a $

$ \text{Gradient of line is} $ $ \text{ undefined } $

If two lines are parallel, they have the same gradient, i.e. $m_1 = m_2$

If two lines are not parallel, the lines will meet at a point. The coordinates of the point of intersection can be found by solving simultaneous equations using the equation of each line.

If two lines are perpendicular, then the gradient of both lines are related by $m_1 \times m_2 = $ $ - 1 $.

$ \text{Area of figure} $ = $ {1 \over 2} \left| \begin{matrix} x_1 & x_2 & ... & x_n & x_1 \\ y_1 & y_2 & ... & y_n & y_1 \end{matrix} \right| $

Steps:

1. Select the points in anti-clockwise order
2. Repeat the first point chosen
3. When calculating the area, go ↘ , then go ↗

Example

Find the area of the quadrilateral $ABCD$.

Solutions

Column vectors is usually taught in Secondary 4 E Maths. It is useful to solve questions like the following!

Example

The points $A(-2, 1)$, $B(0, 2)$ and $C$ lie on a straight line as shown above. Given that $AB: AC = 1:3$, find the coordinates of $C$.

Solutions

$ \text{Equation: } $$ (x - a)^2 + (y - b)^2 = r^2 $

$ \text{Centre: } $$ (a, b) $

$ \text{Radius} = \phantom{.} $$ r \text{ units} $

$ \text{Equation: } $$ x^2 + y^2 + 2gx + 2fy + c = 0 $

$ \text{Centre: } $$ (-g, -f) $

$ \text{Radius} = \phantom{.} $$ \sqrt{ g^2 + f^2 - c } $

If points $A$, $B$ and $C$ lie on the circle such that $ \angle ABC = 90^\circ $, then $AC$ is the diameter of the circle.

Thus, the centre of the circle is the midpoint of $AC$ and the radius of the circle is equals to $ {\text{Length of } AC \over 2} $.

The perpendicular bisector of chord $AB$ passes through the centre of the circle.

If $O$ is the centre of the circle and the line $ABC$ is tangent to the circle at $B$, then $ \angle OBA = \angle OBC = 90^\circ $.

The normal to the circle at $B$ is perpendicular ($m_1 \times m_2 = -1$) to the tangent to the circle at $B$ and passes through the centre of the circle..

$$ y = a \sqrt{x} + {b \over \sqrt{x}} $$

Solutions

$$ y = ax^b $$

Solutions

$ \angle a = \angle b \phantom{0} ($ $ \text{Vertically opposite angles} $ $\text{)}$

$ \angle a = \angle b \phantom{0} ($ $ \text{Alternate angles} $ $\text{)}$

$ \angle a + \angle c = $ $ 180^\circ \text{ (Interior angles} $$\text{)}$

$ \angle a = \angle d \phantom{0} ($ $ \text{Corresponding angles} $ $\text{)}$

If D is the midpoint of AB and E is the midpoint of AC, then

$ DE \phantom{.} // \phantom{.} BC $ $\text{ and } $ $ DE = {1 \over 2} BC $

$ \angle ACB = \angle ADB \text{ (}$ $ \text{Angles in the same segment} $ $ \text{)} $

If O is the centre of the circle,

$ \angle AOB = 2 \times \angle ACB \text{ (}$ $ \text{Angle at centre} = 2 \times \text{Angle at circumference} $ $ \text{)} $

$ \angle ACB + \angle ADB = 180^\circ \text{ (}$ $ \text{Angles in opposite segments} $ $ \text{)} $

If AOB is the diameter of the circle,

$ \angle ACB = 90^\circ \text{ (}$ $ \text{Right-angle in semi-circle} $ $ \text{)} $

If O is the centre of the circle and the line AC is tangent to the circle at B,

$ \angle OBA = 90^\circ \text{ (}$ $ \text{Tangent perpendicular to radius} $ $ \text{)} $

If O is the centre of the circle and tangents to the circle at A and at B meet at an external point T, then

$ AT = BT $ $ \text{ and } $ $ \angle OTA = \angle OTB $

If the line AC is tangent to the circle at B,

$ \angle DBA = $$ \angle DEB $ $ \text{ (Alternate segment theorem)} $

$ \angle EBC = $$ \angle EDB $ $ \text{ (Alternate segment theorem)} $

Side-Side-Side (SSS)

Angle-Side-Angle (ASA)

Angle-Angle-Side (AAS)

Right angle-Hypotenuse-Side (RHS)

Side-Angle-Side (SAS)

Side-Side-Side (SSS)

Angle-angle (AA)

Side-Angle-Side (SAS)

(a) Shape and features of $y = a \sin bx + c$, when $ a > 0 $

Shape

\begin{align*} \text{Center line: } & y = c \\ \\ \text{Amplitude} & = a \\ \\ \text{Period} & = {360^\circ \over b} = {2\pi \text{ radians} \over b} \end{align*}

(b) Shape and features of $y = a \sin bx + c$, when $ a < 0 $

Shape

\begin{align*} \text{Center line: } & y = c \\ \\ \text{Amplitude} & = a \\ \\ \text{Period} & = {360^\circ \over b} = {2\pi \text{ radians} \over b} \end{align*}

(a) Shape and features of $y = a \cos bx + c$, when $ a > 0 $

Shape

\begin{align*} \text{Center line: } & y = c \\ \\ \text{Amplitude} & = a \\ \\ \text{Period} & = {360^\circ \over b} = {2\pi \text{ radians} \over b} \end{align*}

(b) Shape and features of $y = a \cos bx + c$, when $ a < 0 $

Shape

\begin{align*} \text{Center line: } & y = c \\ \\ \text{Amplitude} & = \text{Positive value of } a \\ \\ \text{Period} & = {360^\circ \over b} = {2\pi \text{ radians} \over b} \end{align*}

(a) Shape and features of $y = a \tan bx$, when $ a > 0 $

Shape

Note: The graph cannot pass through the vertical asymptote $x = {90^\circ \over b}$

\begin{align*} \text{Period} & = {180^\circ \over b} = {\pi \text{ radians} \over b} \end{align*}

(b) Shape and features of $y = a \tan bx$, when $ a < 0 $

Shape

Note: The graph cannot pass through the vertical asymptote $x = {90^\circ \over b}$

\begin{align*} \text{Period} & = {180^\circ \over b} = {\pi \text{ radians} \over b} \end{align*}

sin θ is positive in the first and second quadrants and negative in the third and fourth quadrants.

cos θ is positive in the first and fourth quadrants and negative in the second and third quadrants.

tan θ is positive in the first and third quadrants and negative in the second and fourth quadrants.

$ \cot \theta = $ $ \phantom{.} {1 \over \tan \theta} $

$ \sec \theta = $ $ \phantom{.} {1 \over \cos \theta} $

$ \text{cosec } \theta = $ $ \phantom{.} {1 \over \sin \theta} $

$ \tan \theta = $ $ \phantom{.} {\sin \theta \over \cos \theta} $

$ \cot \theta = $ $ \phantom{.} {1 \over \tan \theta} = {\cos \theta \over \sin \theta} $

$a \sin x \pm b \cos x = $ $ R \sin (x \pm \alpha) $

$a \cos x \pm b \sin x = $ $ R \cos (x \mp \alpha) $

$ \text{For both, } R = $ $\sqrt{a^2 + b^2}$ and $\alpha = $ $ \tan^{-1} \left(b \over a\right) $

Principal values of $ \sin^{-1} x $: $ -90^\circ \le \sin^{-1} x \le 90^\circ $

Principal values of $ \cos^{-1} x $: $ 0^\circ \le \cos^{-1} x \le 180^\circ $

Principal values of $ \tan^{-1} x $: $ -90^\circ < \tan^{-1} x < 90^\circ $

If $a$ is a constant, $ {d \over dx} (a) = $ $ \phantom{0} 0 \phantom{0} $

If $a$ is a constant, $ {d \over dx} (ax^n) = $ $ a n x^{n - 1} $

$ {d \over dx} \{ \sin [f(x)] \} = $ $ f'(x) . \cos [f(x)] $

$ {d \over dx} \{ \cos [f(x)] \} = $ $ f'(x) . - \sin [f(x)] $

$ {d \over dx} \{ \tan [f(x)] \} = $ $ f'(x) . \sec^2 [f(x)] $

Note: $f'(x)$ is the derivative of $f(x)$

$ {d \over dx} \left[ e^{f(x)} \right] = $ $ f'(x) . e^{f(x)} $

Note: $f'(x)$ is the derivative of $f(x)$

$ {d \over dx} \{ \ln [f(x)] \} = $ $ {f'(x) \over f(x)} $

Note: $f'(x)$ is the derivative of $f(x)$

$ {d \over dx} [f(x)]^n = $ $ n [f(x)]^{n - 1} . f'(x) $

Note: $f'(x)$ is the derivative of $f(x)$

$ {d \over dx} [\sin^n f(x) ] = $ $ (n) [ \sin^{n - 1} f(x) ] . f'(x) . \cos f(x) $

$ {d \over dx} [\cos^n f(x) ] = $ $ (n) [ \cos^{n - 1} f(x) ] . f'(x) . - \sin f(x) $

$ {d \over dx} [\tan^n f(x) ] = $ $ (n) [ \tan^{n - 1} f(x) ] . f'(x) . \sec^2 f(x) $

$ {d \over dx} (uv) = $ $ u {dv \over dx} + v{du \over dx} $

$ {d \over dx} \left(u \over v\right) = $ $ {v {du \over dx} - u {dv \over dx} \over v^2} $

If $a$ is a constant, $ \int a \phantom{.} dx = $ $ ax + C $

If $a$ is constant and $n \ne -1$, $ \int ax^{n} \phantom{.} dx = $ $ {a \over n + 1} x^{n+ 1} + C $

If $a$ is constant, $ \int ax^{-1} \phantom{.} dx = \int {a \over x} \phantom{.} dx = $ $ a \ln x + C $

If $n \ne -1$, $ \int [f(x)]^n \phantom{.} dx = $ $ { [f(x)]^{n + 1} \over f'(x). (n + 1) } + C $

$ \int [f(x)]^{-1} \phantom{.} dx = \int {1 \over f(x)} \phantom{.} dx = $ $ {\ln [f(x)] \over f'(x)} + C $

Note: $f'(x)$ is the derivative of $f(x)$

$ \int e^{f(x)} \phantom{.} dx = $ $ {e^{f(x)} \over f'(x)} + C $

Note: $f'(x)$ is the derivative of $f(x)$

$ \int \sin [f(x)] \phantom{.} dx = $ $ {- \cos [f(x)] \over f'(x)} + C $

$ \int \cos [f(x)] \phantom{.} dx = $ $ {\sin [f(x)] \over f'(x)} + C $

$ \int \sec^2 [f(x)] \phantom{.} dx = $ $ {\tan [f(x)] \over f'(x)} + C $

Note: $f'(x)$ is the derivative of $f(x)$

Example 1

Given that ${d \over dx} \left[ x \over (x + 1)^2 \right] = {1 - x \over (x + 1)^3}$, find $ \int {2 - 2x \over (x + 1)^3} \phantom{.} dx$.

Solutions

Example 2

Given that ${d \over dx} ( x \ln x) = 1 + \ln x $, find $ \ln x \phantom{.} dx$.

Solutions

The first derivative, ${dy \over dx}$, is the rate of change of $y$ with respect to $x$ (i.e. gradient of the curve). For example, if ${dy \over dx} = 2$, it means that when $x$ increases by $1$ unit, $y$ increases by $2$ units.

The second derivative, ${d^2 y \over dx^2}$, is the rate of change ${dy \over dx}$ with respect to $x$. For example, if ${d^2 y \over dx^2} = 2$, it means that when $x$ increases by $1$ unit, ${dy \over dx}$ increases by $2$ units.

$ \text{Gradient of tangent at } P(x_1, y_1) = $ $ \left. {dy \over dx} \right|_{x = x_1} $

$ \text{Gradient of normal at } P(x_1, y_1) = $ $ {-1 \over \text{Gradient of tangent at } P(x_1, y_1)} $

To form the equation of the line, you need the gradient of the line and the coordinates of a point that the line passes through.

For an increasing function (as $x$ increases, $y$ increases), ${dy \over dx}$ $ > 0 $

For a decreasing function (as $x$ increases, $y$ decreases), ${dy \over dx}$ $ < 0 $

(a) At stationary point(s), ${dy \over dx} = $$ \phantom{.} 0 \phantom{.} $. A stationary point can be:

1. A maximum point (/ ‾ \)
2. A minimum point (\ _ /)
3. A stationary point of inflexion (/ - / or \ - \)

(b) First derivative test (format):

(c) Second derivative test:

• ${d^2 y \over dx^2} > 0 $ means the point is a minimum point
• ${d^2 y \over dx^2} < 0 $ means the point is a maximum point
• ${d^2 y \over dx^2} = 0 $ means the test is inconclusive (use first derivative test instead)

If $y$ and $x$ are related, then

$ {dy \over dt} = $ $ {dy \over dx} \times {dx \over dt} $

$ {dx \over dt} = $ $ {dx \over dy} \times {dy \over dt} $

$ - \int_a^b f(x) \phantom{.} dx = $ $ \int_b^a f(x) \phantom{.} dx $

$ \text{If } k \text{ is a constant, } \int_a^b k f(x) \phantom{.} dx = $ $ k \int_a^b f(x) \phantom{.} dx $

$ \int_a^b f(x) \pm g(x) \phantom{.} dx = $ $ \int_a^b f(x) \phantom{.} dx \pm \int_a^b g(x) \phantom{.} dx $

$ \int_a^b f(x) \phantom{.} dx + \int_b^c f(x) \phantom{.} dx = $ $ \int_a^c f(x) \phantom{.} dx $

$ \text{Area bounded} = $ $ \int_{x_1}^{x_2} f(x) \phantom{.} dx $

$ \text{Area bounded} = $ $ - \int_{x_1}^{x_2} f(x) \phantom{.} dx $

$ \text{Area bounded} = $ $ \int_{y_1}^{y_2} f(y) \phantom{.} dy $

$ \text{Area bounded} = $ $ - \int_{y_1}^{y_2} f(y) \phantom{.} dy $

$ \text{Area of region A} = $ $ \text{Area of triangle} $

$ \text{Area of region B} = $ $ \int_2^3 -x^2 + 2x + 3 \phantom{.} dx $

$ \text{Area of region D} = $ $ \text{Area of triangle} $

$ \text{Area of region C} = $ $ \int_{-1}^2 -x^2 + 2x + 3 \phantom{.} dx - \text{Area of region D} $

Displacement is the shortest distance between the initial position and the current position of an object.

Key terms:

• Initial displacement refers to the value of $s$ when $t = 0$
• Object is at starting/reference point, $O$, implies $s = 0$
• Object reaches maximum displacement means $ {ds \over dt} = 0 \text{ and } {d^2 s \over dt^2} < 0 $
• Object reaches minimum displacement means $ {ds \over dt} = 0 \text{ and } {d^2 s \over dt^2} > 0 $

Velocity, $v = {ds \over dt}$, is the rate of change of displacement.

If $v > 0$, the object is moving in the positive direction. If $v < 0$, the object is moving in the opposite direction.

Key terms:

• Initial velocity refers to the value of $v$ when $t = 0$
• Object is momentarily at rest means $v = 0$ (note: object may change direction)
• Object reaches maximum velocity means $ {dv \over dt} = 0 \text{ and } {d^2 v \over dt^2} < 0 $
• Object reaches minimum velocity means $ {dv \over dt} = 0 \text{ and } {d^2 v \over dt^2} > 0 $

Acceleration, $a = {dv \over dt}$, is the rate of change of velocity.

If $a = 0$, then the object is moving at constant velocity (or at rest). If $a > 0$, the object is accelerating (i.e. velocity is increasing). If $a < 0$, the object is decelerating (i.e. velocity is decreasing).

Key terms:

• Initial acceleration refers to the value of $a$ when $t = 0$
• Object reaches maximum acceleration means $ {da \over dt} = 0 \text{ and } {d^2 a \over dt^2} < 0 $
• Object reaches minimum velocity means $ {da \over dt} = 0 \text{ and } {d^2 a \over dt^2} > 0 $

$ (a + b)^2 = $ $ a^2 + 2ab + b^2 $

$ (a - b)^2 = $ $ a^2 - 2ab + b^2 $

$ a^2 - b^2 = $ $ (a + b)(a - b) $

$ a^2 + b^2 = c^2 $

$ \text{Area of parallelogram} = $ $ \text{Base} \times \text{Height} $

$ \text{Area of trapezium} = $ $ {1 \over 2} \times \text{Sum of parallel sides} \times \text{Height} $

$ \text{Volume of prism} = $ $ \text{Cross-sectional area} \times \text{Height/Length} $

$ \text{Volume of cylinder} = $ $ \pi r^2 h $

$ \text{Curved surface area of cylinder} = $ $ 2 \pi r h $

$ \text{Volume of pyramid} = $ $ {1 \over 3} \times \text{Base area} \times \text{Height} $

$ \text{Volume of cone} = $ $ {1 \over 3} \pi r^2 h $

$ \text{Curved surface area of cone} = $ $ \pi r l, \text{ where } l \text{ is the slant height} $

$ \text{Volume of sphere} = $ $ {4 \over 3} \pi r^3 $

$ \text{Curved surface area of sphere} = $ $ 4 \pi r^2 $

$ \text{Area of sector (degree)} = $ $ {\theta \over 360^\circ} \times \pi r^2 $

$ \text{Area of sector (radian)} = $ $ {1 \over 2} r^2 \theta $

$ \text{Arc length (degree)} = $ $ {\theta \over 360^\circ} \times 2 \pi r $

$ \text{Area of sector (radian)} = $ $ r \theta $