# Polar co-ordinates

## P1 Introduction:

• One co-ordinate system is cartesian, $\left(x,y\right)$
• The polar co-ordinate system contains the following:
- A fixed point $O$, called the pole
- A line coming from this point, usually going to the right. This is the initial line
• Then, positions are defined with $\left(r,\theta \right)$, where:
- $r$ is the distance between the pole and point
- $\theta$ is the angle between the initial line and an imaginary line between the pole and point, going anti-clockwise
• The pole has the co-ordinates $\left(0,\mathrm{u}\mathrm{n}\mathrm{d}\mathrm{e}\mathrm{f}\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{d}\right)$
• If $r$ is negative, then the point will be the opposite side to where $\theta$ actually indicates, i.e. $\left(-r,\theta \right)\equiv \left(r,\theta +\pi \right)$
• If $r>0$, draw a solid line between the pole and point
• If $r<0$, draw a dashed line between the pole and point

## Converting between polar and cartesian co-ordinates:

• To convert from polar co-ordinates to cartesian, use $\mathrm{arctan}$ and Pythagoras' theorem
• You can check with a Casio scientific calculator:
• Cartesian polar:
- Enter $\mathrm{P}\mathrm{o}\mathrm{l}\left(x,y\right)$ where $x$ and $y$ are the cartesian co-ordinates
- $\mathrm{P}\mathrm{o}\mathrm{l}$ is SHIFT and $+$
- Get a comma with SHIFT and the right bracket
• Polar cartesian:
- Enter $\mathrm{R}\mathrm{e}\mathrm{c}\left(r,\theta \right)$ where $r$ is the distance and $\theta$ is the angle
- $\mathrm{R}\mathrm{e}\mathrm{c}$ is SHIFT and $-$
- You do not need to specify the angle units (deg/rad), just set them in the calculator options

## Converting between polar and cartesian equations:

• Use the following relationships:
- ${x}^{2}+{y}^{2}={r}^{2}$
- $x=r\mathrm{cos}\theta$
- $y=r\mathrm{sin}\theta$
• For example, to convert $r=8\mathrm{sin}\theta$ to cartesian:
- Multiply both sides by $r$: ${r}^{2}=8r\mathrm{sin}\theta$
- Use the third relationship listed above: ${r}^{2}=8y$
- Use the first relationship above: ${x}^{2}+{y}^{2}=8y$
- This is a cartesian equation. To simplify further, move the $8y$ to the LHS and complete the square:

$⇒{x}^{2}+{y}^{2}-8y=0$
$⇒{x}^{2}+$ [$\left(y-4{\right)}^{2}-16$] $=0$
$⇒{x}^{2}+\left(y-4{\right)}^{2}=16$

## P2 Sketching polar curves:

• First, create a table of values - for several values of $\theta$ at equal intervals, calculate $r$
• Now use these to sketch the curve, using the rules in P2
• Many graphical calculators can sketch polar curves. For example, in CASIO calculators, change from $Y=$ mode to $r=$, and then enter the RHS using the $\theta$ button

## P3 Finding the area enclosed by a polar curve:

• Use the formula where $r$ is the polar equation (e.g. $r=2\mathrm{cos}\theta$)

## Example:

• The graph of $r=\frac{\theta }{3}$ forms a spiral as shown below:
• We will find the area between the curve and $y$-axis shown in the graph. This is in the range [$0,\frac{\pi }{2}$]
• First, substitute the polar equation into the area formula:
• This simplifies to
• Now we can integrate to $A=\frac{1}{2}$[$\frac{1}{27}{\theta }^{3}$]${}_{0}^{\frac{\pi }{2}}$
• So $A=\frac{1}{2}\left(\frac{\left(\pi ÷3{\right)}^{3}}{27}-0\right)=\frac{{\pi }^{3}}{0.5×8×27}=\frac{{\pi }^{3}}{108}$

# Calculus

## c1 Inverse trig functions:

• $\mathrm{sin}$, $\mathrm{cos}$ and $\mathrm{tan}$ functions all have many-to-one mapping
• Therefore, their inverses would be one-to-many
• A function cannot be one-to-many, so the range is restricted in the inverse trig functions to:
- $-\frac{\pi }{2}\le \mathrm{arcsin}\le \frac{\pi }{2}$ (domain is $-1\le x\le 1$)
- $0\le \mathrm{arccos}\le \pi$ (domain is $-1\le x\le 1$)
- $-\frac{\pi }{2}<\mathrm{arctan}<\frac{\pi }{2}$ (domain is $x\in \mathbb{R}$)

## Inverse reciprocal trig functions in a calculator:

• To enter $\mathrm{arcsec}x$ in a calculator, enter ${\mathrm{cos}}^{-1}\left(\frac{1}{x}\right)$
• For $\mathrm{arccosec}x$, enter ${\mathrm{sin}}^{-1}\left(\frac{1}{x}\right)$
• For $\mathrm{arccot}x$, enter $\frac{\pi }{2}-{\mathrm{tan}}^{-1}x$

## c2 Differentiating inverse trig functions:

• Use one of the following results (in the formula booklet)
- $\frac{\mathrm{d}}{\mathrm{d}x}\left(\mathrm{arcsin}x\right)=$ $\frac{1}{\sqrt{1-{x}^{2}}}$
- $\frac{\mathrm{d}}{\mathrm{d}x}\left(\mathrm{arccos}x\right)=$ $-\frac{1}{\sqrt{1-{x}^{2}}}$
- $\frac{\mathrm{d}}{\mathrm{d}x}\left(\mathrm{arctan}x\right)=$ $\frac{1}{1+{x}^{2}}$
• Also, you should remember the identity $\frac{\mathrm{d}}{\mathrm{d}x}\left(\mathrm{arccot}x\right)=-\frac{\mathrm{d}}{\mathrm{d}x}\left(\mathrm{arctan}x\right)$

## Example: find $\mathrm{d}$/$\mathrm{d}x$$\left(\mathrm{arcsin}x\right)$:

• This is the same as differentiating $y=\mathrm{arcsin}x$
• Find $x$ in terms of $y$: $\mathrm{sin}y=x$
• Now differentiate this implicitly (C3): $\frac{\mathrm{d}y}{\mathrm{d}x}\mathrm{cos}y=1$
• Solve for $\frac{\mathrm{d}y}{\mathrm{d}x}$: $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{\mathrm{cos}y}$
• Now you need to replace the $\mathrm{cos}$ with a $\mathrm{sin}$. The steps are listed below:
- Pick an identity ${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta =1$
- We want this in terms of $\mathrm{cos}$: ${\mathrm{cos}}^{2}\theta =1-{\mathrm{sin}}^{2}\theta$
- We also need to remove the powers: $\mathrm{cos}\theta =±\sqrt{1-{\mathrm{sin}}^{2}\theta }$
• Now substitute: $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{±\sqrt{1-{\mathrm{sin}}^{2}y}}$
• Because of the restricted range of the $\mathrm{arcsin}$ function, the $±$ can be removed
• A similar method can be used to differentiate $\mathrm{arccos}$

## Example: find $\mathrm{d}$/$\mathrm{d}x$$\left(\mathrm{arctan}x\right)$:

• This is the same as differentiating $y=\mathrm{arctan}x$
• So we need to implicitly differentiate $\mathrm{tan}y=x$
• This gives us ${\mathrm{sec}}^{2}y×\frac{\mathrm{d}y}{\mathrm{d}x}=1$ (remember that $\frac{\mathrm{d}}{\mathrm{d}x}\left(\mathrm{tan}a\right)={\mathrm{sec}}^{2}a$)
• Now make this in terms of $\frac{\mathrm{d}y}{\mathrm{d}x}$: $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{{\mathrm{sec}}^{2}y}$
• We said earlier that $y=\mathrm{arctan}x$, so this is the same as $\frac{1}{{\mathrm{sec}}^{2}\left(\mathrm{arctan}x\right)}$
• Now use the identity ${\mathrm{sec}}^{2}a={\mathrm{tan}}^{2}a+1$:
$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{{\mathrm{tan}}^{2}\left(\mathrm{arctan}x\right)+1}$
• We can write this as $\frac{1}{\left(\mathrm{tan}\left(\mathrm{arctan}x\right){\right)}^{2}+1}$
• Because $\mathrm{tan}\left(\mathrm{arctan}x\right)=x$, this simplifies to $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{{x}^{2}+1}$

## c3 Integration:

• From the above differentiation results, the following can be derived for integration; they are given in the formula booklet:
- $\int \frac{1}{\sqrt{{a}^{2}-{x}^{2}}}\phantom{\rule{.1667em}{0ex}}\mathrm{d}x=\mathrm{arcsin}\left(\frac{x}{a}\right)+c$
- $\int \frac{1}{{a}^{2}+{x}^{2}}\phantom{\rule{.1667em}{0ex}}\mathrm{d}x=\frac{1}{a}\mathrm{arctan}\left(\frac{x}{a}\right)+c$
• They will only work when ${x}^{2}$ has no coefficient. Therefore, you must remove coefficients by rearranging to get a fraction outside of the integral. Example:
Find $I$ where $I=\int \frac{1}{\sqrt{16-3{x}^{2}}}\phantom{\rule{.1667em}{0ex}}\mathrm{d}x$.
1) Factorise the bracket to get ${x}^{2}$ on its own: $I=\int \frac{1}{\sqrt{3\left(\frac{16}{3}-{x}^{2}\right)}}\phantom{\rule{.1667em}{0ex}}\mathrm{d}x$
2) Now move the $3$ outside of the square root sign: $I=\int \frac{1}{\sqrt{3}\sqrt{\frac{16}{3}-{x}^{2}}}\phantom{\rule{.1667em}{0ex}}\mathrm{d}x$
3) Now move the $\sqrt{3}$ outside of the integral: $I=\frac{1}{\sqrt{3}}\int \frac{1}{\sqrt{\frac{16}{3}-{x}^{2}}}\phantom{\rule{.1667em}{0ex}}\mathrm{d}x$
4) Now use the relevant result: $I=\frac{1}{\sqrt{3}}\mathrm{arcsin}\frac{x\sqrt{3}}{4}+c$
• The alternative method would be to use the substitution $u=x\sqrt{3}$

## c4 Using identities to integrate functions:

• The below identities can easily be derived from the formula booklet (see C4)
$\mathrm{cos}2x\equiv 1-2{\mathrm{sin}}^{2}x$
$\mathrm{cos}2x\equiv 2{\mathrm{cos}}^{2}x-1$
• If you rearrange these to make ${\mathrm{sin}}^{2}x$ and ${\mathrm{cos}}^{2}x$ the subject, you get the following identities, which are particularly useful (since it's very easy to integrate $\mathrm{sin}2x$ and $\mathrm{cos}2x$):
${\mathrm{sin}}^{2}x=\frac{1}{2}\left(1-\mathrm{cos}2x\right)$
${\mathrm{cos}}^{2}x=\frac{1}{2}\left(1+\mathrm{cos}2x\right)$

## Example: find $\int \mathrm{sin}$$2$:

• Start with ${\mathrm{sin}}^{2}x=\frac{1}{2}\left(1-\mathrm{cos}2x\right)$ from above
• Substitute this into the integral: $I=\int \frac{1}{2}\left(1-\mathrm{cos}2x\right)\phantom{\rule{.1667em}{0ex}}\mathrm{d}x$
• Move the $\frac{1}{2}$ to the left of the $\int$ and integrate to $I=\frac{1}{2}\left(x-\frac{1}{2}\mathrm{sin}2x\right)+c$
• Expand the brackets to $I=\frac{x}{2}-\frac{1}{4}\mathrm{sin}2x+c$

## Example: find $\int \mathrm{sin}$$3$:

• Rearrange to get a quadratic:
• Rearrange ${\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x=1$ to ${\mathrm{sin}}^{2}x=1-{\mathrm{cos}}^{2}x$
• Combine the two:
• Let $u=\mathrm{cos}x$. Use integration by substitution to get $I=\int \frac{\mathrm{sin}x\left(1-{u}^{2}\right)}{-\mathrm{sin}x}\phantom{\rule{.1667em}{0ex}}\mathrm{d}u$
• Integrate this to $I=\frac{1}{3}{u}^{3}-x=\frac{1}{3}{\mathrm{cos}}^{3}x-x+c$

## Example: find $\int \mathrm{sin}$$4$:

• Rearrange to get the integral in terms of ${\mathrm{sin}}^{2}x$:
• Rearrange $\mathrm{cos}2x=1-{\mathrm{sin}}^{2}x$ to ${\mathrm{sin}}^{2}x=\frac{1}{2}\left(1-\mathrm{cos}2x\right)$
• Combine the two:
• Expand the brackets: $I=\frac{1}{4}\int \left(1-2\mathrm{cos}2x+{\mathrm{cos}}^{2}2x\right)$
• The third term needs to be simplified further to be integrated:
- Rearrange $2{\mathrm{cos}}^{2}a-1=\mathrm{cos}2a$ to ${\mathrm{cos}}^{2}a=\frac{1}{2}\left(1+\mathrm{cos}2a\right)$
- We have ${\mathrm{cos}}^{2}2x$ and not ${\mathrm{cos}}^{2}x$, so let $a=2x$: ${\mathrm{cos}}^{2}2x=\frac{1}{2}\left(1+\mathrm{cos}4x\right)$
- This simplifies to $\frac{1}{2}+\frac{1}{2}\mathrm{cos}4x$
• Now substitute this in: $I=\frac{1}{4}\int 1-$ $2\mathrm{cos}2x+\frac{1}{2}+$ $=\frac{1}{4}\int \frac{3}{2}-$
• Now integrate to $I=\frac{1}{4}\left(\frac{3}{2}x-$ $\mathrm{sin}2x+\frac{1}{8}\mathrm{sin}4x\right)+c$
• Simplify to $I=\frac{3}{8}x-\frac{1}{4}\mathrm{sin}2x$ $+\frac{1}{32}\mathrm{sin}4x+c$

## Example: find :

• Let $u=\mathrm{cos}x$.
• Use integration by substitution:
• Simplify:
• Use the result :
$I=-\mathrm{ln}\left(|u|\right)+c=-\mathrm{ln}\left(|\mathrm{cos}x|\right)+c$
• This can be simplified further, since $-\mathrm{ln}\left(a\right)=\mathrm{ln}\left({a}^{-1}\right)$. So $I=\mathrm{ln}\left(|\mathrm{sec}x|\right)+c$

# Series

## s1 Maclaurin series:

• A Maclaurin series is an approximation of a function as a sum of infinite terms
• They are centred around $x=0$ (unlike Taylor series) so the approximation will be the most accurate near to $x=0$
• The more terms you use, the more accurate the approximation will become
• The formula (given in the formulae booklet) is $\frac{{x}^{2}}{2!}$ $\frac{{x}^{r}}{r!}$
• To find an expansion to a specified degree of accuracy (e.g. 5 decimal places), keep adding terms until adding another does not affect your answer to the degree of accuracy required

## Example expansion:

• We'll be finding the expansion of $f\left(x\right)=\mathrm{cos}x$
• The first term is $f\left(0\right)$ which is 1
• The second term is $x{f}^{\prime }\left(0\right)$. We don't know $x$ so we'll leave that as it is. This is multiplied by the derivative of $f\left(x\right)$ at $x=0$ which is $-\mathrm{sin}0=0$. So the second term is $x×0=0$
• The third term is $\frac{{x}^{2}}{2!}$${f}^{″}\left(0\right)$. ${f}^{″}\left(0\right)$ is $-\mathrm{cos}0=-1$ so the third term is $-$$\frac{{x}^{2}}{2}$
• With the same method, this continues. So the fourth term is $\frac{{x}^{3}}{3!}$$×\mathrm{sin}0=0$, and fifth is $\frac{{x}^{4}}{4!}$$×\mathrm{cos}0=$ $\frac{{x}^{4}}{4!}$
• This trend continues. So our expansion up to the fourth power would be $\frac{{x}^{2}}{2!}$ $\frac{{x}^{4}}{4!}$ (which can be further simplified)

## s2 Convergence:

• With infinite terms, some series such as the one above for $f\left(x\right)=\mathrm{cos}x$ will be accurate for all values of $x$
• But for some series, the approximation will only be valid for a small domain, and outside of this, adding more terms will decrease the accuracy
• For example, $f\left(x\right)=$ $\frac{1}{1-x}$ and the first 8 terms of its Maclaurin series are shown below. You can see that it is only valid for a limited domain

With more terms, the values for $x$-coordinates outside of the valid domain would be even less accurate. For example, the Maclaurin series with only one term is just $y=1$. At $x=4$, this is much more accurate than with more terms

## s3 Maclaurin series for standard functions:

• You need to be famililar with the following Maclaurin expansions from the Infinite series section of the formula booklet
${e}^{x}=1+x+\frac{{x}^{2}}{2!}+...+$ $\frac{{x}^{r}}{r!}+...$, all $x$
$\mathrm{ln}\left(1+x\right)=x-\frac{{x}^{2}}{2}+$ $\frac{{x}^{3}}{3}-...+\left(-1{\right)}^{r+1}\frac{{x}^{r}}{r}+...$, $-1
$\mathrm{sin}x=x-\frac{{x}^{3}}{3!}+\frac{{x}^{5}}{5!}$ $-...+\left(-1{\right)}^{r}\frac{{x}^{2r+1}}{\left(2r+1\right)!}+...$, all $x$
$\mathrm{cos}x=1-\frac{{x}^{2}}{2!}$ $+\frac{{x}^{4}}{4!}-...+$ $\left(-1{\right)}^{r}\frac{{x}^{2r}}{\left(2r\right)!}+...$, all $x$
• You also need to be familiar with the $\left(1+x{\right)}^{n}$ binomial expansion from the C4 module (in the Binomial expansions section of the formula booklet). This is also a Maclaurin series; you can derive it by using the chain rule and the Maclaurin series general formula in s1
$\left(1+x{\right)}^{n}=1+nx+$ $\frac{n\left(n-1\right)}{2!}{x}^{2}+...+$ $\frac{n\left(n-1\right)...\left(n-r+1\right)}{1\cdot 2...r}{x}^{r}+...$, $|x|<1$

# Complex numbers

## j1 Polar form:

• Polar form or modulus-argument form is a way of writing complex numbers with the notation $z=r\left(\mathrm{cos}\theta +j\mathrm{sin}\theta \right)$ where $r=|z|$ (the modulus) and $\theta =\mathrm{arg}z$ (the argument, usually written in radians), as seen in FP1
• Rectangular form ($z=a+bj$) and polar form can be converted on fx-991 series calculators, see here for more details
• Ensure that $r>0$ and $-\pi <\theta \le \pi$
• To get the argument within this range, keep adding/subtracting $2\pi$

## j2 Multiplying and dividing numbers in polar form:

• To multiply two complex numbers in polar form, multiply their moduli and add their arguments
• To divide two complex numbers in polar form, divide their moduli and subtract their arguments

## j3 de Moivre's theorem:

• For any integer $n$, $\left(\mathrm{cos}\theta +j\mathrm{sin}\theta {\right)}^{n}=\mathrm{cos}n\theta +j\mathrm{sin}n\theta$
• When the modulus of a complex number is not 1, de Moivre's theorem is often useful if you treat it separately:
$\left[r\left(\mathrm{cos}\theta +j\mathrm{sin}\theta \right){\right]}^{n}$ $={r}^{n}\left(\mathrm{cos}\theta +j\mathrm{sin}\theta {\right)}^{n}$ $={r}^{n}\left(\mathrm{cos}n\theta +j\mathrm{sin}n\theta \right)$

## j4 Applying de Moivre's theorem:

Example: express $\mathrm{cos}3\theta$ in terms of $\mathrm{cos}\theta$ only:

• 1) Write this as a complex number ($\mathrm{cos}3\theta$ is real so $j=0$): $\mathrm{cos}3\theta =\left(\mathrm{cos}\theta +j\mathrm{sin}\theta {\right)}^{3}$
• 2) Expand this with binomial expansion (see the C1 module). (I use $c$ to mean $\mathrm{cos}\theta$ etc):
$\left(\mathrm{cos}\theta +j\mathrm{sin}\theta {\right)}^{3}$
$={c}^{3}+3{c}^{2}js+3c{j}^{2}{s}^{2}+\left(js{\right)}^{3}$
$={c}^{3}+3{c}^{2}js-3c{s}^{2}+\left(js{\right)}^{3}$
• 3) Now set $j=0$ and simplify, to ${c}^{3}-3c{s}^{2}$
• 4) Because ${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta \equiv 1$, the $s$ can be eliminated, to ${c}^{3}-3c\left(1-{c}^{2}\right)=$ ${c}^{3}-3c+3{c}^{3}=$ $4{c}^{3}-3c$
• 5) So $\mathrm{cos}3\theta =4{\mathrm{cos}}^{3}\theta -3\mathrm{cos}\theta$

Similarly, if expressing a $\mathrm{sin}$ function in terms of $\mathrm{sin}\theta$ only, you would instead eliminate the real part in step 3

For a $\mathrm{tan}$ function, remember that $\mathrm{tan}\theta =$ $\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }$

## Useful deductions:

• Let $z=\mathrm{cos}\theta +j\mathrm{sin}\theta$
• The following deductions can be made from the main theorem:
• ${z}^{-n}=\mathrm{cos}n\theta -j\mathrm{sin}n\theta$
• $\mathrm{cos}n\theta =$ $\frac{{z}^{n}+{z}^{-n}}{2}$
• $\mathrm{sin}n\theta =$ $\frac{{z}^{n}-{z}^{-n}}{2j}$

## Example: express $\mathrm{cos}$$5$$\phantom{\rule{.1667em}{0ex}}\theta$ in terms of multiple angles:

• Let $z=\mathrm{cos}\theta +j\mathrm{sin}\theta$
• With the first identity above, we can say that $2\mathrm{cos}\theta =z+{z}^{-1}$
• If we raise this to the 5th power, we get ${2}^{5}{\mathrm{cos}}^{5}\theta =\left(z+{z}^{-1}{\right)}^{5}$
• Now expand and simplify the right side to ${z}^{5}+5{z}^{3}+10z$ $+10{z}^{-1}+5{z}^{-3}+{z}^{-5}$
• By factorising into groups, RHS $=\left({z}^{5}+{z}^{-5}\right)+5\left({z}^{3}+{z}^{-3}\right)$ $+10\left(z+{z}^{-1}\right)$
• Using the definition that $z=\mathrm{cos}\theta +j\mathrm{sin}\theta$, this simplifies to $2\mathrm{cos}5\theta +10\mathrm{cos}3\theta$ $+20\mathrm{cos}\theta$
• The left hand side is still ${2}^{5}co{s}^{5}\phantom{\rule{.1667em}{0ex}}\theta$, so we need to divide both sides by ${2}^{5}$ to get the final answer, ${\mathrm{cos}}^{5}\phantom{\rule{.1667em}{0ex}}\theta =$ $\frac{2\mathrm{cos}5\theta +10\mathrm{cos}3\theta +20\mathrm{cos}\theta }{32}$. This can still be simplified - divide the numerator and denominator by 2

## Summing series:

• Aim to simplify the series to a form where reverse binomial expansion or a sum of series formula can be applied
• To get to this stage, it is often useful to find a way to use de Moivre's theorem
• For example, when simplifying a $\mathrm{cos}$ series, try introducing a $\mathrm{sin}$ series called $S$ and try adding $jS$ to the original sequence. de Moivre's theorem can now be used

## Summing series example:

• Here is an exam question, exactly as it was written in the paper I took it from:

The infinite series $C$ and $S$ are defined as follows.
$C=\mathrm{cos}\theta +\frac{1}{3}\mathrm{cos}3\theta +\frac{1}{9}\mathrm{cos}5\theta +\dots$
$S=\mathrm{sin}\theta +\frac{1}{3}\mathrm{sin}3\theta +\frac{1}{9}\mathrm{sin}5\theta +\dots$

By considering $C+jS$, show that
$C=\frac{3\mathrm{cos}\theta }{5-3\mathrm{cos}2\theta }$

and find a similar expression for $S$

• First, you need to add the two series, multiplying each term in the $S$ series by $j$:
$C+jS=\left(\mathrm{cos}\theta +j\mathrm{sin}\theta \right)$ $+\frac{1}{3}\left(\mathrm{cos}3\theta +j\mathrm{sin}3\theta \right)$ $+\frac{1}{9}\left(\mathrm{cos}5\theta +j\mathrm{sin}5\theta \right)$ $+\dots$
• Now de Moivre's theorem can be used:
$C+jS=\left(\mathrm{cos}\theta +j\mathrm{sin}\theta {\right)}^{1}$ $+\frac{1}{3}\left(\mathrm{cos}\theta +j\mathrm{sin}\theta {\right)}^{3}$ $+\frac{1}{9}\left(\mathrm{cos}\theta +j\mathrm{sin}\theta {\right)}^{5}$ $+\dots$
• Convert to exponential form (this is covered in j5):
$C+jS={e}^{j\theta }+\frac{1}{3}{e}^{3j\theta }$ $+\frac{1}{9}{e}^{5j\theta }$
• You can tell that this is a geometric series where
$a={e}^{j\theta }$ and
$r=\frac{1}{3}{e}^{2j\theta }$
• From the formula booklet, ${S}_{\mathrm{\infty }}=$ $\frac{a}{1-r}$, so ${S}_{\mathrm{\infty }}=C+jS=$ $\frac{{e}^{j\theta }}{1-\frac{1}{3}{e}^{2j\theta }}$
• It's generally best to get rid of fractions at this stage:
$C+jS=$ $\frac{3{e}^{j\theta }}{3-{e}^{2j\theta }}$
• Now multiply the top and bottom of the fraction by the conjugate of the denominator:
$C+jS=$ $\frac{3{e}^{j\theta }}{3-{e}^{2j\theta }}$ $×$ $\frac{3-{e}^{-2j\theta }}{3-{e}^{-2j\theta }}$ $=$ $\frac{9{e}^{j\theta }-3{e}^{-j\theta }}{9-3{e}^{-2j\theta }-3{e}^{2j\theta }+1}$
• Convert back into $\mathrm{cos}\theta +j\mathrm{sin}\theta$ form:
$C+jS=$ $\frac{9\left(\mathrm{cos}\theta +j\mathrm{sin}\theta \right)-3\left(\mathrm{cos}\theta -j\mathrm{sin}\theta \right)}{10-3\left(\mathrm{cos}2\theta -j\mathrm{sin}2\theta \right)-3\left(\mathrm{cos}2\theta +j\mathrm{sin}2\theta \right)}$
• Simplify to $C+jS=$ $\frac{6\mathrm{cos}\theta +12j\mathrm{sin}\theta }{10-6\mathrm{cos}2\theta }$
• $C$ will be the real part of the numerator, with the entire denominator, so
$C=$ $\frac{6\mathrm{cos}\theta }{10-6\mathrm{cos}2\theta }$
• $S$ will be whatever you need to add to this to get your simplified $C+jS$, so
$S=$ $\frac{12\mathrm{sin}\theta }{10-6\mathrm{cos}2\theta }$
• (you can simplify the above fractions by dividing numerator and denominators by 2)

## j5 Exponential form:

• If $z=r\left(\mathrm{cos}\theta +j\mathrm{sin}\theta \right)$, then:
$z=r{e}^{j\theta }$
• de Moivre's theorem is simply $\left({e}^{j\theta }{\right)}^{n}={e}^{jn\theta }$
• The following can now be derived from the relationships in the previous section:
$\mathrm{cos}\theta =$ $\frac{{e}^{j\theta }+{e}^{-j\theta }}{2}$,
$\mathrm{sin}\theta =$ $\frac{{e}^{j\theta }-{e}^{-j\theta }}{2j}$

## j6$n$th roots of a complex number:

• All non-zero complex numbers have $n$ $n$th roots. For example, ${z}^{3}=5+5j$ has three roots
• On an Argand diagram, these form the vertices of a $n$-gon (e.g. the roots of ${z}^{4}=2+2j$ would form a square)

## j7 Finding $n$th roots:

• The roots of $r{e}^{j\theta }$ are ${}^{\frac{\theta +2k\pi }{n}}$ where $k$ is the root number (starting at $k=0$ for the first root, and finishing at $k=n-1$ for the last)

## Roots of unity:

• A root of unity is a complex number which gives 1 when raised to the power of a positive integer
• For example, the polynomial ${z}^{5}-1$ has 5 roots of unity (values of $z$ which multiply together to make 1)
• The argument of 1 is 0, so you should use the formula above with $\theta =0$
• The first root of unity is therefore always 1
• The symbol $\omega$ is used to represent the root of unity with the smallest positive argument
• ${\omega }^{2}$ is used to represent the next root, ${\omega }^{3}$ for the next and so on

## j8 Sum of roots of unity:

• $1+\omega +{\omega }^{2}+{\omega }^{3}+\dots +{\omega }^{n-1}=$ $\frac{a\left(1-{r}^{n}\right)}{1-r}$ $=$ $\frac{1\left(1-{\omega }^{n}\right)}{1-\omega }$ $=$ $\frac{0}{1-\omega }$ $=0$ (since ${\omega }^{n}={\omega }^{0}=1$)
• Therefore, the sum of roots of unity is always 0. You need to memorise the derivation above
• You can also prove this geometrically - the shape formed when you connect the roots is always centred about $\left(0,0\right)$

## j9 Multiplication by a complex number:

• In an Argand diagram, if you multiply by $r{e}^{j\theta }$, there will be an enlargement of scale factor $r$ and a rotation of $\theta$ about the origin
• For example, if you multiply by $j$ (which can be written as $1{e}^{j×\frac{\pi }{2}}$), there will be a rotation of $\frac{\pi }{2}$ about the origin (and no enlargement, since $r=1$)

# Matrices

## m1 3×3 determinants with the cofactor method:

• If $\mathbf{M}$ is the matrix $\left(\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right)$, $det\mathbf{M}$ = $a|\begin{array}{cc}e& f\\ h& i\end{array}|-$ $b|\begin{array}{cc}d& g\\ f& i\end{array}|+c|\begin{array}{cc}d& e\\ g& h\end{array}|$
• The second component of this sum is subtracted because $b$ is negative on the sign matrix $\left(\begin{array}{ccc}+& -& +\\ -& +& -\\ +& -& +\end{array}\right)$
• The 2×2 determinants here are called minors; the minor of $a$ is the determinant of the 2×2 matrix when the row and column of $a$ is crossed out
• A minor with the correct sign from the sign matrix above is called a cofactor
• You don't have to use the sum of the top row element cofactors to calculate the determinant, you can use any row or column in the matrix. Therefore, it's easier to pick a row/column with some zeroes

## 3×3 determinants with the Sarrus method:

• Create two new rows under the matrix, and copy into them the first two rows of the matrix, e.g.: $\left(\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\\ a& b& c\\ d& e& f\end{array}\right)$
• Multiply the 3 elements in each of the 6 diagonals
• The determinant is the sum of the ↙ subtracted from the sum of the ↘ products; ↘ $-$
• So the determinant would be $\left(aei+dhc+gbf\right)-\left(ceg+fha+ibd\right)$

## Finding the inverse of a (non-singular) 3×3 matrix:

• Find the cofactor of each element and create a 3×3 matrix of cofactors
• Then transpose it (the first column becomes the first row, the second column becomes the second row, etc) to get the adjugate, $\mathrm{adj}\mathbf{M}$
• Finally, multiply this by $\frac{1}{det\mathbf{M}}$ to get ${\mathbf{M}}^{-1}$

## m2, m6 Eigenvectors:

• If $s$ is a non-zero vector and $\mathbf{M}\mathbf{s}=\lambda \mathbf{s}$, then $\mathbf{s}$ is called an eigenvector and $\lambda$ is called an eigenvalue
• An eigenvector is a vector which is mapped to a multiple of itself by $\mathbf{M}$
• Each eigenvector can be multiplied by any integer and it will still work. Therefore, the elements are simplified as much as possible. For example, the eigenvectors $\left(\begin{array}{c}2\\ 4\end{array}\right)$ and $\left(\begin{array}{c}1\\ 2\end{array}\right)$ are the same
• The trace of a matrix is the sum of its eigenvalues. It's also the sum of the top-left to bottom-right diagonal on the matrix. You don't need to know this, but it's useful for checking that you haven't made an error

## Finding eigenvalues and eigenvectors:

• Start with $\mathbf{M}\mathbf{s}=\lambda \mathbf{s}$ from above
$⇒\mathbf{M}\mathbf{s}-\lambda \mathbf{s}=0$
$⇒\mathbf{M}\mathbf{s}-\lambda \mathbf{I}\mathbf{s}=0$ (multiplying by the identity matrix)
$⇒\left(\mathbf{M}-\lambda \mathbf{I}\right)\mathbf{s}=0$ (factorising)
$⇒det\left(\mathbf{M}-\lambda \mathbf{I}\right)=0$
• This gives the equations $det\left(\mathbf{M}-\lambda \mathbf{I}\right)=0$ and $\left(\mathbf{M}-\lambda \mathbf{I}\right)\mathbf{s}=0$
• Use the first to find the eigenvalues; find the determinant of the matrix with $\lambda$ subtracted from all values on the top-left to top-right diagonal. Set this equal to 0 and solve for $\lambda$ to get a polynomial
• The polynomial you get is called the characteristic equation. It will have the same order as the matrix (e.g. it'll be a cubic if the matrix is 3×3). Solve it to find the eigenvalues
• Substitute each eigenvalue into the second equation [$\left(\mathbf{M}-\lambda \mathbf{I}\right)\mathbf{s}=0$] to find the eigenvectors for each eigenvalue

## m3 Diagonal form:

• The matrix $\mathbf{P}$ is created by writing the eigenvectors next to each other. For example, with the eigenvectors $\left(\begin{array}{c}-1\\ 1\end{array}\right)$ and $\left(\begin{array}{c}2\\ 1\end{array}\right)$, $\mathbf{P}=\left(\begin{array}{ccc}-1& & 2\\ 1& & 1\end{array}\right)$
• The matrix $\mathbf{D}$ is created by diagonally writing the eigenvalues onto a matrix of zeroes. For example, with the eigenvalues 2 and 5, $\mathbf{D}=\left(\begin{array}{ccc}2& & 0\\ 0& & 5\end{array}\right)$
• If you construct $\mathbf{P}$ and $\mathbf{D}$ correctly, $\mathbf{M}={\mathbf{P}\mathbf{D}\mathbf{P}}^{-1}$

Note: the symbols $\phantom{\rule{.2778em}{0ex}}\mathbf{S}$ and $\phantom{\rule{.2778em}{0ex}}\mathbf{\Lambda }$ are used in the textbook instead of $\phantom{\rule{.2778em}{0ex}}\mathbf{P}$ and $\phantom{\rule{.2778em}{0ex}}\mathbf{D}$ respectively. I used these because they're what the exams use

## m4 Finding powers of matrices:

• The formula ${\mathbf{M}}^{n}={\mathbf{P}\mathbf{D}}^{n}{\mathbf{P}}^{-1}$ can be used to raise a matrix $\mathbf{M}$ to a power $n$
• Calculating ${\mathbf{D}}^{n}$ is very simple - because each eigenvalue in the matrix only shares its row and column with zeroes, you can simply raise each eigenvalue by the power $n$

## m5 Simultaneous equations:

• If the determinant of a matrix is not 0, you can solve three variable simultaneous equations with a 3×3 matrix as in FP1
• If it is 0, there are either no solutions or infinite solutions
• To find out which of the above is the case, you need to eliminate one of the variables. Add or substract a multiple of an equation to/from another equation to do this. If the answer is the same as the remaining equation, then the equations are consistent and there are infinite solutions
• To solve a set of consisten simultaneous equations, introduce a new variable $\lambda$. Let it equal one of your variables and then find the solutions in terms of $\lambda$

## m7 The Cayley–Hamilton Theorem:

• The Cayley–Hamilton Theorem says that a matrix satisfies its own characteristic equation
• For example, let $\mathbf{M}=\left(\begin{array}{ccc}1& & 2\\ 1& & 4\end{array}\right)$. This has the characteristic equation ${\lambda }^{2}-5\lambda +2=0$
Let's say we want to calculate ${\mathbf{M}}^{4}$. First, substitute $\mathbf{M}$ into the characteristic equation, replacing $\lambda$. Multiply the integer by $\mathbf{I}$. This gives us ${\mathbf{M}}^{2}-5\mathbf{M}+2\mathbf{I}=0$
We now need to rearrange this in terms of ${\mathbf{M}}^{2}$, leaving us ${\mathbf{M}}^{2}=5\mathbf{M}-2\mathbf{I}$
This relationship can be used to calculate ${\mathbf{M}}^{4}$:
${\mathbf{M}}^{4}=\left(5\mathbf{M}-2\mathbf{I}{\right)}^{2}$

Therefore,
${\mathbf{M}}^{4}=105\left(\begin{array}{ccc}1& & 2\\ 1& & 4\end{array}\right)$ $-46\left(\begin{array}{ccc}1& & 0\\ 0& & 1\end{array}\right)$
$-\left(\begin{array}{ccc}46& & 0\\ 0& & 46\end{array}\right)$

# Hyperbolic functions

## a4 The hyperbolic functions:

• The three main hyperbolic trig functions are $\mathrm{sinh}$, $\mathrm{cosh}$ and $\mathrm{tanh}$
• There's no 'correct' pronunciation, but they are usually pronounced 'shine', 'cosh' (rhyming with 'posh') and 'than' (with the 'th' being said as it is in 'theta')
• These also have inverses; $\mathrm{arsinh}$, $\mathrm{arcosh}$ and $\mathrm{artanh}$ (note: it's $\mathrm{a}\mathrm{r}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h}$, not $\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h}$; the same applies for $\mathrm{arcosh}$ and $\mathrm{artanh}$)
• And there are $\mathrm{cosech}$, $\mathrm{sech}$ and $\mathrm{coth}$ and their inverses. These are the reciprocals of their related standard hyperbolic trig function

## Identities involving hyperbolic trig functions:

• You can use the standard trigonometric identities. However, you must flip the sign if you are squaring $\mathrm{sinh}$
• For example, if you apply the identity ${\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta \equiv 1$ to $\mathrm{sinh}\theta$ and $\mathrm{cosh}\theta$, you must flip the sign in front of ${\mathrm{sin}}^{2}\theta$ since you are squaring $\mathrm{sin}$. So this identity would become ${\mathrm{cosh}}^{2}\theta -{\mathrm{sinh}}^{2}\theta \equiv 1$

## $y=\mathrm{sinh}x$:

• Has an infinite domain and range
• Is an odd function
• $\mathrm{sinh}x=\frac{1}{2}\left({e}^{x}-{e}^{-x}\right)$

## $y=\mathrm{cosh}x$:

• Has an infinite domain. Range is limited: $y\ge 1$
• Is an even function
• $\mathrm{cosh}x=\frac{1}{2}\left({e}^{x}+{e}^{-x}\right)$

## $y=\mathrm{tanh}x$:

• Has an infinite domain. Range is $-1
• Is an odd function
• Use the identity $\mathrm{tanh}x=$ $\frac{\mathrm{sinh}x}{\mathrm{cosh}x}$
• Therefore, $\mathrm{tanh}x=$ $\frac{{e}^{x}-{e}^{-x}}{{e}^{x}+{e}^{-x}}$

## a5 Differentiating hyperbolic functions:

• $\frac{\mathrm{d}}{\mathrm{d}x}\left(\mathrm{sinh}x\right)=\mathrm{cosh}x$
• $\frac{\mathrm{d}}{\mathrm{d}x}\left(\mathrm{cosh}x\right)=\mathrm{sinh}x$
• Sometimes you need to convert to $\frac{1}{2}\left({e}^{x}±{e}^{-x}\right)$ form, simplify, then convert back to hyperbolic functions

## Integrating hyperbolic functions:

• The above applies for integrating

## a6 Inverse hyperbolic functions:

• These functions are the same as the above, except reflected in the line $y=x$
• $y=\mathrm{arsinh}x$:
• $y=\mathrm{arcosh}x$:
• $y=\mathrm{artanh}x$:
• To prevent it being a one-to-many function, the domain of $y=\mathrm{arcosh}x$ is restricted to $x\ge 1$
• Similarly, $y=\mathrm{artanh}x$ is restricted to $-1
• The functions $\mathrm{arcosech}x$, $\mathrm{arsech}x$ and $\mathrm{arcoth}x$ are equivalent to $\mathrm{arsinh}\left(\frac{1}{x}\right)$, $\mathrm{arcosh}\left(\frac{1}{x}\right)$ and $\mathrm{artanh}\left(\frac{1}{x}\right)$ respectively

## a7 Logarithmic form of inverse hyperbolic functions:

• The formula book contains the following, which can be used to differentiate and integrate inverse hyperbolic functions:
• $\mathrm{arsinh}x=\mathrm{ln}\left(x+\sqrt{{x}^{2}+1}\right)$
• $\mathrm{arcosh}x=\mathrm{ln}\left(x+\sqrt{{x}^{2}-1}\right)$, $x\ge 1$
• $\mathrm{artanh}x=\frac{1}{2}\mathrm{ln}$$\left(\frac{1+x}{1-x}\right)$, $|x|<1$

## a8 Integrating functions of the form $\left(x$$2$$2$$\right)$$-1/2$:

• For these, you just need to use integration by substitution with the results in the formula booklet. These are:
and