# MandysNotes

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## Trigonometric Formulae

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Sunday, 01 May 2011 00:00

### A Brief Introduction to Complex Numbers

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Take a pencil and place it along the edge of a ruler, so that the ruler and the pencil are parallel, and so that the end of the eraser is at zero, and the point of the pencil points towards increasing numbers on the ruler.

Tuesday, 13 November 2012 18:53

### Derivatives of Trigonometric Functions

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$\frac{d}{dx}{ \color{Goldenrod}\sin}\ {z} = {\color{Blue}\cos}\ {z}$

$\frac{d}{dx}{\color{Blue} \cos}\ {z} = -{\color{Goldenrod} \sin}\ {z}$

Sunday, 01 May 2011 17:14

### Complex Numbers and Quadratic Polynomials

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Recall that a polynomial of degree two is called a Quadratic Polynomial.

A general quadratic polynomial (over the real numbers) is of the form:

$az^{2} + bz + c = 0.$

The formula for solving for the roots of this equation is:

$z = \frac{-b \pm \sqrt{b^{2} -4ac}}{2a} .$

Tuesday, 13 November 2012 18:03

### Primitives of Trigonometric Functions

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$\int{ \color{Goldenrod}\sin}\ {z} = - {\color{Blue}\cos}\ {z}$

$\int{\color{Blue} \cos}\ {z} ={\color{Goldenrod} \sin}\ {z}$

Sunday, 01 May 2011 20:58

### The Complex Conjugate

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Two complex numbers of the form:

$z_+ = x + iy ,$

and,

$z_- = x - iy ,$

are called Complex Conjugates.

Tuesday, 13 November 2012 19:19

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${\color{Goldenrod}\sin}\ \left( {\color{MidnightBlue}z_{1}} + {\color{MidnightBlue}z_{2}} \right) = {\color{Goldenrod}\sin}\ {\color{MidnightBlue}z_{1}} \ {\color{Blue}\cos}\ {\color{MidnightBlue}z_{2}} \ + \ {\color{Blue}\cos}\ {\color{MidnightBlue}z_{1}}\ {\color{Goldenrod}\sin}\ {\color{MidnightBlue}z_{2}} \$

${\color{Blue}\cos}\ \left( {\color{MidnightBlue}z_{1}} + {\color{MidnightBlue}z_{2}} \right) = {\color{Blue}\cos}\ {\color{MidnightBlue}z_{1}} \ {\color{Blue}\cos}\ {\color{MidnightBlue}z_{2}} \ - \ {\color{Goldenrod}\sin}\ {\color{MidnightBlue}z_{1}}\ {\color{Goldenrod}\sin}\ {\color{MidnightBlue}z_{2}} \$

Sunday, 01 May 2011 23:16

### The Modulus of a Complex Number

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For any complex number $z = x + iy ,$

$z \bar{z} = (x + iy)(x -iy)$

$= x^{2} + ixy - ixy + y^{2} = x^{2} + y^{2} .$

That is, any complex number times its complex conjugate equals the square of its real part plus the square of its imaginary part.

Because both the real and imaginary parts of any complex number are real:

$z \bar{z} = x^{2} + y^{2} \geq 0, \forall z \in \mathbb{C}.$

We define the positive square root of $z \bar{z}:$

$+\sqrt{z\bar{z}} : = |z| ,$

to be the Modulus of the complex number $z .$

Tuesday, 13 November 2012 19:13

### Half-Angle Formulae

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${ {\color{Goldenrod}\sin}\ \frac{{\color{MidnightBlue}z} }{2} } = \pm \left( \frac{1- {\color{Blue}\cos}\ {\color{MidnightBlue}z} }{2} \right)^{\frac{1}{2} }$

${ {\color{Blue}\cos}\ \frac{{\color{MidnightBlue}z} }{2} } = \pm \left( \frac{1+ {\color{Blue}\cos}\ {\color{MidnightBlue}z} }{2} \right)^{\frac{1}{2} }$

### sin(a+b) = (cos a)(sin b) + (sin a)(cos b)

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{module [79]}

Disclaimer: I did not create nor do I own these videos. I have simply embedded them, courtesy of YouTube.

Wednesday, 11 May 2011 23:16

### The Exponential Function

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(Note: this is a quick and dirty introduction to exponentials. I will not bother with any epsilons or deltas, but for those of you who are concerned about such things, be assured that all convergence is uniform, all sequences are Cauchy, etc. The series that defines the exponential is about as well behaved as an infinite series can be.)

The function $e^{x}$is defined to be the limit as $$n$$ approaches infinity of $( 1 + \frac{x}{n})^{n}:$

exp(x) = $e^{x} = \lim_{n \to \infty} (1 + \frac{x}{n})^{n} .$

### cos(a+b) = (cos a)(cos b)-(sin a)(sin b)

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{module [79]}

Disclaimer: I did not create nor do I own these videos. I have simply embedded them, courtesy of YouTube.

Friday, 13 May 2011 23:01

### Multiplication of Exponentials

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Consider first the exponential of a constant, a:

$e^{a} = \lim_{n \to \infty} (1 + \frac{a}{n})^{n} = \sum_{k=0}^{\infty} \frac{a^{k}}{k!}.$

If b is another constant, what is $e^{a}e^{b} ?$

Let's go back to the definition of exponential:

$e^{a}e^{b} \ = \lim_{n \to \infty} f(n) .$

Where:

$f(n) = (1 + \frac{a}{n})^{n} (1 + \frac{b}{n})^{n} = ( 1 + \frac{ a + b}{n} + \frac{ab}{n^{2}})^{n}$

$= \sum_{k=0}^{n} \binom{n}{k} c^{k},$

Friday, 13 May 2011 23:50

### The Exponential of a Purely Imaginary Number

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Consider the exponential of a purely imaginary number:

$z = i\alpha$

with $\alpha \in \mathbb{R}\ .$

$e^{i\alpha} = \lim_{n \to \infty } (1 + \frac{i\alpha}{n})^{n}$

$= \sum_{k=0}^{\infty }\frac{i\alpha^{k}}{k!} .$

### Addition and Double Angle Formulae

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From the definition of cosine:

$\cos{\theta} =\frac{e^{i\theta} + e^{-i\theta}}{2}$

we find:

$2\cos{(\alpha + \beta)} = e^{i(\alpha + \beta)} + e^{-i(\alpha + \beta)} .$

### Euler’s Formula

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The real part of $e^{i\alpha} \$

is:

$\text{Re}(e^{i \alpha}) = \frac{ e^{i\alpha} + e^{-i\alpha}}{2}$

$=\sum_{k=0}^{\infty} (-1)^{k}\frac{\alpha^{2k}}{(2k)!} .$

The imaginary part of $e^{i\alpha} \$

is:

$\text{Im}(e^{i\alpha}) = \frac{ e^{i\alpha} - e^{-i\alpha}}{2i}$

$= \sum_{k=1}^{\infty}(-1)^{k -1}\frac{\alpha^{2k -1}}{(2k -1)!} .$

We define:

$\text{Re}(e^{i\alpha}) = \frac{ e^{i\alpha} + e^{-i\alpha}}{2} := \cos{\alpha} ;$

$\text{Im}(e^{i\alpha}) = \frac{ e^{i\alpha} - e^{-i\alpha}}{2i} := \sin{\alpha}.$

Therefore:

$e^{i\alpha} = \text{Re} (e^{i\alpha}) + i\text{Im}(e^{i\alpha}) =\cos{\alpha} + i \sin{\alpha}.$

This is Euler's Formula.

### Sum of Squares Using Euler's Formula

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Euler's formula makes life much easier when it comes to deriving trigonometric identities.

Note, first of all, that $|e^{i\alpha}| = 1,$gives us:

$\left( \text{Re} (e^{i\alpha}) + i\text{Im}(e^{i\alpha}) \right) \left( \text{Re}(e^{i\alpha}) – i\text{Im}(e^{i\alpha}) \right)$

$= \left( \cos{\alpha} + i\sin{\alpha} \right) \left( \cos{\alpha} - i\sin{\alpha} \right)$

$= \cos^{2}{\alpha} + \sin^{2}{\alpha} = 1.$

### The Pythagorean Theorem from Euler’s Formula

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Consider any complex number $z, \$with modulus

$\sqrt{z\bar{z} } = |z| = r \ ,$

where $r \in \mathbb{R^{+}}\ .$

Note that $\frac{z}{r} \$would be a complex number with modulus one:

$| \frac{z}{r} | = | \frac{z}{|z|} | = 1.$

### Half Angle Formulae

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We can find the half angle formula for cosine by starting with the double angle formula:

$\cos{(2\alpha)} = \cos^{2}{\alpha} - \sin^{2}{\alpha} \ .$

Making the substitution:

$\gamma = 2\alpha ,$

$\cos{\gamma} = \cos^{2}{(\frac{\gamma}{2})} - \sin^{2}{(\frac{\gamma}{2})} .$