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Trig Identities

Started by Vixus, Feb 20 2006 10:16 PM

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#1 Vixus

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Posted 20 February 2006 - 10:16 PM

I want to build up a list of trig identities here (and any other identities for that matter) that may be of help to people in the exam. Here're the ones I have so far.

$\begin{align*} \sin^2 x + \cos^2 x = 1 \\ 2\sin^2 x - 1 = \cos(2x) \\ 2\sin x\cos x = \sin(2x) \end{align*}$

$\begin{align*} \tan x = \frac{\sin x}{\cos x} \\ \sec x = \frac{1}{\cos x} \\ \textrm{cosec} x = \frac{1}{\sin x} \\ \cot x = \frac{1}{\tan x} \end{align*}$

$\begin{align*} \tan^2 x + 1 = \sec^2 x \\ \frac{d}{dx}\tan x = \sec^2 x \\ \frac{d}{dx}\sin2x = 2\cos2x \end{align*}$

$\begin{align*} \frac{d}{dx} \textrm{cosec} x = -\textrm{cosec} x \textrm{cot} x \\ \int \frac{1}{\sqrt{a^2 - x^2}} = \frac{1}{a} \sin^{-1} \frac{x}{a} \\ \int \frac{1}{a^2 + x^2} = \frac{1}{a} \tan^{-1} \frac{x}{a} \end{align*}$

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#2 The Wedge Effect

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Posted 20 February 2006 - 10:21 PM

I've got an entire list of identities on my formula sheet, if you want me to put them up?

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#3 Vixus

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Posted 20 February 2006 - 10:22 PM

QUOTE(Wedge37 @ Feb 20 2006, 10:21 PM)

I've got an entire list of identities on my formula sheet, if you want me to put them up?

Yeah! I wish we got one in the exam, myself. XD

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#4 George

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Posted 20 February 2006 - 10:35 PM

I edited your post to fix the tex markup

Just a couple of pointers:

you can put a slash in front of trig functions (eg \sin rather than sin) to get them in roman text (ie $\sin x$ rather than
I think there's a problem with having two newlines eg:
code

more code
Basically, if you want a new line you should use " \\ " in the code.

These aren't specific to you either; they're just things I've noticed over the past few weeks.

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#5 The Wedge Effect

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Posted 20 February 2006 - 10:38 PM

$\begin{align*} \cos^2 x +\sin^2 x=1\\ 1+\tan^2x=sec^2x\\ 1+cot^2x=cosec^2x\\ \sin (x+y)=\sin x\cos y+\cos x\sin y\\ \sin (x-y)=\sin x\cos y-\cos x\sin y\\ \cos (x+y) = \cos x \cos y - \sin x \sin y\\ \cos (x-y) = \cos x \cos y + \sin x \sin y\\ \tan (x+y) = \frac {\tan x + \tan y}{1-\tan x \tan y}\\ \, \\ \tan (x-y)=\frac {\tan x -\tan y}{1+\tan x \tan y}\\ \, \\\end{align*}$
$\begin{align*} \sin (2x)=2\sin x \cos x\\ \cos (2x)= \cos^2x-\sin^2x\\ \cos (2x)=1-2\sin^2x\\ \cos (2x)=2\cos^2x-1\\ \tan (2x)= \frac {2 \tan x}{1-\tan^2x}\\ \cos^2x=\frac{1}{2} [\cos (2x) - 1]\\ \sin^2x=-\frac{1}{2} [\cos (2x) -1]\\ \, \end{align*}$

That should come out correctly. I'll type in the rest later on when I've got time (the derivatives and integrals of trig functions).

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#6 Paul

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Posted 20 February 2006 - 10:49 PM

I also have a post with a list of trig identities to download....its somewhere in this thread!

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#7 Ally

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Posted 21 February 2006 - 06:40 PM

QUOTE(Wedge37 @ Feb 20 2006, 10:38 PM)

$\begin{align*} \tan (x+y) = \frac {\tan x + \tan y}{1-\tan x \tan y}\\ \, \\ \tan (x-y)=\frac {\tan x -\tan y}{1+\tan x \tan y}\\ \, \\\end{align*}$

That should come out correctly. I'll type in the rest later on when I've got time (the derivatives and integrals of trig functions).

You don't need to know them for AH IIRC.

PS - Paul, it is here.

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#8 Vixus

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Posted 23 February 2006 - 05:40 PM

QUOTE(George @ Feb 20 2006, 10:35 PM)

I edited your post to fix the tex markup

Just a couple of pointers:

you can put a slash in front of trig functions (eg \sin rather than sin) to get them in roman text (ie $\sin x$ rather than
I think there's a problem with having two newlines eg:
code

more code
Basically, if you want a new line you should use " \\ " in the code.

These aren't specific to you either; they're just things I've noticed over the past few weeks.

Thanks. I was wondering why it was all crushed.

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#9 The Wedge Effect

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Posted 08 March 2006 - 05:22 PM

$\begin{align*} \cot x=\frac {\cos x}{\sin x}\end{align*}$

Just remembered another one that needed added to the list.

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#10 Vixus

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Posted 08 March 2006 - 07:45 PM

I think this is a useful one:

$\begin{align*}\frac{d}{dx}\textrm{cosec}x = -\textrm{cosec}x \cotx\end{align*}$

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#11 Pringles

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Posted 01 May 2006 - 06:21 PM

Here is quite an important result that is useful to know

$\int secx = ln (secx + tanx) = \int secx \frac {secx + tanx}{secx + tanx} dx\\ \\ = \int \frac {sec{^2}x+ sectanx}{secx + tanx} dx\\$

Heres how to prove it for nobody that knows:

Firstly the derivative of secx = secxtanx

$set u = secx + tanx\\ \frac{du}{dx} = secxtanx + sec{^2}x \\ du = (secxtanx + sec{^2}x) dx$

input these values into the equation above

$\int \frac{du}{u} \\ =ln(u) \\ =ln(secx + tanx)$

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