# Auxilary Method

• Sep 5th 2006, 02:52 AM
Nick744
Auxilary Method
G'day,

Firstly, i'd like to say hello to everyone, also, not to mention what a great service is being run here and how many people benefit from it.

I live in Australia and am in year 11.

I have been having problems understanding the auxilary method for some strange reason. Is there anyone that could maybe explain it to me, (and hope I understand it) or point me to a website?

Thanks,
Nick
• Sep 5th 2006, 05:45 AM
ThePerfectHacker
What is that supposed to be?
I search Wikipedia, MathWorld, and Planetmath all results are negative.
• Sep 5th 2006, 02:07 PM
Nick744
Sorry. You probably use different terms in the U.S.

The Auxilary method contains the following:

$\displaystyle r = SQUAREROOT of a^2 + b^2$

sorry...forgot the code for square root...
• Sep 5th 2006, 02:12 PM
Quick
Quote:

Originally Posted by Nick744
Sorry. You probably use different terms in the U.S.

The Auxilary method contains the following:

$\displaystyle r = \sqrt{a^2 + b^2}$

sorry...forgot the code for square root...

What is the purpose of the auxilary method? (what does it do or solve)
• Sep 5th 2006, 02:14 PM
Nick744
A trig equation
• Sep 5th 2006, 02:21 PM
Quick
Quote:

Originally Posted by Nick744
A trig equation

very specific :rolleyes:

could you maybe give us one of your problems that uses the method? Because that would probably clear things up.
• Sep 5th 2006, 02:35 PM
JakeD
Quote:

Originally Posted by Nick744
Sorry. You probably use different terms in the U.S.

The Auxilary method contains the following:

$\displaystyle r = SQUAREROOT of a^2 + b^2$

sorry...forgot the code for square root...

See "5. Introducing an auxiliary angle" here.
• Sep 5th 2006, 03:55 PM
ThePerfectHacker
It is the distance from point $\displaystyle (a,b)$ to the origin.
• Sep 5th 2006, 04:17 PM
Quick
Quote:

Originally Posted by ThePerfectHacker
It is the distance from point $\displaystyle (a,b)$ to the origin.

than why $\displaystyle r$ instead of $\displaystyle d$?
• Sep 5th 2006, 04:22 PM
topsquark
Quote:

Originally Posted by Quick
than why $\displaystyle r$ instead of $\displaystyle d$?

Because when we switch from rectangular coordinates (x, y) to polar coordinates we use the form $\displaystyle (r, \theta)$, where r is the distance from the point to the origin.

-Dan
• Sep 5th 2006, 06:31 PM
CaptainBlack
Since nobody seems to have followed the link in JakeD's post here
is the guts of what it says:
• Sep 5th 2006, 11:33 PM
chancey
Quote:

Originally Posted by Nick744
I live in Australia and am in year 11.

I also live in Australia and in yr 11

What level of maths you doing? i've never heard of this auxilary method but it sounds like your talking about $\displaystyle (r, \theta)$

Do you mean something like this:
Convert the coordinates (3, 4) to polar form:

1. Find distance using distance formula $\displaystyle = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$

2. Find the angle
$\displaystyle \tan \theta = \frac{4}{3}$

$\displaystyle \therefore \theta \approx 53^o$

$\displaystyle \therefore (3,4) => (5, 53^o)$

EDIT: OK, I just saw the post above, i've never seen that method before, what does it do?
• Sep 5th 2006, 11:49 PM
chancey
Ahhh, I found what your looking for in my 3 unit textbook. I sort of remmeber doing it, we didnt spend alot of time on it.

Its for solving trig equations, it some crazy method that didnt really make sense, so we didnt use it. But heres a question;

$\displaystyle a \sin \theta + b \sin \theta = r \sin (\theta + \alpha)$ where $\displaystyle r = \sqrt{a^2 + b^2}$ and $\displaystyle \tan \alpha = \frac{b}{a}$

Solve $\displaystyle 2 \sin \theta = \cos \theta$ for $\displaystyle 0^o <= \theta <= 360^o$

Solution:

$\displaystyle 2 \sin \theta = \cos \theta$

Dividing both sides by $\displaystyle \cos \theta$

$\displaystyle \frac{2 \sin \theta}{\cos \theta} = \frac{\cos \theta}{\cos \theta}$

$\displaystyle 2 \tan \theta = 1$

$\displaystyle \tan \theta = 0.5$

Since tan is positive in the first and forth quadrants:
$\displaystyle \theta \approx 26^o, 206^o$

How that has anything to do with this fancy equation solving thing I dunno? Thats probly why we never used it