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#1 2009-12-20 00:29:32

Mahamaya
Member
Registered: 2009-11-21
Posts: 6

Still on trig

Am new to trig and this identity thing just screws me silly.

My question is: how do i know which identity to use in solving a particular trig problem?

What are the series of questions i need to ask myself the moment i see a trig problem?

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#2 2009-12-20 01:09:30

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Still on trig

Hi Mahamaya;

My question is: how do i know which identity to use in solving a particular trig problem?

You won't at first, you have to acquire a feel for which identities to use. This will come with practice.

Please go here, to see this very question being discussed.

http://www.mathisfunforum.com/viewtopic … 70#p124470


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#3 2009-12-20 02:44:49

Mahamaya
Member
Registered: 2009-11-21
Posts: 6

Re: Still on trig

Thanks a lot bobbym for your prompt response. Got some links from that page i'll like to go check out.
But i'll like to know please if u do not mind: were all identities derived from this simple one

? Because for starters i know this
was derived from it. However to get to the latter formula we divided through by
. Why?

Last edited by Mahamaya (2009-12-20 02:46:59)

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#4 2009-12-20 10:52:57

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Still on trig

Hi Mahamaya;

I don't know whether all the trig identitites can be derived from the pythagorean theorem. You can certainly get some of them.

For that particular one yes you can reason like this: Divide the pythagorean formula by c^2

So looking at the picture:

sin(A) = a / c and cos(A) = b / c

we can substitute this into B:


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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