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Topic: Proving trigonometric identities
Replies: 32   Last Post: Jan 27, 2009 1:59 AM

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Re: Proving trigonometric identities
Posted: Jan 20, 2009 10:14 AM

On Jan 20, 12:19 am, Angus Rodgers <twir...@bigfoot.com> wrote:
> On Tue, 20 Jan 2009 00:11:54 -0800 (PST), Albert
>
>
>
> <albert.xtheunkno...@gmail.com> wrote:

> >On Jan 20, 6:40 pm, Angus Rodgers <twir...@bigfoot.com> wrote:
> >> On Mon, 19 Jan 2009 23:31:43 -0800 (PST), Albert
>
> >> <albert.xtheunkno...@gmail.com> wrote:
> >> >On Jan 20, 6:25 pm, Angus Rodgers <twir...@bigfoot.com> wrote:
> >> >> Are you familiar with the algebraic identity (a + b)(a - b) = a^2 - b^2?
>
> >> >Yes I am - are you implying the following steps:
> >> >(cosx)^2 = (1 + sinx)(1 - sinx)
> >> >..``..``.= 1 - (sinx)^2
> >> >?

>
> >> Yes, I was, but you hardly need to know what I was thinking,
> >> because the result is clear now, isn't it?

>
> >I don't think that it's a proof though. Please enlighten me.
>
> Well, assuming that B and D are both non-zero (which in this case
> reduces to cos(x) != 0), an identity between fractions, A/B = C/D,
> is equivalent to AD = BC.  So, the identity you have to prove is
> equivalent to the one you have just written out - which follows
> simply from the given algebraic identity.  That's a proof (unless
> I'm even more groggy than I thought!).
>
> --
> Angus Rodgers

Hello,

Curiously enough, the (of course correct) strategy you mention above
would have been marked as being wrong on a British Columbia Math 12
Provincial Exam!

One "accepted" way of proving the identity is to multiply "top" and
"bottom" of
(1 - sin x)/(cos x) by 1 + sin x. And full marks are given even if
one does not worry about whether 1 + sin x = 0.

Date Subject Author
1/20/09 Albert
1/20/09 Angus Rodgers
1/20/09 Albert
1/20/09 Angus Rodgers
1/20/09 Albert
1/20/09 Angus Rodgers
1/20/09 Angus Rodgers
1/20/09 Guest
1/20/09 Angus Rodgers
1/20/09 Albert
1/20/09 Angus Rodgers
1/20/09 matt271829-news@yahoo.co.uk
1/21/09 Albert
1/21/09 matt271829-news@yahoo.co.uk
1/23/09 Albert
1/23/09 Angus Rodgers
1/23/09 Angus Rodgers
1/23/09 Passerby
1/23/09 Dave Dodson
1/24/09 Albert
1/24/09 Angus Rodgers
1/24/09 Albert
1/26/09 Albert
1/26/09 Driveby
1/26/09 Albert
1/26/09 A N Niel
1/27/09 Albert