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Posts:
153
Registered:
10/3/07


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 nonzero (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.

