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
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.