
Re: Proving trigonometric identities
Posted:
Jan 20, 2009 3:41 AM


On Jan 20, 7:22 pm, Angus Rodgers <twir...@bigfoot.com> wrote: > On Tue, 20 Jan 2009 08:19:58 +0000, I groggily wrote: > >[...] 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 [...] > > ... in conjunction, of course, with the known trigonometric > identity: > > cos(x)^2 + sin(x)^2 = 1 > >  > Angus Rodgers
Are you implying that (1+sinx)(1sinx) = 1  (sinx)^2 = 1  (sin^2)x = 1  sin(x)^2 ?
My booklet writes square relations (sin^2)x, you seem to write sin(x) ^2 and I've always thought that sinx * sinx = (sinx)^2 which was not equal to (sin^2)x. Which means I MUST have been wrong because = 1  (sinx)^2 WOULD result in no further steps possible. But as I've clearly seen, it gets rearranged (with some correct stuff on the left hand side of the equals sign) to be one of the trigonometric identities and therefore my original problem is proven!... :

