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Kendra22 · May 16, 2009, 10:35 AM

(1-cos2x +sin2x) / (1+cos2x +sin2x) = tanx

Unknown008 · May 16, 2009, 10:56 AM

$\frac{1-cos2x +sin2x}{1+cos2x +sin2x} = tanx$

$\frac{1-(1-2sin^2x) +2sinxcosx}{1+(1-2sin^2x) +2sinxcosx} = tanx$

$\frac{2sin^2x +2sinxcosx}{2-2sin^2x +2sinxcosx} = tanx$

$\frac{2sinx(sinx +cosx)}{2(1-sin^2x) +2sinxcosx} = tanx$

$\frac{2sinx(sinx +cosx)}{2cos^2x+2sinxcosx} = tanx$

$\frac{2sinx(sinx +cosx)}{2cosx(cosx+sinx)} = tanx$

$\frac{2sinx(sinx +cosx)}{2cosx(sinx +cosx)} = tanx$

Can you take that from here, it's already finished! :) Just remember, you have a single angle on the RHS, convert everything into single angles first. Then everything is simplifying.

Kendra22 · May 16, 2009, 05:35 PM

I get everything up till where you went but I'm not to sure how to reduce to single form and simplify I'm reallly really confused with trig identities they are uber difficult!:(

Unknown008 · May 16, 2009, 11:35 PM

But... but...

OK, I'll give you the answer. I'm really surprised you solved everything up to what I've done and yet, you couldn't do the rest, the actually simplest part.

the 2s are eliminated by division, as well as the (sinx+cosx)

$\frac{2sinx(sinx+cosx)}{2cosx(sinx+cosx)}= tanx$

$\frac{sinx}{cosx}= tanx$

Well, that's true, since $\frac{sinx}{cosx}$ does equal to $tanx$

Kendra22 · May 17, 2009, 10:10 AM

Wow I'm such an idiot I totally understand this stuff know! Just got a little brain fart yesterday

Unknown008 · May 17, 2009, 10:13 AM

LOL, OK, that may happen. Just try to remember these if you have an assessment or a test or exams, k? :)

Kendra22 · May 17, 2009, 10:58 AM

Now**