TRIG HELP D:

[ninbam1k]

heyy guys, how do i simply this? been trying to figure this out for agess

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

HELP  

[pi]

1 + 2sin2x

= 1 + 4 sinx cosx


(or was there supposed to be brackets on the denominator?)

[ninbam1k]

my bad
brackets on the denominator
so the whole top on the whole bottom

[/0]

It's a matter of simplifying the top and bottom into a product of factors so that you can later cancel out.



Extra hint 1:
The 1s on the top and bottom stick out... can you expand one of the sin(2x) or cos(2x)s to get rid of them?


Extra hint 2:
You have (2x)'s and (x)'s as angles, but it's better to make them all the same. So expand the rest of the (2x)'s.

[pi]

I only got this far:

(sin^2 x + cos^2 x + 2 sinx cox - sin^2 x + cos^2)/(sin^2 x + cos^2 x + 2 sinx cox + sin^2 x - cos^2)

= (2 cos^2 x + 2 sinx cosx)/(2 sin^2 x + 2 sinx cosx)

= (2 cosx (cosx+ sinx))/(2 sinx (cosx+ sinx))

= 2 cotx (cosx + sinx)/(sinx + cosx)

= 2 cotx


Is this right? Now it is right!



(I REALLY need to learn LaTeX these hols!)

[ninbam1k]

wooo i got it thanks alot guys :}

[pi]

wooo i got it thanks alot guys :}

What the answer? I think I might have been wrong...

[ninbam1k]

i got cotx buddy

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

= (2(cosx)^2 + 2sinxcosx) / (2(sinx)^2 +2sinxcosx)

kinda skipped a step here but all i did was let cos2x = 2(cosx)^2 - 1 for the numerator and cos2x = 1 - (sinx)^2 for the denominator

FACTORISE

= (2cosx)(cosx + sinx) / (2sinx)(sinx + cosx)

CANCEL OUT (sinx+cosx)

= 2cosx / 2 sinx

= cotx :}

[xZero]

@Rohitpi the 2 cancles out so its cot x

[pi]

lol, forgot to cancel the 2  :idiot2:

Yep, you are right!

[TrueTears]

i got cotx buddy

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

= (2(cosx)^2 + 2sinxcosx) / (2(sinx)^2 +2sinxcosx)

kinda skipped a step here but all i did was let cos2x = 2(cosx)^2 - 1 for the numerator and cos2x = 1 - (sinx)^2 for the denominator

FACTORISE

= (2cosx)(cosx + sinx) / (2sinx)(sinx + cosx)

CANCEL OUT (sinx+cosx)

= 2cosx / 2 sinx

= cotx :}

another way, i haven't checked it all but try multiplying by [1-(sin2x -cos2x)]/[1-(sin2x-cos2x)], this way is definitely longer but it's just for fun haha