some trig questions

prove the following....

(1) sin2x/1+cos2x = tanx

ive done this so far...2sinxcosx/2cosx^2-1
sinx/cosx - 1
dno bout the rest..if thats right lol

(2) 1-cos2x/1+cos2x= tan^2x
1-(1-2sin^2x)/1+(2cos^2x-1)
2sin^2x/2cos^2x
= tan^2x

(3) sin2x + sinx/ 1 + cos2x + cosx = tanx

(4) 1-cos2x + sin2x/1+cos2x+sin2x = tanx

ne help would be appreciated...trig aint too hard..jus the provin that pissin me off

Reply 1

danniella

Hello!

First one:
sin2x/(1+ cos2x) = 2sinxcosx/(1+cos^2x - sin^2x)
= 2sinxcosx/(1-sin^2x + cos^2x)
= 2sinxcosx(cos^2x + cos^2x)
= 2sinxcosx/2cos^2x
= sinx/cosx
=tan x

love danniella

Reply 2

danniella

Second one:
(1-cos 2x)/(1+cos 2x) = [1-(cos^2x - sin^2x)]/[1+ cos^2x-sin^2x]
= [1-cos^2x + sin^2x]/[1-sin^2x + cos^2x]
= [sin^2x + sin^2x]/[cos^2x + cos^2x]
=[2 sin^2x]/[2 cos^2x]
=sin^2x/cos^2x
=tan^2x

love danniella

Reply 3

danniella

Number 3:
(sin 2x + sin x) / (1 + cos 2x + cos x)
= [2sinxcosx + sin x] /[1-sin^2x + cos^2x + cos x]
= [sin x(2cos x + 1)]/[cos^2x + cos^2x + cos x]
= [sin x (2 cos x + 1)]/[2 cos^2x + cos x]
= [sin x(2 cos x + 1)]/[cos x(2 cos x + 1)]
= sin x / cos x
= tan x

love danniella

Reply 4

danniella

Last one:
[1-cos 2x + sin 2x]/[1+ cos 2x + sin 2x]
= [1-cos^2x + sin^2x + 2sinxcosx]/[1+cos^2x - sin^2x + 2sinxcosx]
=[sin^2x + sin^2x + 2sinxcosx]/[1-sin^2x + cos^2x + 2sinxcosx]
=[2sin^2x + 2sinxcosx]/[cos^2x + cos^2x + 2sinxcosx]
= [2sinx(sinx + cosx)]/[2 cos^2x + 2 sinxcosx]
= [2sinx(sinx + cosx)]/[2cosx(sinx + cosx)]
= 2sinx/2cosx
=sinx/cosx
=tan x

love danniella

Reply 5