Question

how would i begin to prove the equation 2/1+cosx - tan^2 x/2=1

Answer 1

so it is \[\frac{1}{1+\cot x}-\tan^2x=1\]

Answer 2

okay i understand

Answer 3

the first thing im thinking about is multiplying by 1-cotx top and bottom for the 1/1+cotx

Answer 4

i haven't started yet i was just asking is that the right thing?
because the way you wrote it can be misleading

Answer 5

oh hold on i missed 2 on top

Answer 6

well it's like \[(2/1+cosx) - \tan^2 x/2=1\]

Answer 7

\[\frac{2}{1+\cot x}-\frac{\tan^2x}{2}=1\]

Answer 8

may bad! cos not cot

Answer 9

right right i got it

Answer 10

i must be blind or something lol

Answer 11

\[\frac{2}{1+\cos x}-\frac{\tan^2x}{2}=1\]
looking at the left side
\[\frac{2}{1+\cos x}-\frac{\tan^2x}{2}=\frac{2(1-\cos x)}{(1+\cos x)(1-\cos x)}-\frac{\tan^2x}{2}\]

Answer 12

let's go from there

Answer 13

\[\frac{2(1-\cos x)}{1-\cos^2x}-\frac{\tan^2x}{2}\]
\[\frac{2-2\cos x}{\sin^2x}-\frac{\tan^2x}{2}\]
\[\frac{2-2\cos x}{\sin^2x}-\frac{\sin^2x}{2\cos^2x}\]

Answer 14

can you go further your self

Answer 15

actually that's not an identity

Answer 16

i didnt think so lol

Answer 17

oh no i take it back it is!
when i plugged in 0
i forgot something lol

Answer 18

hmm for pi/4 it is not good!

Answer 19

are you sure you typed the right thing!

Answer 20

i will show you exactly what's in the textbook but how do you get the horizontal division line in the equation like you did in all your formulas?

Answer 21

use drawing tool if you have a mouse you will do find drawing the problem

Answer 22

snap shot your problem don't you have a phone?

Answer 23

what you have there is not an identity
i verified it with couple values and it fails miserably
if it is an identity it should not feel even one

Answer 24

fail*

Answer 25

\[\frac{ 2 }{ 1 + cosx } - \tan^2\frac{ x }{ 2 } = 1\]

Answer 26

ah wow why you didn't correct me above
i was dividing tan^2 over 2
2 was supposed to be within tan

Answer 27

darn you made me work really hard lol

Answer 28

im sorry i wouldve if i known it was wrong but i was confused on this entire problem hence why i am here asking for help

Answer 29

here what you can do
\[\frac{2}{1+\cos x}=1+\tan^2(x/2)=\sec^2(x/2)\]
if you prove the left is sec^2(x/2) then you are done!

Answer 30

that is the same as proving
\[\frac{1+\cos x}{2}=\cos^2(x/2) \]

Answer 31

isn't that some familiar identity to yoou?

Answer 32

the last line a i wrote should be obvious to you

Answer 33

i get it i think the sec and tan identity was what i was thinking i would get to to prove the identity i just wasnt sure how to get there

Answer 34

1+cosx/2=cos^2(x/2)
is called half angle identity
which can be proven easily
but you don't need to draw it all the way
as long as that is an identity we know of
so the initial problem is an identity as well

Answer 35

right right i remember that now

Answer 36

let's prove that half angle identity for the sake of argument
we know that cosx=2cos^2(x/2)-1
so 1+cosx=2cos^2(x/2)
cos^2(x/2)=(1+cosx)/2 hence proved
[cos(2a)=2cos^2a-1 is another useful identity]

Answer 37

they are all linked!

Answer 38

oh well gotta leave!

Answer 39

how do i prove it's sec though?

Answer 40

@zepdrix I was wondering if you can demonstrate this identity
2/1+cosx - tan^2 x/2=1

Answer 41

I'm pretty new to this myself, so I'm trying to understand what @xapproachesinfinity did
but I think I'm kinda lost...

Answer 42

Mmm can we do it in a new thread? :D
I don't feel like reading through all of these comments XD lol

Answer 43

mm I pretty much a noob in trig identity LOL
but ok xD

Answer 44

@Learner1298 we already proved it?
why you need to prove sec lol
read through what i did

Answer 45

\[\dfrac{2}{1+cos (x) }-tan^2(x/2)\] \[= \dfrac{2}{1+cos(x)} -\dfrac{sin^2(x)}{(1+cos (x))^2}\]
because \(tan (x/2) = \dfrac{sin (x) }{1+cos x}\)

Answer 46

hence, it turns to
\(\dfrac{2(1+cos(x)-sin^2(x)}{(1+cos(x))^2}\)
Now, numerator only:

Answer 47

\(2(1+ cos(x) ) -sin^2(x)\\=2+2cosx-(1-cos^2(x) =2+2cos(x)-1+cos^2(x) \\=1+2cos(x) +cos^2(x) = (1+cos(x))^2 \)

Answer 48

Put it back to the fraction above, we can see that numerator = denorminator , hence the fraction =1 = R H S

Answer 49

@xapproach.... hey kid, the second reader didn't understand your stuff. Let old man helps you out. hahahaha.....................

Answer 50

you did a different version lol
this trig stuff there are often many diff ways of doing stuff