Question

Prove the identity. Help? :)

Answer 1

@Astrophysics ...I'm going to bug you just a little more :P can you help? :)

Answer 2

@sdfgsdfgs

Answer 3

dats gonna take some work...

1st step is to use this formula \[\tan ^{2}x +1 = \sec ^{2}x\]
wat will u get?

Answer 4

um, just a moment. I'm not sure o.0

Answer 5

tan^2x=sec^2x-1

Answer 6

\[\sec^2(\frac{ x }{2}+\frac{ \pi }{ 4 })-1\]

Answer 7

is that correct?

Answer 8

dats correct - good job!

next step is to simplify the expression further...

\[\sec ^{2}(\frac{ x }{ 2 }+\frac{ \pi }{ 4 }) - 1 = \frac{ 1 }{ \cos ^{2}(\frac{ x }{ 2 }+\frac{ \pi }{ 4 } )} - 1\]

Answer 9

then use cos(A+B) = cosA*cosB - sinA*sinB

cos (x/2 + pi/4) = cos(x/2)cos(pi/4) - sin(x/2)sin(pi/4)
=(cos(x/2)-sin(x/2))/sqrt(2) because cos(pi/4)=sin(pi/4)=1/sqrt(2)

so far so good?

Answer 10

Hm, I was good up till the second line :P

Answer 11

which part of second line u dont understand?

Answer 12

The sum difference identity I assume

Answer 13

\[\cos(A+B) = cosA*cosB - sinA*sinB\]

Answer 14

\[\frac{ \cos \frac{ x }{ 2 }-\sin \frac{ x }{ 2 }}{ \sqrt{2} }\]
why does it equal that?

Answer 15

@Babynini sorry i need to leave soon :(

@Astrophysics if u dont mind, would u please take over? actually im not sure what I did is the best way. it may be simpler to use the eqn for tan(A+B) instead...

Answer 16

Sure @sdfgsdfgs I will try :)

Answer 17

\[\tan(A+B) = \frac{ tanA + tanB }{1-tanA*tanB }\]

Answer 18

Thanks so much @sdfgsdfgs ! :)

Answer 19

A=x/2 and B=pi/4 and tan(pi/4)=1 so that will simplify the expression by quite a bit....

thanks @Astrophysics

good luck @Babynini

Answer 20

Yes that will work I believe, if you use \[\tan(A+B) = \frac{ tanA + tanB }{1-tanA*tanB }\]

Answer 21

even with tan^2?

Answer 22

Yup, it should work out let a = x/2 b =pi/4\[\tan(x/2+\pi/4) = \frac{ \tan(x/2)+\tan(\pi/4) }{ 1-\tan(x/2)\tan(\pi/4) }\] and we note that \[\tan(\pi/4) = 1\] so we have \[\frac{ \tan(x/2)+1 }{ 1-\tan(x/2) \times 1 }\] and then we can use \[tanx = \frac{ sinx }{ cosx }\] simplify it a bit and square the whole expression after and it should work out :)!

Answer 23

Now we just need to do the math :P, so try it out and don't be afraid to make mistakes we can go over it together after ^.^

Answer 24

k, so that would simplify to:

((sinx/cos2)+1)/1-(sinx/cos2)(1)

and since they're both over cos2 I can cross those out, correct?

so it becomes
(sinx+1)/(sinx-1)

Answer 25

wait wrong post lel

Answer 26

Waait, the answer is supposed to be (1-sinx)/(1+sinx) o.0

Answer 27

Nnesha, it's all good :P

Answer 28

oh oops, yeah it does come out to positive on top, negative on the bottom. I just wrote it wrong in here.

Answer 29

ai ai ai, that was very wrong. Sorry. I randomly added 1 to the top when it wasn't over cosine.

Answer 30

yeah you did it wrong :(
tan x = sin(x)/cos(x)
so tan (x/2) = sin(x/2)/cos(x/2)

Answer 31

-.- the whole thing is wrong because I split cosine and sin haha aii. K, starting over.

Answer 32

Thank you @Nnesha

Answer 33

Hey did you get it? :)

Answer 34

...no hehe I got stuck because there's that ^2 that we just took out o.0

Answer 35

I was lost right when you left xD I'm hopeless lol

Answer 36

@Babynini looks like u have a long but productive nite ;)

im back n can see the solution now....let me know if u wanna continue w this....or not! lol

Answer 37

bleeh dude, i'm so ready to be done with this homework -.- haha em, are you up for finishing it right now??

Answer 38

Though I should warn you i'm not completely functional ;P you went out for dinner, right? how did it go? :)

Answer 39

hahahaa its afternoon tea, movie, dinner n then some more lol thanx 4 asking :)

btw - nice white board at home! do u always do ur math on a white board? lol

OK Astrophysics actually got pretty far down...where u want to pick up from?

Answer 40

hahah white board isn't that good
lol i like it

Answer 41

that sounds so nice!!
thanks :D yes i do lol, it helps for some reason xD

Answer 42

main risk is that a random swipe on the white board could lose you a few key steps!

Answer 43

^ so true.

Answer 44

Em, i'm not sure where to pick up on o.0 I need to refresh on what the problem was even haha let me look it over.

Answer 45

i wuld suggest we start with tan[...] = (tan(x/2) + 1) / (1 - tan(x/2))
and then use tan(x/2)=sin(x/2) / cos(x/2) and multiply cos(x/2) to top n bottom...

dont worry we will put the ^2 back in the next step :)

Answer 46

Sorry, guys, I'm in the middle of another problem now my mind is completely on something else. I'll be with yall in like 5 mins.

Answer 47

OK

Answer 48

aww that's fine go and eat chocolates plz

Answer 49

http://prntscr.com/73jxz1

Answer 50

Bleh, actually. Can we do this tomorrow? Unless yall wanna solve it for me xD

Answer 51

well...seeing how hard u have worked tonite....hmmm.....sure why the hell not :P

Answer 52

wait, that you'll solve it for me? I will love you forever if you do!

Answer 53

for simplicity, lets replace x/2 by A

last we have (1+tanA)/(1-tanA) right?

Answer 54

ok

Answer 55

tanA=sinA/cosA n mutiply top n bottom by cosA...it becomes
(1+tanA)/(1-tanA) = (cosA + sinA) / (cosA - sinA)

since the original eqn is for tan^2[.....] so it becomes
(cosA+sinA)^2 / (cosA-sinA)^2

Answer 56

@Babynini heheee how brain-dead are u? cuz' u really should be able to see the ans from here ;)

Answer 57

I can't...even... #.#

Answer 58

do some things cancel out or something? xP

Answer 59

it would be (1)/(cos^2a -sin^2a) ?

Answer 60

(sinA)^2 + (cosA)^2 =1 so.....

Answer 61

nope it should be (1+2sinAcosA)/(1-2sinAcosA)

Answer 62

wait why. If the ^2 factors out to the top row it should all equal 1

Answer 63

do u get this:
"
since the original eqn is for tan^2[.....] so it becomes
(cosA+sinA)^2 / (cosA-sinA)^2"

Answer 64

yep

Answer 65

(a+b)^2=a^2+b^2+2ab
(a-b)^2=a^2+b^2-2ab
so wat wuld u get?

Answer 66

haha nah man I can't. I can't even get that right now D:

Answer 67

hmmmm u wuld hate urself if i solve it 4 u right now....

how about this - take a look again tmw n if u cant figure out from here:

(cosA+sinA)^2 / (cosA-sinA)^2

ping me n i will solve it 4 u then. deal? :)

Answer 68

a=x/2, b=pi/4 right? just so we're on the same page at least there hah

Answer 69

nope u use
(a+b)^2=a^2+b^2+2ab
(a-b)^2=a^2+b^2-2ab

in

(cosA+sinA)^2 / (cosA-sinA)^2

Answer 70

oh gosh. Ok, deal :) sorry I collapsed on you haha I need sleep @.@ Nothing is making sense xP (but that's not your fault at all!!)

Answer 71

its quite alright @Babynini u had a long nite. get sum rest n pick it up tmw - i guarantee u wuld see it right away ;)

Answer 72

I sure hope so! Thanks for everything, Sleep happy :)

Answer 73

Ai this is going to be difficult to do on mothers day :P
Em... @sdfgsdfgs would you mind writing out all the steps we've taken so far to get where we are? Because i'm not even sure. Like write it on paper and put a pic up on here. That would be soo helpful. If you don't want to that's fine though :)

Answer 74

@Babynini my suggestion would be to stroll up and read the last post by Astrophysics. That is as good a starting pt as any :)

Answer 75

Hear ye, hear ye! I have solved it! With much help from you guys and also siblings , but in the end it was solved by help from a tutor at school and running on negative amounts of sleep.
Thank you all soo so much:)

If anyone is interested in how this question from hell was solved I can put a pic up but if you're all just sick of it I understand that too xD
Thanks again!!

Answer 76

Good job, @Babynini :) now go get some zzzzz lol

Answer 77

^^ I wish haha math homework today, math exam tomorrow, and then a 10 pg essay due Thusday. =.=