Need help with a Trig Problem!!!!!

[Kerouactivist]

It is extra credit for my trig class(and I really need the extra credit)!!!


I must verify the identitiy of:

sec^2 x=(sec2x)(2-sec^2 x)


The people at the math lab can't even figure out how to work this one.

I need help
Please

 

[Kerouactivist]

I gotta get this one back up there so I don't fail trig

 

[shiner]

Ok...here's the solution. Got get yourself some beer....then start drinking...when you have had enough to say "ahh....screw this trig crap I need some Cheetos." then you have found the correct solution.

 

[Kerouactivist]

Actually I was planning on having a few drinks this afternoon.

But, first the trig problem must be solved....

 

[Pundit]

You're gonna want to jump out the window after you see how easy this is. Well maybe not, but jump out the window anyways. j/k

Here:
sec²x = (sec2x)(2-sec²x)
sec²x = 2*sec2x - sec2x*sec²x
sec²x = 2/cos2x - 1/(cos2x*cos²x)
sec²x = (2*cos²x - 1)/(cos2x*cos²x)
sec²x = (2*cos²x - 1)/(2*cos²x - 1)(cos²x)
sec²x = 1/cos²x

...and remember: ALT+ 0178 is your friend

 

[jpsj82]

hmm - don't remember all my trig.

we know
(1/cos^2(x)) = ( 1/cos(x))*(2- (1/cos^2(x)) )
1 = (cos^2(x)/cos(2x))
cos(2x) = cos^2(x)
2sin(x)cos(x) = cos^2(x)
2sin(x) = cos(x)
tan (x) = 1/2 - which of course is wrong. so i either made a mistake or this is not possible.

 

[Capn]

<< cos(2x) = cos^2(x) >>




No....

 

[TuffGirl]

pundit, how did you get from

<< sec²x = (2*cos²x - 1)/(cos2x*cos²x) >>

to...

<< sec²x = (2*cos²x - 1)/(2*cos²x - 1)(cos²x) >>

?

 

[Pundit]

<< pundit, how did you get from

<< sec²x = (2*cos²x - 1)/(cos2x*cos²x) >>

to... >>

sec²x = (2*cos²x - 1)/(2*cos²x - 1)(cos²x)
sec²x = 1/cos²x [/i] >>

?[/i] >>



The thingy in the numerator cancels with the thingy in the denominator.

 

[jpsj82]

<<

<< cos(2x) = cos^2(x) >>




No.... >>

what are you saying - that this is where i have my mistake or that this is not possible (because this does have solutions; 0).

 

[TuffGirl]

Ahh I should clarify, I edited my post to exclude the last step where you obviously canceled the numerator and denominator thingies. But in the one step I pointed out, the numerators remain the same on the right side but I don't see how they equate: (cos2x*cos²x) = (2*cos²x - 1)(cos²x) ?

My trig is a bit rusty from 10 yrs ago

 

[Kerouactivist]

You are my hero.

Thanks for the help....

I am a Political Science major, I am graduating this semester, and for some reason I have to take Trig. I really don't see how it's applicability

Anyway once again thank you

and with Trig I always wanna jump out the window lol

Also what is ALT+ 0178

 

[Pundit]

<< cos2x= (2*cos²x - 1) >>


This is a trig identity.

 

[RSMemphis]

<<

<<

<< cos(2x) = cos^2(x) >>




No.... >>

what are you saying - that this is where i have my mistake or that this is not possible (because this does have solutions; 0). >>


Yeah, but we are looking for all x, not just x =0

 

[TuffGirl]

<<

<< cos2x= (2*cos²x - 1) >>


This is a trig identity. >>

Ahh that makes sense. Nifty proof!


bthorny, ALT+ 0178 is the key sequence for the ² symbol.

 

[Pundit]

<< You are my hero.

Thanks for the help....

I am a Political Science major, I am graduating this semester, and for some reason I have to take Trig. I really don't see how it's applicability >>


The more you know... something, something (can't remember)


<<
Anyway once again thank you

and with Trig I always wanna jump out the window lol

Also what is ALT+ 0178 >>


No problem. Hold down ALT key and then press 0+1+7+8 in succession, then let go of the ALT key and you will get the ² character.

 

[Kerouactivist]

Thanks for the info, I don't know HTML code, anybody know a good website or book that I can use to learn HTML

I did a google search but, I got a lot of crap.

 

[TuffGirl]

<< Thanks for the info, I don't know HTML code, anybody know a good website or book that I can use to learn HTML

I did a google search but, I got a lot of crap. >>

I googled for keywords "HTML primer" 'cause I remember they used to have a good clearly-written basic one. HTML Primer

 

[BooneRebel]

<< You're gonna want to jump out the window after you see how easy this is. Well maybe not, but jump out the window anyways. j/k

Here:
sec²x = (sec2x)(2-sec²x)
sec²x = 2*sec2x - sec2x*sec²x
sec²x = 2/cos2x - 1/(cos2x*cos²x)
sec²x = (2*cos²x - 1)/(cos2x*cos²x)
[/b]sec²x = (2*cos²x - 1)/(2*cos²x - 1)(cos²x)[/b]
sec²x = 1/cos²x

...and remember: ALT+ 0178 is your friend >>

I've been trying to puzzle through this one, too, but it's been a long time since trig. In your third step it looks like you're using secx=1/cosx to solve. I've been looking and can find an identity sec^2x=1/cos^2x, but can't find secx=1/cosx. Are you sure this is valid? I'm not sure about cos2x=2xcos^2x-1 either. I'm not saying your wrong, I just don't recognize the identities that support these.

 

[Kerouactivist]

Thanks Klee58 that site looks very informative. I'll have to check it out when time permits

 

[Pundit]

<< I've been trying to puzzle through this one, too, but it's been a long time since trig. In your third step it looks like you're using secx=1/cosx to solve. I've been looking and can find an identity sec^2x=1/cos^2x, but can't find secx=1/cosx. Are you sure this is valid? I'm not sure about cos2x=2xcos^2x-1 either. I'm not saying your wrong, I just don't recognize the identities that support these. >>



Well, I found this site through Google.

 

[Kerouactivist]

I think that will come in handy for me too.