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Need help with a Trig Problem!!!!!

Kerouactivist

Kerouactivist

Diamond Member
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
 
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.
 
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 😉
 
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.
 
pundit, how did you get from

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

to...

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

?
 


<< 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. 😀
 


<<

<< 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).
 
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😛
 
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
 


<<

<<

<< 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

 


<<

<< 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.🙂
 


<< 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. 😀
 
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.
 


<< 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
 


<< 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.
 


<< 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.
 
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