

12132009, 09:06 PM

#1

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Join Date: Mar 2006
Posts: 5,574

sec(tan^1(4/3))^2=25/9
Could someone explain why that is?
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12132009, 09:30 PM

#2

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Join Date: Jul 2004
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Re: sec(tan^1(4/3))^2=25/9
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12132009, 10:14 PM

#3

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Re: sec(tan^1(4/3))^2=25/9
You mean tan(4/3*pi)?
hmm, nvm, 4/3pi doesnt get me that answer.
I have no idea, I think that might be false.
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12132009, 10:48 PM

#4

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Join Date: Jan 2008
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Re: sec(tan^1(4/3))^2=25/9
It checks correctly.
The inverse tangent of 4/3, denoted as tan^1(4/3), is equal to approximately .9273 radians. This value is a little less than pi/3. Then, when you take the square of the secant of that value, denoted by (sec(.9237))^2, you get 25/9. What that statement says is the following:
The square of the secant whose angle gives a slope of 4/3 is 25/9.
Remember that secant is 1/cos. Make sure to type all of the parenthesis correctly in your calculator.



12132009, 10:56 PM

#5

Fully Optimized
Join Date: Oct 2008
Location: USA
Posts: 2,404

Re: sec(tan^1(4/3))^2=25/9
Quote:
Originally Posted by JogaBonito1502
It checks correctly.
The inverse tangent of 4/3, denoted as tan^1(4/3), is equal to approximately .9273 radians. This value is a little less than pi/3. Then, when you take the square of the secant of that value, denoted by (sec(.9237))^2, you get 25/9. What that statement says is the following:
The square of the secant whose angle gives a slope of 4/3 is 25/9.
Remember that secant is 1/cos. Make sure to type all of the parenthesis correctly in your calculator.

I got the the problem by myself up until the "sec".
What is Secant?
EDIT: Google is my friend.
Hm. Okay so it's hypotenuse over adjacent. Alrighty.
Haven't learned that yet I only have worked with sin, tan, and cos.
and of course also the inverse of all of those.
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12132009, 11:16 PM

#6

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Join Date: May 2006
Posts: 7,903

Re: sec(tan^1(4/3))^2=25/9
Quote:
Originally Posted by JogaBonito1502
It checks correctly.
The inverse tangent of 4/3, denoted as tan^1(4/3), is equal to approximately .9273 radians. This value is a little less than pi/3. Then, when you take the square of the secant of that value, denoted by (sec(.9237))^2, you get 25/9. What that statement says is the following:
The square of the secant whose angle gives a slope of 4/3 is 25/9.
Remember that secant is 1/cos. Make sure to type all of the parenthesis correctly in your calculator.

tan(x)^1 would be 1/tan(x) correct? And 1/tan(4/3) is not .9273....
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12132009, 11:21 PM

#7

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Re: sec(tan^1(4/3))^2=25/9
Quote:
Originally Posted by superman22x
tan(x)^1 would be 1/tan(x) correct? And 1/tan(4/3) is not .9273....

You're mistaken.
1/tan(x) = cot(x); where x is an angle measurement
tan^1(x) = y; where x is an number and y is an angle measurement such that tan (y) = x



12132009, 11:26 PM

#8

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Join Date: May 2006
Posts: 7,903

Re: sec(tan^1(4/3))^2=25/9
Oh, yeah, mixed up, lol.
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12142009, 12:36 AM

#9

Golden Master
Join Date: Apr 2009
Location: USA
Posts: 5,438

Re: sec(tan^1(4/3))^2=25/9
ya, i sure as hell have no idea what your all talking about lol



12142009, 12:52 PM

#10

Daemon Poster
Join Date: Mar 2005
Posts: 1,144

Re: sec(tan^1(4/3))^2=25/9
Banned for Spam!
P.S What am I looking at?
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