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Old 12-14-2009, 02:00 PM   #11
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Default Re: sec(tan^-1(4/3))^2=25/9

Quote:
Originally Posted by JogaBonito1502 View Post
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.
Yea, I got the right answer in my calc, but I'm wondering if there was a theorem of something I missed that kept it out of decimal form.

Quote:
Originally Posted by AmericanSensei View Post
Banned for Spam!

P.S What am I looking at?
Part of a calculus problem, more specifically, trigonometry.
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