so i have to find the exact value of this:

sec(2pi/3) and answer in the form -A/B.

how do i do this?

so far what i THINK i have to do is 1/cos(2pi/3).

1 / cos(2pi/3) = 1.000668473.

but how do i get it in the "-A/B" form?




your help is much appreciated

11/1/2007 8:30:06 PM

look at the unit circle

11/1/2007 8:32:36 PM

it won't be available to us during the exam. neither will a calculator.

11/1/2007 8:35:12 PM

but if you memorize the first quadrant you can get the other three quadrants from it fairly easily. The numbers in the ordered pairs are the same, only their signs change.

[Edited on November 1, 2007 at 8:47 PM. Reason : jlkjhlihj]

11/1/2007 8:47:35 PM

You should memorize the special triangles and what they mean for trig functions. Otherwise this is a very difficult problem.

cos(pi/4) = 1/sqrt(2) = sqrt(2)/2

sin(pi/4) = 1/sqrt(2) = sqrt(2)/2

sin(2pi/3) = sqrt(3)/2 = cos(pi/3)

sin(pi/3) = 1/2 = cos(2pi/3)

sin(pi/2) = 1 = cos(0)

cos(pi/2) = 0 = sin(0)

also:
pi/3 = 60deg.
pi/4 = 45deg.
2pi/3 = 60deg.
pi/2 = 90deg.

11/1/2007 8:56:32 PM

wow

I haven't done that stuff since like 11th grade, I forgot what a secant was there for a second

yeah and you're gonna need that unit circle, you might as well memorize it now

[Edited on November 1, 2007 at 10:13 PM. Reason : ]

11/1/2007 10:12:00 PM

alright good deal. worked backwards from the solutions manual for like 6 problems and i finally know what i'm doin.


thanks guys

11/2/2007 9:50:00 AM

You should sign up for a free account on cramster.com for questions like these.

11/2/2007 12:16:30 PM