Question

Hi, I have a quick question. I am real rusty on trig, how do I find an answer for 4sec(4)? Needs to be exact, so I think pi is a part of the answer. Thanks

Answer 1

4 is not a standard angle and would not be able to be done by hand that readily

Answer 2

sec = 1/cos; but a calculator is not going to give an exact answer either ....

Answer 3

I am trying to follow an example and then apply it to my problem. In this particular example, they say that \[\sec \sqrt{2}\] = \[\pi/4\]

Answer 4

what's the original material?

Answer 5

It is part of a MUCH longer calculus problem. In order to change the bounds on an integral, I have to find \[4\sec (2\sqrt{2})\]

Answer 6

hmm, all I can think of is as amistre64 said, \(\bf sec(\theta) = \cfrac{1}{cos(\theta)}\\ \quad \\
sec(2\sqrt{2}) = \cfrac{1}{cos(2\sqrt{2})}\)

Answer 7

I can't think for what angle cos is \[2\sqrt{2}\]..... Brain no work good

Answer 8

you're right.. the number has to be between -1 and 1
are you .. sure it isn't \(\bf sec^{-1}(2\sqrt{2})\quad ?\)

Answer 9

well. wait. a second... that wouldn't work dohh

Answer 10

This last statement has no sense :).

Answer 11

hehhe

Answer 12

\(\bf sec(2\sqrt{2}) = \cfrac{1}{cos(2\sqrt{2})}\)

that just means the cosine function, with such angle of \(\bf 2\sqrt{2} \) radians I'd assume

Answer 13

May be with complex numbers (I remember something similar to find the aureum number with the roots of a polinomy). But it will be interesting to solve this.

Answer 14

the cosine function, not the arcCosine, and such is there

Answer 15

Yes, this is true.

Answer 16

I'm afraid I will have to close this. Thanks for ya'lls time and effort.

Answer 17

yw

Answer 18

;)