Question

What is arccos(-1/6)?
Please explain how to solve it without a calculator.
Thanks!

Answer 1

I don't know. But what if you graphed it and figured it out from that?

Answer 2

Yes, I understand how to solve it using graphing or a calculator.
The full problem is: tan[arccos(-1/6)] which I had to solve by hand.
Like, for: arcsin[sin(5pi/4)] I can easily just flip it around because it contains a pi. This one doesn't though, so I'm a bit confused.

Answer 3

You know more about it than I do--right now

I did find this:

Since none of the six trigonometric functions are one-to-one, they are restricted in order to have inverse functions. Therefore the ranges of the inverse functions are proper subsets of the domains of the original functions

So, that gives me a hint.

Answer 4

Just asking: doesn't that mean you have to turn one-sixth into radians?

Answer 5

Cosine 45° = square root of 2

The arc cosine of (1/6) = 80.4059317731 degrees = 1.4033482476 radians

Answer 6

Well, for arcsin[sin(5pi/4)].

sin(5pi/4) = \[-\sqrt{2}/2\]

Then, arcsin\[-\sqrt{2}/2\]

would be \[-\sqrt{\pi}/4\]

Answer 7

I'm looking for the exact values. P:. Oops, forgot to mention that.

Answer 8

I'd say it doesn't have to bee converted to radians, since it is a ratio of two sides and not an angle.

Answer 9

Good point

Answer 10

BUT, if it has to be an "exact value" she does have to use the square root in this case, right?

Answer 11

Yes, it can have a square root.

Answer 12

I think it has to have a square root. So, wouldn't you just graph it with the square root location, just as you would a specific number. Just curious.

Answer 13

Well the arc tangent of sqrt(3) = 60° (2 exact values)

Answer 14

Ah, I see. Thanks.

Answer 15

This was just a really odd question, but I think I've got it now. Thank you guys. :).

Answer 16

:)

Answer 17

okay :-)