Question

without using a calculator find the exact value of cos^-1 (cos 17pi/5)

Answer 1

hmmm, what does \(\bf cos^{-1}(whatever)\) means?

Answer 2

the inverse

Answer 3

Yes, now think about that. The inverse cosine gives the ANGLE that has, as its cosine value, the INPUT of the inverse function, right?

Answer 4

the inverse? well, ahemm...
\(\bf cos^{-1}(whatever) \) means, what's the angle WHOSE COSINE is "whatever"

Answer 5

so, what would be \(\bf cos(whoever) \)?

Answer 6

or I should say what would be \(\bf cos(someangle)\)

Answer 7

the cosine function is just a ratio, a number, a value
you give it an angle, it gives you the ratio value, some number

\(cos^{-1}\) takes THAT NUMBER and spits out the angle it came from

Answer 8

17pi/5, but aren't there restrictions on the inverse of cosine like it only is in the first two quadrants from 0 to pi? so then wouldn't 17pi/5 not work?

Answer 9

yes, when you have the actual value, yes the range restriction will show up
when you know the source, then you don't have to do anything with the restrictions

Answer 10

Yes, now you have the find the function that is IN the output of the inverse cosine function, that wold have the SAME cosine as 17pi/5

Answer 11

\(\bf cos(2\pi)=1, cos(50\pi)=1\\
cos^{-1}(1) = 0\\
\textit{but 0 is a coterminal with }50\pi\\
cos^{-1}(cos(50\pi)) = 50\pi\)

Answer 12

17pi/5 - 2pi=7pi/5 so is in the 3rd quadrant, so not in the range of inverse cosine (which is limited to 1st and 2nd quadrant). But we know that it has reference angle of 7pi/5-pi=7pi/5-5pi/5=2pi/5. Since in Q3 it has negative cosine. So the angle between 0 and pi that has the same cosine is.....?

Answer 13

Sorry but I will disagree with @jdoe0001 on this point. The output range for the function of 0 to 2pi, 50pi is not in that range, so can't be the value of the inverse cosine. I have given my students problems designed exactly like this to test their understanding of this. cos(50pi) is a NUMBER that is in the domain of inverse cosine, but it gives the result =0, not =50pi, because 50pi is not in the range of inverse cosine.

Answer 14

@DebbieG ok i understand how to get the 7pi/5, but then how do you know to subtract pi?

Answer 15

The angle 7pi/5 is in quadrant 3, right?
Do you know what I mean by "reference angle"? (Different texts may have different terminology.) That is the acute, positive angle between the terminal side of any (non-quadrantal) angle and the x-axis. So, there is one angle in each quadrant that has a given (acute) reference angle, and each of those 4 angles has the same value for cosine, EXCEPT POSSIBLY for the sign of it.

E.g., COS(pi/3)=1/2, COS(2pi/3)=-1/2, COS(4pi/3)=-1/2, COS(5pi/3)=1/2. All have the same reference angle.

So now think about 7pi/5. where is it? it is 2pi/5 "past" pi (7pi/5-pi=2pi/5). So that's the reference angle.

So COS(7pi/5)=-COS(2pi/5), and COS(7pi/5)=COS(3pi/5).... do you see why? (Think about which quadrant each is in, and what the reference angle is.)

Answer 16

@DebbieG oh ok i was confused i thought the answer you were saying was 2pi/5, but that's the reference angle and the actual answer is 3pi/5 thank you so much for your help!!!

Answer 17

yes indeed, sorry for the confusion.... :)