Question

p

Answer 1

do you know what \(\bf \large cos^{-1}\) stand for?

Answer 2

sec?

Answer 3

well, is easy to mixed up a reciprocal or inverse with the ARC functions... no quite, \(\bf ^{-1}\) in trig stands for, "WHAT IS THE ANGLE WHOSE VALUE IS THIS"

Answer 4

say for example... what's the cosine of \(\bf \cfrac{\pi}{3}\) ?

Answer 5

1/2

Answer 6

\(\bf cos\left(\frac{\pi}{3}\right) = \cfrac{1}{2}\qquad \qquad cos^{-1}\left(\frac{1}{2}\right) = \cfrac{\pi}{3}\)

Answer 7

once you get the value for the trig function, the ARC function with the \(\bf ^{-1}\), will give you back the angle that has that value

Answer 8

So tan 3pi/4= -1.. yeah? So then...I...?

Answer 9

ahemm, well \(\bf tan\left(\cfrac{3\pi}{4}\right) = \textit{some value}
\\
tan^{-1}(\textit{some value}) = \cfrac{3\pi}{4}\quad \\
\textit{some value} = tan\left(\cfrac{3\pi}{4}\right) \implies tan^{-1}(\textit{some value}) = tan\left(\cfrac{3\pi}{4}\right)\\
\quad \\\quad \\
tan^{-1}\left(tan\left(\cfrac{3\pi}{4}\right)\right) = \cfrac{3\pi}{4}\)

Answer 10

hmm, missed something, but anyhow,

Answer 11

\(\bf tan\left(\cfrac{3\pi}{4}\right) = \textit{some value}
\\
tan^{-1}(\textit{some value}) = \cfrac{3\pi}{4}\quad \\
\textit{some value} = tan\left(\cfrac{3\pi}{4}\right) \implies tan^{-1}(\textit{some value}) = tan^{-1}\left(tan\left(\cfrac{3\pi}{4}\right)\right)\\
\quad \\\quad \\
tan^{-1}\left(tan\left(\cfrac{3\pi}{4}\right)\right) = \cfrac{3\pi}{4}\)

Answer 12

Wait so it simplifies out to 3pi/4?

Answer 13

yes => \(\bf \large{ cos^{-1}(\quad cos(\theta)\quad ) = \theta\\
sin^{-1}(\quad sin(\theta)\quad ) = \theta\\
tan^{-1}(\quad tan(\theta)\quad ) = \theta} \)

Answer 14

So essentially there is no work at all to do/show, its just the inverse?

Answer 15

yeap, pretty much

Answer 16

so if the problem is cos^-1[(-1)/2], the answer would be (1/2)?

Answer 17

did you mean \(\bf cos^{-1}\left(cos\left(-\frac{1}{2}\right)\right) \quad ?\)

Answer 18

No the numerator is only negative

Answer 19

... ok... then \(\bf cos^{-1}\left(\frac{-1}{2}\right)\)

Answer 20

yes

Answer 21

it's a bit different... one sec

Answer 22

And isn't 3pi/4 out of range so shouldnt it be -pi/4?

Answer 23

\(\bf cos^{-1}\left(\frac{-1}{2}\right)\\
\frac{-1}{2} \textit{is the cosine value obtained from an angle, say }\theta\\
\textit{so we can say that } cos(\theta) = \frac{-1}{2}\)

if you look at your Unit Circle, it'll be the angles, whose cosine is \(\bf \cfrac{-1}{2}\)

Answer 24

I understand that, but on the previous question isnt it -pi/4 answer?

Answer 25

hmm, no, why not, well, the inverse functions have a range from where they return their angle value
for the \(\bf cos^{-1}\) the range is \(\bf 0 \ to \ \pi\)

Answer 26

-pi/4 will be in the 4th quadrant, out of range from it

Answer 27

if its tan^-1(tan) why use the cos restrictions?

Answer 28

isnt tan^-1 range -pi/2 to pi/2?

Answer 29

so 3pi/4 -> -pi/4?

Answer 30

yes, you're correct, those are the restrictions for inverse tangent
what's the difference?

with -1/2 we don't have an angle we can look at, as source what angle it came from, it could have been from who know what revolution or cycle
thus the inverse cosine restrictions apply

when you have the case of \(\bf cos^{-1}(cos(\textit{some angle here}))\) we have a source to see for the angle, thus we can safely return it, without being restricted

Answer 31

So for the problem tan^-1[tan(3pi/4)] the restrictions make the final answer neg pi/4 because of that

Answer 32

if you apply the restrictions, yes

Answer 33

so how do i dtermine the angle based off the unit circle for the cos^-1(-1/2)? I just use 1/2?

Answer 34

yes, and since we dunno the source right off, the range restrictions will apply when you get the angle from the inverse function

usually Unit Circles have the cos/sin pair for the angles

Answer 35

so thats pi/3?

Answer 36

hmmm thought you said ...-1/2 that'd be \(\bf \cfrac{2\pi}{3}\)

Answer 37

making the answer sq rt 3/2?

Answer 38

or 120 degrees?

Answer 39

\(\bf sin\left(\cfrac{2\pi}{3}\right) = \cfrac{\sqrt{3}}{2}\)

Answer 40

cosine of \(\bf \cfrac{2\pi}{3}\) is -1/2

Answer 41

So cos^-1[(-1)/2] = 2pi/3 or 120 degrees?

Answer 42

yeap \(\bf cos^{-1}\left(\cfrac{-1}{2}\right) = \cfrac{2\pi}{3} = 120^o\)

Answer 43

Wow thanks. A lot. are you any good with identities?

Answer 44

depends, you may want to post anew :)

Answer 45

k