Question

Can someone check my answers?
14. cos^-1(1)=0
18. tan^-1(-1)=7pi/4
Will fan and medal

Answer 1

The first one is correct.

The 2nd one is also correct, but I'm also sure that 135 degrees is also an answer (2nd quadrant, tan,cot, is also negative)

Answer 2

well usually y=tan(x) is restricted on (-pi/2,pi/2) so that y=arctan(x) can exist as a function.
arctan(x) should therefore have a unique output in the interval (-pi/2,pi/2)

Answer 3

7pi/4 is not in that interval

Answer 4

So then what would the answer be for 2?

Answer 5

Would it be -pi/4 then? @freckles

Answer 6

yeah

Answer 7

yes
arccos(1)=0
and
arctan(-1)=-pi/4

Answer 8

22. sin^-1 \[-\sqrt{3}/2\]
is the answer -pi/3

Answer 9

sounds great

Answer 10

cos^-1(cos4pi/5) can you help me with this one?

Answer 11

arccos( ) should give you in output in the interval [0,pi]
and 4pi/5 is in that interval
so your work is basically cut out for you

if x is in the interval [0,pi] then arccos(cos(x))=x

Answer 12

cos^-1 [cos(-5pi/3)], is it pi/3? @freckles

Answer 13

\[\cos(\frac{-5\pi}{3})=\cos(\frac{5\pi}{3}) \text{ since } \cos(x) \text{ is an even function } \\ \text{ that is } \cos(x)=\cos(-x) \\ \cos(\frac{-5\pi}{3})=\cos(\frac{5\pi}{3})\]
Well we know 5pi/3 is in the 4th quadrant
And its reference angle is pi/3 as you said since 2pi-5pi/3=pi/3
Now
in both the 4th and 1st quadrant cos(x) is positive
so we know the following:
\[\cos(\frac{-5\pi}{3})=\cos(\frac{5\pi}{3})=\cos(\frac{\pi}{3}) \\ \text{ so yeah } \arccos(\cos(\frac{-5\pi}{3}))=\arccos(\cos(\frac{5\pi}{3}))=\arccos(\cos(\frac{\pi}{3}))=\frac{\pi}{3}\]

Answer 14

Sorry for asking so many questions
cos[cos^-1(-2/3)] It wouldn't be -2/3 would it because it isnt in the range of (0, pi), so is it just 2/3 @freckles

Answer 15

cos(x) should out put numbers between -1 and 1
and yes -2/3 is between -1 and 1
so you cool

Answer 16

i mean cos(arccos(-2/3)) is -2/3 since -2/3 is between -1 and 1

Answer 17

sorry I missed the n't part of your would

Answer 18

\[\cos(\arccos(x))=x \text{ iff } x \text{ is between } -1 \text{ and }1 \text{ (inclusive)}\]

Answer 19

Okay, last one maybe, sin[sin^-1(-2)] sin goes from -pi/2 to pi/2 and I don't it's x,y coordinates so I don't know where -2 would fit in @freckles

Answer 20

well just like cos(x), sin(x) will have outputs between -1 and 1
-2 is not in that interval
so guess what?

Answer 21

what? its not defined?

Answer 22

yep to that question

you could also look at it this way:
arcsin(-2) is not defined
the number inside the arcsin( ) has to be between -1 and 1

Answer 23

https://math.la.asu.edu/~surgent/mat170/invtrig.pdf
from what I can tell this is a good site for some notes

Answer 24

does tan also have -1,1?

Answer 25

@freckles

Answer 26

y=tan(x) is restricted on (-pi/2,pi/2) so that the inverse function we call
y=arctan(x) can exist

So looking at the above restriction:
\[y=\tan(x) \text{ with domain } (-\frac{\pi}{2},\frac{\pi}{2}) \text{ and range} (-\infty,\infty) \\ y=\arctan(x) \text{ has domain } (-\infty,\infty) \text{ and range} (-\frac{\pi}{2},\frac{\pi}{2})\]
which means:
\[\tan(x) \text{ can be any real number } \\ \arctan(x) \text{ can be any real number between } \frac{-\pi}{2} \text{ and } \frac{\pi}{2} \text{ (exclusive })\]
So
\[\tan(\arctan(x))=x \text{ for all } x \]

examples with tan and arctan

\[\tan(\arctan(-4))=-4 \\ \arctan(\tan(\frac{3\pi}{4}))=\arctan(\tan(\frac{-\pi}{4}))=\frac{-\pi}{4}\]

Answer 27

@minisweet4 I'm going to leave here in a second for awhile.
Good luck with your problems.

Answer 28

Thanks