Question

arccot(0)

Answer 1

the number whose cotangent is zero
since \[\cot(x)=\frac{\cos(x)}{\sin(x)}\] it is also the number whose cosine is zero

Answer 2

pi/2

Answer 3

that will do it

Answer 4

so arccot(-1)

Answer 5

the number whose cosine is -1?

Answer 6

no, it is the number whose COTANGENT is \(-1\)

Answer 7

the reason it worked for the first answer is because a fraction is only zero if the numerator is zero
so since
\[\cot(x)=\frac{\cos(x)}{\sin(x)}\] it is only zero if \(\cos(x)=0\)

Answer 8

but for that one you need to know where
\[\frac{\cos(x)}{\sin(x)}=-1\]

Answer 9

it is where cosine and sine are the same number, but of opposite sign

Answer 10

3pi/4

Answer 11

7pi/4

Answer 12

careful here

Answer 13

it is \(-1\) and an infinite number of places, you have to find the one in the right range

Answer 14

eek what is the range?

Answer 15

ok you have a function that is periodic, so it is certainly not one to one
in order to make the inverse a function, you have to restrict the range of the inverse so that everyone gets the same answer
notice that you provided two answers, but as a function it can only have one output per input

Answer 16

i believe traditionally the range of arccotangent is \([0,\pi]\)

Answer 17

so your first answer of \(\frac{3\pi}{4}\) is the right answer, the second answer is not right, because it is not in the range or arccotangent

Answer 18

okay yes.. I remember that.

Answer 19

so that claculators can work :-)

Answer 20

you should help me:)))

Answer 21

lol
also so you have a well defined function

Answer 22

so, how would I approach arccot(sqrt3)

Answer 23

it is where you have
\[\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}\] for
\[\frac{\cos(x)}{\sin(x)}\]

Answer 24

pi/6

Answer 25

i.e. where
\[\cos(x)=\frac{\sqrt{3}}{2}\] and
\[\sin(x)=\frac{1}{2}\] a very familiar pair

Answer 26

yes

Answer 27

you make it seem so easy!

Answer 28

last one..

Answer 29

arctan(tan(5pi/4))

Answer 30

then I have about 50 more to do on my own :-\