Question

Write an algebraic expression equivalent to cot(arctanx). WILL MEDAL AND FAN

Answer 1

draw a right triangle

Answer 2

ok

Answer 3

there is a picture of a triangle with an angle labelled, the tangent of that angle is \(x\)

Answer 4

i was just drawing it lol

Answer 5

opposite over adjacent

Answer 6

right

Answer 7

and the cotangent of that angle should be pretty obvious too

Answer 8

so im trying to find cotangent

Answer 9

so 1/x?

Answer 10

yup

Answer 11

all righty. what about sin(arctanx)?

Answer 12

same thing

Answer 13

except this time you need the hypotenuse of the triangle, which you get (in your head) via pythagors

Answer 14

pythagoras
something like that

Answer 15

ok

Answer 16

then take "opposite over hypotenuse"

Answer 17

so i get x^2+1^2=c^2

Answer 18

yikes

Answer 19

you are solving for the hypotenuse correct?

Answer 20

yeah

Answer 21

i mean you are not wrong, but you should have it instantly
one side is 1, one side is x, the hypotenuse is \(\sqrt{1+x^2}\)

Answer 22

which is what you get when you solve \[x^2+1=c^2\] for \(c\) but you should really just know it

Answer 23

oh my bad.

Answer 24

so xsqrt1+x^2/x^2+1??

Answer 25

my teacher does not like having square roots as the denominator so i put it as the numerator

Answer 26

ok then pick that one
stupid, but ok

Answer 27

all right so sin(arccos(x+2). i had trouble with this.

Answer 28

\[x+2^2+B=1\]

Answer 29

how do i solve that?? @satellite73

Answer 30

i meant \[x+2^2+B^2=1^2\]

Answer 31

\[b=\sqrt{1-(x+2)^2}\] by pythagoras

Answer 32

you can clean that up a bit if you like or just leave it

Answer 33

if it were to be cleaned up, would i just simplify the (x+2)^2??

Answer 34

no

Answer 35

you would have to expand \((x+2)^2\) and subtract it from \(1\) and leave it inside the radical

Answer 36

so (x+2)(x+2)-1 all inside the radical?

Answer 37

i meant 1-(x+2)(x+2)

Answer 38

yeah or \[\sqrt{1-(4+4x+x^2)}\] or whatever you like

Answer 39

ok cool. u think you can give me a mini lesson on inverse functions?

Answer 40

what do you need to know?

Answer 41

like problems such as arctan(cos1/2). Problems similar to these confuse me.

Answer 42

most of it is probably right in the book
you really need to understand what the range of the inverse functions are (that is what makes them functions)
i do those by drawing a triangle,labelling the sides appropriately and then finding the missing side via pythagoras
there are other ways, but that is the simplest
easiest to answer if you ask a specific question

Answer 43

EXACTLY. The ranges for the problems confuse me.

Answer 44

memorize them

Answer 45

What are the ranges though.

Answer 46

for arcsine and arctangent it is \[[-\frac{\pi}{2},\frac{\pi}{2}]\]

Answer 47

I will write that down.

Answer 48

for arccosine it is \[[0,\pi]\]

Answer 49

anything else?

Answer 50

hmm that is the major stuff
it will be more complicated when you have to compute something like \[\sin(\tan^{-1}(x)+\cos^{-1}(x))\]

Answer 51

so what about arcsec or arccosecant or maybe arccotangent. what are the ranges of those?

Answer 52

yeah im still here lol