Question

verify the identity cot(x-pi/2)=-tan x

Answer 1

one side

Answer 2

oh it's fine, take your time, im just glad ur helping

Answer 3

I think I am going the wrong direction. I haven't done these for a hundred years...

Answer 4

so cot is and odd function
that is
cot(-x)=-cot(x)
so we have that cot(x-pi/2)=-cot(pi/2-x)
i love to draw triangles...

Answer 5

the other angle is pi/2-x

Answer 6

find tan(x) and cot(pi/2-x)

Answer 7

and compare the values

Answer 8

oh ok, thank you for trying

Answer 9

that is my favorite way to pick when doing cofunctions

Answer 10

it's cot(x-pi/2)=-tan x :)

Answer 11

but you could also expand the cot(x-pi/2) using other identities

Answer 12

that is correct @SolomonZelman
but cot(x-pi/2)=-cot(pi/2-x)

Answer 13

since cot is odd

Answer 14

yes

Answer 15

oh i thought you were saying something about what i was saying

Answer 16

how do you get one side to equal the other side?

Answer 17

You can prove the identity by using the fact that cot is odd
and then construct a right triangle if you didn't know that the following is a co-function identity
cot(pi/2-x)=tan(x)

Answer 18

so the answer would be because it's a cofunction identity and it's odd?

Answer 19

cot(x-pi/2)=cot(-(pi/2-x))=-cot(pi/2-x) since cot(x) is an odd function
cot(pi/2-x)=tan(x) by co function identity
-cot(pi/2-x)=-tan(x)

honestly I love to show the cofunction identity but if you have already shown them in class then we don't really need to do it over again

Answer 20

on the last step, the two equations aren't equal to each other

Answer 21

how so

Answer 22

if 5=5
then -5=-5

Answer 23

after -cot(pi/2-x) = -tan x, would i write -tan x = -tan x?

Answer 24

cot(pi/2-x)=tan(x)
so yeah

Answer 25

ok thank you so much :D

Answer 26

you could also go the long way about this
\[\cot(x-\frac{\pi}{2})=\frac{\cos(x-\frac{\pi}{2})}{\sin(x-\frac{\pi}{2})} \\ =\frac{\cos(x)\cos(\frac{\pi}{2})+\sin(x)\sin(\frac{\pi}{2})}{\sin(x)\cos(\frac{\pi}{2})-\sin(\frac{\pi}{2})\cos(x)} \\ =\frac{\cos(x) \cdot 0+\sin(x) \cdot 1}{\sin(x) \cdot 0 -1 \cdot \cos(x) }\]

Answer 27

i'm sure you can finish this way