Question

Verify the identity.

cot(x-pi/2 = -tan x

I have one explaination...but it doesnt particularly PROVE

Answer 1

I feel like this is too litterary and not enough MATH basis...

The original cofunction identity read:

cot(pi/2-x)=tan x

But by "flipping" or inverting what is in the parenthesis, you have now inverted tan, forming -tan.

This is in conjunction with the Odd/Even Identities.

Answer 2

im looking for a more numbery....kind of way of putting it...or well you understnad

Answer 3

$$\bf
cot\left(x-\cfrac{\pi}{2}\right) \implies
\cfrac{
cos\left(x-\cfrac{\pi}{2}\right)
}{
sin\left(x-\cfrac{\pi}{2}\right)
}\\
\cfrac{
cos(x)cos\left(\cfrac{\pi}{2}\right)+sin(x)sin\left(\cfrac{\pi}{2}\right)
}{
sin(x)cos\left(\cfrac{\pi}{2}\right)-cos(x)sin\left(\cfrac{\pi}{2}\right)
}
$$

Answer 4

now you'd just need to keep in mind
what is the \(\bf sin\left(\cfrac{\pi}{2}\right)\)?
what is the \(\bf cos\left(\cfrac{\pi}{2}\right)\)?

Answer 5

sin pi/2 is cospi/2 and inverted for cos...

Answer 6

i cant use pictures in my explaination @cwrw238 ...

Answer 7

@jdoe0001 ..was i correct?

Answer 8

well, yes
look at your unit circle, sine of pi/2 and cos of pi/2 are pretty conspicuous

Answer 9

haha yes

Answer 10

ok

Answer 11

(0) and 1

Answer 12

well 1 and 0

Answer 13

anything there, multiplying for \(\bf cos\left(\cfrac{\pi}{2}\right)\)
will be multiplied by 0 and 0 times anything = 0

anything there multiplying for \(\bf sin\left(\cfrac{\pi}{2}\right)\)
will be multiplied by 1, and 1 times anything is "anything"

Answer 14

ok..wait a second i need to catch up a little ok?

Answer 15

ok

Answer 16

so you get one over one?...

Answer 17

wait no

Answer 18

sinx/-cosx

Answer 19

OOOH

Answer 20

\(\bf \cfrac{
cos(x)cos\left(\cfrac{\pi}{2}\right)+sin(x)sin\left(\cfrac{\pi}{2}\right)
}{
sin(x)cos\left(\cfrac{\pi}{2}\right)-cos(x)sin\left(\cfrac{\pi}{2}\right)
}\\

\cfrac{
0+sin(x)(1)
}{
0-cos(x)(-1)
}\)

Answer 21

THANK YOUUU

Answer 22

I see now :3

Answer 23

fanks

Answer 24

yw

Answer 25

woops I even put a -1 for ... anyhow \(\bf \cfrac{
0+sin(x)(1)
}{
0-cos(x)(1)
}\)
better

Answer 26

haha I understand :)

Answer 27

just had a typo

Answer 28

thanks a heap. im glad you explainedstep by step!

Answer 29

np