Question

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

Answer 1

\[\Large \cot\left(x - \frac{\pi}{2}\right)=\frac{\cos\left(x - \frac{\pi}{2}\right)}{\sin\left(x - \frac{\pi}{2}\right)}\]

Answer 2

@Jdina97 make your presence felt :3

Answer 3

im here

Answer 4

How do you break down the inside. What's in the parentheses

Answer 5

Okay, so you know that \[\Large \cos(x) = \cos(-x) \]right? :)
also, \[\Large \sin(x) =-\sin(-x)\]

These will make things easier :)

Answer 6

yup

Answer 7

So the numerator can be written as:
\[\Large \cot\left(x - \frac{\pi}{2}\right)=\frac{\cos\left(\frac{\pi}2 - x\right)}{-\sin\left(\frac{\pi}2 - x\right)}\]

Answer 8

but doesn't the Sin(x) =-sin(x) not -sin(-x)?

Answer 9

Oh no :)
sin(-x) = -sin(x)

So it only follows that
sin(x) = -sin(-x)

Answer 10

You want me to prove it? :3

Answer 11

Oh okay my bad i saw it wrong

Answer 12

all righty then

Answer 13

I misunderstood what my Formula sheet said:P

Answer 14

Okay, so another thing that's useful is the fact that
\[\Large \sin\left(\frac{\pi}{2} - x\right)= \cos(x) \]\[\Large \cos\left(\frac{\pi}{2} - x\right) = \sin(x)\]

Answer 15

Hmm. How did you get that?

Answer 16

Thank for answering by the way

Answer 17

:)
\[\Large \sin(\alpha-\beta) = \sin(\alpha)\cos(\beta) - \cos(\alpha)\sin(\beta) \]

Answer 18

Now if we let
\[\Large \alpha = \frac\pi 2\]\[\Large \beta = x\]

Answer 19

Oh okay

Answer 20

We get
\[\Large \sin\left(\frac{\pi}2\right)\cos(x) -\cos\left(\frac{\pi}2\right)\sin(x)\]

Answer 21

\[\Large \sin\left(\frac{\pi}2\right) = 1\]\[\Large \cos\left(\frac{\pi}2\right)=0\]
Therefore...
\[\Large = \cos(x) \]

Understood ? :)

Answer 22

Sort of

Answer 23

Well, that's not good enough :)
What part don't you get? :D

Answer 24

I understood how you got sin(π2)cos(x)−cos(π2)sin(x)

Answer 25

Oh wait. After that you go to the unit circle correct?

Answer 26

Yes, if only to find out that
\[\large \sin\left(\frac{\pi}2\right)= 1\]and\[\large \cos\left(\frac{\pi}2\right)=0\]

Answer 27

Okay yea that makes sense now.

Answer 28

so, similarly, we get
\[\large \cos\left(\frac{\pi}2-x\right)=\cos\left(\frac{\pi}2\right)\cos(x)+\sin\left(\frac{\pi}2\right)\sin(x)=\sin(x)\]

Answer 29

So, no more doubts that
\[\sin\left(\frac{\pi}2-x\right)=\cos(x)\]and\[\cos\left(\frac{\pi}2-x\right)=\sin(x)\]

Answer 30

Yup

Answer 31

Okay, so back to this
\[\Large \cot\left(x - \frac{\pi}{2}\right)=\frac{\cos\left(\frac{\pi}2 - x\right)}{-\sin\left(\frac{\pi}2 - x\right)}\]

Answer 32

And then Sin/Cos = Tan and Since Sin is Negative it equals -tanx correct?

Answer 33

Very observant :)

And your observation is correct ^_^

Answer 34

Thank you so much for your help :D

Answer 35

\[\Large \cot\left(x - \frac{\pi}{2}\right)=\frac{\cos\left(\frac{\pi}2 - x\right)}{-\sin\left(\frac{\pi}2 - x\right)}=\frac{\sin(x)}{-\cos(x)}=-\tan(x)\]

Answer 36

Now do you mind helping someone else :P i started helping them but i realized i was saying the wrong thing. They needed help with a "train station problem" from geometry but i forget how to do it

Answer 37

I'll see... link/tag me :)