Question

How do I prove that 2 cot 2x = cot x-tan x?

Thank you! I'd appreciate it!

Answer 1

use the identity:\[\cot(2x)=\frac{\cot^2(x)-1}{2\cot(x)}\]

Answer 2

Where did you get cot\[^{2}\] at?

Answer 3

cot^2

Answer 4

it comes from the re-arrangement of:\[\tan(2x)=\frac{2\tan(x)}{1-\tan^2(x)}\]

Answer 5

you know:\[\cot(x)=\frac{1}{\tan(x)}\]

Answer 6

In fact, just take the reciprocal of the last given identity...

Answer 7

The tan one, the answer drops out...

Answer 8

jennifer473: are you aware of these identities? you need to know them in order to prove your question.

Answer 9

Yes I know the identities. It's just hard for me to really see where to put the identity. I know that I should only work on one side to match the other. But when I tried working through this problem on my exam a few days ago, I did not arrive at the answer. I kept going around in circles.

Answer 10

ok, so if we start from:\[\tan(2x)=\frac{2\tan(x)}{1-\tan^2(x)}\]and invert this identity, we get:\[\frac{1}{\tan(2x)}=\frac{1-\tan^2(x)}{2\tan(x)}\]follow so far?

Answer 11

I am able to do that even though the problem starts with 2cot2x?

Answer 12

yes - just see if you can follow along - it will all become clear soon...

Answer 13

so next, we know:\[\frac{1}{\tan(2x)}=\cot(2x)\]agreed?

Answer 14

Correct

Answer 15

so we substitute this into my last result to get:\[\frac{1}{\tan(2x)}=\frac{1-\tan^2(x)}{2\tan(x)}\]therefore:\[\cot(2x)=\frac{1-\tan^2(x)}{2\tan(x)}\]now multiply both sides by 2 to get:\[2\cot(2x)=\frac{1-\tan^2(x)}{2\tan(x)}\]make sense?

Answer 16

sorry, last line should be:\[2\cot(2x)=\frac{1-\tan^2(x)}{\tan(x)}\]

Answer 17

making sense?

Answer 18

We are working on just the left side of the equation correct?

Answer 19

we started with an identity for tan(2x) and are re-arranging it to get the desired result.

Answer 20

Oh I see. It's making a little sense to me.

Answer 21

ok, so final step is:\[2\cot(2x)=\frac{1-\tan^2(x)}{\tan(x)}=\frac{1}{\tan(x)}-\frac{\tan^2(x)}{\tan(x)}=\cot(x)-\tan(x)\]

Answer 22

I hope my explanation made sense

Answer 23

So for 2cot2x, I bring the 2 from the 2x to the front? What happened to the 2 that was in front of 2cot2x?

Answer 24

sorry - I don't understand your question?

Answer 25

The original problem is to prove that 2cot2x = cotx-tanx.
So what happened to the "2" that is in front of cot2x?

Answer 26

look at the last line of my result - it is there

Answer 27

we started with the identity:\[\tan(2x)=\frac{2\tan(x)}{1-\tan^2(x)}\]which we inverted to get:\[\frac{1}{\tan(2x)}=\frac{1-\tan^2(x)}{2\tan(x)}\]and then used the fact that \(\frac{1}{\tan(2x)}=\cot(2x)\) to get:\[\cot(2x)=\frac{1-\tan^2(x)}{2\tan(x)}\]then multiplied both sides by 2 to get:\[2\cot(2x)=\frac{1-\tan^2(x)}{\tan(x)}=\cot(x)-\tan(x)\]

Answer 28

Oh! I see now! I feel dumb for not noticing that earlier. Lol. Thank you so much. I really appreciate you taking the time out to explain this step by step. It has really helped me understand how I got it wrong on the exam. So thank you!

Answer 29

no problem - we all have to start somewhere - so don't feel as if you are "dumb" - you definitely are not. :)

Answer 30

Thank you :) Have a lovely rest of the day!

Answer 31

I will thanks - you too :)