Question

Is tan(pi-2x)=cot(2x)?

Answer 1

Is this worked alright?\[\Large \tan(\pi-2x)=\tan2(\frac{\pi}{2}-x)=\cot2(x)=\cot(2x)\]

Answer 2

Hmm

Answer 3

The definition of \(\text{co}\) in trigonometry, is that it takes any function, for example \(\sin\), and it replaces \(x\) with \(\pi/2-x\).

Answer 4

Yeah, that's what I'm saying so I got it in that form, so does it continue in that way, but you keep the 2?

Answer 5

In this case you have \[
\tan(2(\color{red}{\pi/2-x})) = \cot(2(\color{red}{x}))
\]

Answer 6

Well, does that actually work? Let's think about it.

Answer 7

Ok, thanks, I'm trying to take a derivative of \[\Large {\left(\tan{\left(\pi-2\theta\right)}\right)}^4\]
I know how to do it, I just wanted to clean it up in any way that I could

Answer 8

Let \(y = \pi-2x\). Now we replace \(y\) with \(\pi/2-y\).\[
\pi/2-y = \pi/2-(\pi-2x) = -\pi/2 + 2x = 2x-\pi/2

\]

Answer 9

OK, what are you getting at there?

Answer 10

Which is not \(2x\). So I believe there may be an issue here.

Answer 11

You're correct

Answer 12

\[
\tan(\pi-2x) = \tan(y) = \cot(\pi/2-y) = \cot(\pi/2-(\pi-2x)) = \cot(2x-\pi/2)
\]

Answer 13

It's cot(2x-pi/2) I just plugged in a problem into my calc and 2x-pi/2 works

Answer 14

That's a nice approach. I'll use that from now on

Answer 15

Okay, cool.