- #1

astronomia84

- 20

- 0

why

2p=ln[(1-cost)/(1+cost)] => t=(1/coshp)^2

?

?

2p=ln[(1-cost)/(1+cost)] => t=(1/coshp)^2

?

?

Last edited by a moderator:

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter astronomia84
- Start date

- #1

astronomia84

- 20

- 0

why

2p=ln[(1-cost)/(1+cost)] => t=(1/coshp)^2

?

?

2p=ln[(1-cost)/(1+cost)] => t=(1/coshp)^2

?

?

Last edited by a moderator:

- #2

theperthvan

- 184

- 0

why not?

- #3

murshid_islam

- 442

- 17

i think you wrote it wrong. it should be

[tex]\sin^{2}t = \left(\frac{1}{\cosh p}\right)^2[/tex]

[tex]\sin^{2}t = \left(\frac{1}{\cosh p}\right)^2[/tex]

- #4

cristo

Staff Emeritus

Science Advisor

- 8,140

- 74

why

2p=ln[(1-cost)/(1+cost)] => t=(1/coshp)^2

?

?

What is the actual question you are asking here? Are you asking why, or if, the left implies the right. Have you done anything to attempt to show this? I doubt anyone will help unless you firstly specify your question, and secondly show some effort!

Share:

- Replies
- 1

- Views
- 206

- Replies
- 1

- Views
- 266

- Replies
- 1

- Views
- 130

- Replies
- 5

- Views
- 460

- Replies
- 4

- Views
- 310

- Last Post

- Replies
- 6

- Views
- 593

- Replies
- 3

- Views
- 239

- Replies
- 1

- Views
- 34

- Last Post

- Replies
- 1

- Views
- 185

- Replies
- 2

- Views
- 267