Question

How do you prove that the wavelength of \(|\sin x|\) is \(\pi\)?

Answer 1

@experimentX @ChristopherToni @uri ?

Answer 2

|sin(x+L)| - |sin(x)| = 0

Answer 3

find the value of L

Answer 4

\(n\pi\)?

Answer 5

yes ... that it the minimum value of n?

Answer 6

practically it's \(-\infty\) but I bet you want me to answer \(0\)

Answer 7

one has to be different from another, n=1

Answer 8

wait sorry negative answers are invalid so yes it's 0

Answer 9

But what does it have to do with the wavelength

Answer 10

wait it's symmetrical so negative answers work too...

Answer 11

anyway what does it have to do with the wavelength

Answer 12

http://www.wolframalpha.com/input/?i=plot+%7Csin%28x%29%7C

Answer 13

But how do you prove that...

Answer 14

f(x+mutiple of wavelength) = f(x)

Answer 15

that is the definition of wavelength

Answer 16

I mean how do you prove that \(|\sin(x+n\pi)|=|\sin x|\)

Answer 17

sin(x+npi) = -sin(x) is n is odd and
= sin(x) if n is even

Answer 18

How do you prove this?

Answer 19

use sin(A+B) = ...

Answer 20

oh i see thanks so much

Answer 21

yw

Answer 22

@Kainui lol why medal me

Answer 23

You ask good questions.

Answer 24

lol thx