Hi!

Itīs been some time since I last did trigonometric equations.

I dont remember if I should know this or not, but anyway, right now I dont.

f(x)= sin 2x - sin x/2 is periodic with what period?

Just point me in a direction, thanks!

Results 1 to 8 of 8

- Apr 8th 2008, 01:06 PM #1

- Apr 8th 2008, 01:12 PM #2
Hello,

We know that sinus is 2pi-periodic.

sin(X+2pi)=sin(X)

sin(2x) -> find t such as sin(2(x+t))=sin(2x)

sin(2(x+t))=sin(2x+2t)=sin(2x)

This means that 2t=2pi -> t=pi

Do the same for sin(x/2) and finding t' such as sin((x+t')/2)=sin(x/2)

The period of the function will be gcd(t,t')

- Apr 8th 2008, 01:16 PM #3

- Apr 8th 2008, 01:17 PM #4

- Apr 8th 2008, 01:23 PM #5

- Apr 8th 2008, 02:04 PM #6
No, you can't...

so sin 2x -> sin 2(x + n*2pi)

The periodicity of a function is defined as : h(x+t)=h(x)

Suppose that f(x)=sin(x) and g(x)=2x

Here, sin(2x)=f(g(x))=h(x)

So you have to find t such as h(x+t)=h(x)

If you replace, it makes : $\displaystyle \underbrace{f(g(x+t))}_{\sin(2(x+t))}=\underbrace{ f(g(x))}_{\sin(2x)}$

So you have to find t as i told you above : such as $\displaystyle \sin(2x+2t)=\sin(2x)$

sin(2x+2pi)=sin(2x)

So t=pi works

- Apr 8th 2008, 02:08 PM #7

- Apr 8th 2008, 02:25 PM #8