# Using the concept of finite difference

1. Aug 2, 2005

### irony of truth

I am trying to find the ▲cos x. By using its definition.

It simply turned out to be cos(x+1) - cos(x). How do I express this in
terms of sine? (and as only one term)?

2. Aug 2, 2005

### HallsofIvy

Staff Emeritus
cos(x+1)= cos(x)cos(1)- sin(x) sin(1) so cos(x+1)- cos(x)= cos(x)(cos(1)-1)- sin(x)sin(1). What makes you think you can write it in terms of sine only?

3. Aug 2, 2005

### lurflurf

That is a completely correct equation for the difference. It is also not a particularly useful form. The standard form for such things in finite calculus is
$${\Delta}^n\cos(x+a)=(2\sin(\frac{h}{2}))^n\cos(x+a+\frac{h}{2}n+\frac{\pi}{2}n)$$
in this specific case where a=0 n=1 h=1
$$\Delta\cos(x)=-2\sin(\frac{1}{2})\sin(x+\frac{1}{2})$$
This is one term additive, but two terms multiplicative.

Last edited: Aug 2, 2005
4. Aug 2, 2005

### Antiphon

$$sin(x+1 - \frac{\pi}{4}) - sin(x-\frac{\pi}{4})$$

5. Aug 2, 2005

### lurflurf

little off
you may have been attempting to use the cofunctional identity
$$\cos(x)=\sin(\frac{\pi}{2}-x)$$
getting
$$\sin(\frac{\pi}{2}-x-1) - \sin(\frac{\pi}{2}-x)$$
or equivalently
$$\sin(x-\frac{pi}{2})-\sin(x+1-\frac{\pi}{2})$$
but this form is not perfered for most purposes that would be
$$-2\sin(\frac{1}{2})\sin(x+\frac{1}{2})$$

Last edited: Aug 2, 2005
6. Aug 3, 2005

### irony of truth

Excuse me, how was $$-2\sin(\frac{1}{2})\sin(x+\frac{1}{2})$$
derived? My derivation just ended with that of HallsofIvy.

7. Aug 3, 2005

### lurflurf

A well known identity in trigonometry is
$$\cos(a)-\cos(b)=-2\sin(\frac{a-b}{2})\sin(\frac{a+b}{2})$$
The result is mode clear by taking a=x+1; b=x
This identity can be derived by writing
a=(a+b)/2+(a-b)/2
b=(a+b)/2-(a-b)/2
Then using the identity
cos(a+b)=cos(a)cos(b)-sin(a)sin(b)
and then adding the like terms

8. Aug 4, 2005

### irony of truth

Thank you very much.