Using the concept of finite difference

  • #1
irony of truth
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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)?
 

Answers and Replies

  • #2
HallsofIvy
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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
lurflurf
Homework Helper
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irony of truth said:
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)?
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
[tex]{\Delta}^n\cos(x+a)=(2\sin(\frac{h}{2}))^n\cos(x+a+\frac{h}{2}n+\frac{\pi}{2}n)[/tex]
in this specific case where a=0 n=1 h=1
[tex]\Delta\cos(x)=-2\sin(\frac{1}{2})\sin(x+\frac{1}{2})[/tex]
This is one term additive, but two terms multiplicative.
 
Last edited:
  • #4
Antiphon
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irony of truth said:
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?

[tex] sin(x+1 - \frac{\pi}{4}) - sin(x-\frac{\pi}{4}) [/tex]
 
  • #5
lurflurf
Homework Helper
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Antiphon said:
[tex] sin(x+1 - \frac{\pi}{4}) - sin(x-\frac{\pi}{4}) [/tex]
little off
you may have been attempting to use the cofunctional identity
[tex]\cos(x)=\sin(\frac{\pi}{2}-x)[/tex]
getting
[tex]\sin(\frac{\pi}{2}-x-1) - \sin(\frac{\pi}{2}-x)[/tex]
or equivalently
[tex]\sin(x-\frac{pi}{2})-\sin(x+1-\frac{\pi}{2})[/tex]
but this form is not perfered for most purposes that would be
[tex]-2\sin(\frac{1}{2})\sin(x+\frac{1}{2})[/tex]
 
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  • #6
irony of truth
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Excuse me, how was [tex]-2\sin(\frac{1}{2})\sin(x+\frac{1}{2})[/tex]
derived? My derivation just ended with that of HallsofIvy.
 
  • #7
lurflurf
Homework Helper
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irony of truth said:
Excuse me, how was [tex]-2\sin(\frac{1}{2})\sin(x+\frac{1}{2})[/tex]
derived? My derivation just ended with that of HallsofIvy.
A well known identity in trigonometry is
[tex]\cos(a)-\cos(b)=-2\sin(\frac{a-b}{2})\sin(\frac{a+b}{2})[/tex]
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
irony of truth
90
0
Thank you very much.
 

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