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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)?

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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)?

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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 isirony of truth said:

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)?

[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.

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[tex] sin(x+1 - \frac{\pi}{4}) - sin(x-\frac{\pi}{4}) [/tex]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?

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little offAntiphon said:[tex] sin(x+1 - \frac{\pi}{4}) - sin(x-\frac{\pi}{4}) [/tex]

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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derived? My derivation just ended with that of HallsofIvy.

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A well known identity in trigonometry isirony of truth said:

derived? My derivation just ended with that of HallsofIvy.

[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

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Thank you very much.

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