What Is the Identity for |sinx - siny| and |cosx - cosy| in Trigonometry?

[dglee]

does anybody know the identity for |sinx-siny| and |cosx-cosy|?

 

[StatusX]

What exactly do you want these identities to contain? sin(x+y)'s and cos(x+y)'s? I don't see how you could make these expressions much simpler.

 

[VietDao29]

Are you looking for some Sum-to-product identities?
If yes, then here are the four identities:
\cos \alpha + \cos \beta = 2 \cos \left( \frac{\alpha + \beta}{2} <br /> \right) \cos \left( \frac{\alpha - \beta}{2} \right).
\cos \alpha - \cos \beta = -2 \sin \left( \frac{\alpha + \beta}{2} \right) \sin \left( \frac{\alpha - \beta}{2} \right).
\sin \alpha + \sin \beta = 2 \sin \left( \frac{\alpha + \beta}{2} \right) \cos \left( \frac{\alpha - \beta}{2} \right).
\sin \alpha - \sin \beta = 2 \cos \left( \frac{\alpha + \beta}{2} \right) \sin \left( \frac{\alpha - \beta}{2} \right).
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From the 4 identities above, one can easily show that:
| \sin \alpha - \sin \beta | = 2 \left| \cos \left( \frac{\alpha + \beta}{2} \right) \sin \left( \frac{\alpha - \beta}{2} \right) \right|.
and:
| \cos \alpha - \cos \beta | = \left| -2 \sin \left( \frac{\alpha + \beta}{2} \right) \sin \left( \frac{\alpha - \beta}{2} \right) \right| = 2 \left| \sin \left( \frac{\alpha + \beta}{2} \right) \sin \left( \frac{\alpha - \beta}{2} \right) \right|.
Is that what you are looking for?
And that's not any simpler than your original expressions.

 

[matt grime]

It is simpler, because you can drop with abs value signs using the odd or evenness of sin and cos respectively.

 

[arildno]

As has been stated, it is crucial that you specify what sort of identity you'
re after.

For example, the following identity holds (for all x,y):
|sin(x)-sin(y)|=|sin(x)-sin(y)|+0

 

[Ali 2]

Hi,

This is an inequality ..


| \sin x - \sin y | \leq | x - y |

 

[spacetime]

Maybe this is what you're looking for...

|sinx - siny| = 2 * |{sin(x-y)/2} * {cos(x+y)/2}|

|cosx-cosy| = 2 * |{sin(x+y)/2} * {sin(x-y)/2}|


Spacetime
Physics