inverse trig identity

Mar 2017
8
0
North Carolina
Is there an identity for the expression: [arccos (x/z) - arccos (y/z)] ? or can the expression be otherwise simplified?
 
Mar 2012
654
11
Belgium
Let's write \(\displaystyle x/z=u\) and \(\displaystyle y/z=w\) for simplicity.

Suppose \(\displaystyle \arccos(u)-\arccos(w)=t\)
then \(\displaystyle \cos(\arccos(u)-\arccos(w))=\cos(t)\)
and thus by sum formula for cosinus we have \(\displaystyle uw+\sin(\arccos(u))\sin(\arccos(w))=\cos(t)\)
and now because \(\displaystyle \sin(\arccos(u))=\sqrt{1-u^2}\)
we have \(\displaystyle uw+\sqrt{1-u^2}\sqrt{1-w^2}=\cos(t)\)
so \(\displaystyle t = \arccos(uw+\sqrt{1-u^2}\sqrt{1-w^2})\)

Now replace u and w by their original values again...
 
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