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March 29th, 2017, 03:41 AM   #1
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inverse trig identity

Is there an identity for the expression: [arccos (x/z) - arccos (y/z)] ? or can the expression be otherwise simplified?
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March 29th, 2017, 04:16 AM   #2
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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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Last edited by skipjack; March 29th, 2017 at 02:10 PM.
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