My Math Forum inverse trig identity

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 March 29th, 2017, 03:41 AM #1 Newbie   Joined: Mar 2017 From: North Carolina Posts: 8 Thanks: 0 inverse trig identity Is there an identity for the expression: [arccos (x/z) - arccos (y/z)] ? or can the expression be otherwise simplified?
 March 29th, 2017, 04:16 AM #2 Senior Member   Joined: Mar 2012 From: Belgium Posts: 654 Thanks: 11 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... Thanks from rww88 Last edited by skipjack; March 29th, 2017 at 02:10 PM.

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