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 $\displaystyle \sin(\arccos(u))=\sqrt{1u^2}$ we have $\displaystyle uw+\sqrt{1u^2}\sqrt{1w^2}=\cos(t)$ so $\displaystyle t = \arccos(uw+\sqrt{1u^2}\sqrt{1w^2})$ Now replace u and w by their original values again... Last edited by skipjack; March 29th, 2017 at 02:10 PM. 

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identity, inverse, trig 
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