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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. Tags identity, inverse, trig Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post heinzelmannchen Trigonometry 4 December 9th, 2013 01:44 AM najaa Trigonometry 6 May 14th, 2012 03:30 PM elim Trigonometry 2 September 12th, 2011 11:10 AM jsmith613 Algebra 25 January 6th, 2011 07:52 PM Siria Abstract Algebra 9 November 14th, 2008 05:25 AM

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