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: 653 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. 

Tags 
identity, inverse, trig 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Trig Identity  heinzelmannchen  Trigonometry  4  December 9th, 2013 02:44 AM 
trig identity  najaa  Trigonometry  6  May 14th, 2012 03:30 PM 
A trig identity  elim  Trigonometry  2  September 12th, 2011 11:10 AM 
Trig identity  jsmith613  Algebra  25  January 6th, 2011 08:52 PM 
Same identity and inverse element in a field  Siria  Abstract Algebra  9  November 14th, 2008 06:25 AM 