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April 18th, 2016, 08:20 PM  #1 
Senior Member Joined: Apr 2008 Posts: 193 Thanks: 3  a horrible looking trigonometric identity
I am really stuck in the identity below. Prove Using sums and differences, I can simplify the terms with Next, putting these results in the expression on the left side gives I don't know what to do now. Could someone please help me? Thank you very much. Last edited by skipjack; April 18th, 2016 at 11:51 PM. 
April 18th, 2016, 08:36 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,815 Thanks: 1458  
April 18th, 2016, 11:20 PM  #3 
Member Joined: Mar 2016 From: Nepal Posts: 37 Thanks: 4 
(cosAcosB)/(sinBsinA) =(2.sin(A+B/2).sin(AB/2))/((2cosB+A/2.sinBA/2)) [This is the product form of sin and cos for differences of different angles] upon cancelling terms by adjusting signs you get (sin(A+B/2))/(cosA+B/2)) =tan(A+B/2) Thank me. 
April 19th, 2016, 12:46 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,262 Thanks: 1958 
More parentheses would be needed; it's clearer if formatted neatly. $\displaystyle \frac{\cos A  \cos B}{\sin B  \sin A} = \frac{2\sin{\small\dfrac{A+B}{2}} \sin{\small\dfrac{AB}{2}}}{2\cos{\small\dfrac{B+A}{2}} \sin{\small\dfrac{BA}{2}}} = \frac{\sin{\small\dfrac{A+B}{2}}} {\cos{\small\dfrac{A+B}{2}}} = \tan\frac{A+B}{2}$ 

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