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 Trigonometry Trigonometry Math Forum

 April 18th, 2016, 07:20 PM #1 Senior Member   Joined: Apr 2008 Posts: 194 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 10:51 PM. April 18th, 2016, 07:36 PM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 3,016 Thanks: 1600 Go to the link ... Look at #1, (a) and (c) Sum and Product of sine and cosine April 18th, 2016, 10:20 PM #3 Member   Joined: Mar 2016 From: Nepal Posts: 37 Thanks: 4 (cosA-cosB)/(sinB-sinA) =(-2.sin(A+B/2).sin(A-B/2))/((2cosB+A/2.sinB-A/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. Thanks from Pacioli April 18th, 2016, 11:46 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,966 Thanks: 2216 More parentheses would be needed; it's clearer if formatted neatly. $\displaystyle \frac{\cos A - \cos B}{\sin B - \sin A} = \frac{-2\sin{\small\dfrac{A+B}{2}} \sin{\small\dfrac{A-B}{2}}}{2\cos{\small\dfrac{B+A}{2}} \sin{\small\dfrac{B-A}{2}}} = \frac{\sin{\small\dfrac{A+B}{2}}} {\cos{\small\dfrac{A+B}{2}}} = \tan\frac{A+B}{2}$ Thanks from fysmat, Pacioli and evs Tags horrible, identity, trigonometric 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 Lobinho Trigonometry 7 March 31st, 2015 02:13 PM OriaG Trigonometry 2 August 13th, 2012 03:35 PM OriaG Trigonometry 10 August 12th, 2012 04:37 PM chnixi Algebra 1 August 19th, 2009 03:52 AM TG Algebra 2 March 7th, 2008 03:22 PM

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