My Math Forum a horrible looking trigonometric identity

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 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 $\frac{\cos A + \cos(120^\circ + B) + \cos(120^\circ - B)}{\sin B + \sin(120^\circ + A) - \sin(120^\circ - A)}= \tan\frac{A + B}{2}$ Using sums and differences, I can simplify the terms with ${120^\circ}$ $\cos(120^\circ + B) + \cos(120^\circ - B)=2\cos(120^\circ)\cos B=-\cos B$ $\sin(120^\circ + A) - \sin(120^\circ - A)=2\cos(120^\circ)\sin A=-\sin A$ Next, putting these results in the expression on the left side gives $\frac{\cos A-\cos B}{\sin B-\sin A}$ 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

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