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 August 12th, 2012, 10:34 AM #1 Senior Member   Joined: Jul 2012 Posts: 225 Thanks: 0 Another trigonometric identity Can anyone help me prove this http://s137.photobucket.com/albums/q225 ... trigo4.jpg please?
 August 12th, 2012, 10:39 AM #2 Senior Member   Joined: Jul 2012 Posts: 225 Thanks: 0 Re: Another trigonometric identity I should mention - you're not allowed to use the tanx+tany identity. You must solve by representing tanx as sinx/cosx.
 August 12th, 2012, 10:52 AM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,204 Thanks: 511 Math Focus: Calculus/ODEs Re: Another trigonometric identity $\tan$$\alpha+15^{\circ}$$+\tan$$\alpha+75^{\circ}\ )=\frac{\sin\(\alpha+15^{\circ}$$\cos$$\alpha+75^{ \circ}$$+\sin$$\alpha+75^{\circ}$$\cos$$\alpha+15^ {\circ}$$}{\cos$$\alpha+15^{\circ}$$\cos$$\alpha+7 5^{\circ}$$}=$ $\frac{\sin$$2\alpha+90^{\circ}$$}{\frac{\cos$$60^{ \circ}$$+\cos$$2\alpha+90^{\circ}$$}{2}}=\frac{2\c os$$2\alpha$$}{\frac{1}{2}-\sin$$2\alpha$$}=$ $\frac{4\cos$$2\alpha$$}{1-2\sin$$2\alpha$$}$
 August 12th, 2012, 12:27 PM #4 Senior Member   Joined: Jul 2012 Posts: 225 Thanks: 0 Re: Another trigonometric identity I so fail at this subject, donno what I'm gonna do. http://s137.photobucket.com/albums/q225 ... trigo5.jpg Can't solve this now. Can't solve 2 problems in a row.
 August 12th, 2012, 12:44 PM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,204 Thanks: 511 Math Focus: Calculus/ODEs Re: Another trigonometric identity Try using a double-angle identity for cosine on $\cos$$4\alpha$$$ then factor the resulting quadratics. Then use double-angle identities for cosine again.
 August 12th, 2012, 12:45 PM #6 Senior Member   Joined: Jul 2012 Posts: 225 Thanks: 0 Re: Another trigonometric identity I did, tried that several times, didn't get tan^4x.
 August 12th, 2012, 12:52 PM #7 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,204 Thanks: 511 Math Focus: Calculus/ODEs Re: Another trigonometric identity $\frac{\cos4\alpha-4\cos2\alpha+3}{\cos4\alpha+4\cos2\alpha+3}=\frac{ 2\cos^22\alpha-1-4\cos2\alpha+3}{2\cos^22\alpha-1+4\cos2\alpha+3}=$ $\frac{2\cos^22\alpha-4\cos2\alpha+2}{2\cos^22\alpha+4\cos2\alpha+2}=\fr ac{\cos^22\alpha-2\cos2\alpha+1}{\cos^22\alpha+2\cos2\alpha+1}=$ $$$\frac{\cos2\alpha-1}{\cos2\alpha+1}$$^2=$$\frac{1-2\sin^2\alpha-1}{2\cos^2\alpha-1+1}$$^2=$$-\frac{\sin^2\alpha}{\cos^2\alpha}$$^2=$ $$$\tan^2\alpha$$^2=\tan^4\alpha$
 August 12th, 2012, 01:03 PM #8 Senior Member   Joined: Jul 2012 Posts: 225 Thanks: 0 Re: Another trigonometric identity Man, you are the best. I have no words. Just one question - I did the same as you did, double angle identity on cosine4? but how did you know not to use that same identity on cos2?? because I did, and it made the problem more complicated.
 August 12th, 2012, 01:06 PM #9 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,204 Thanks: 511 Math Focus: Calculus/ODEs Re: Another trigonometric identity There are several double-angle identities for cosine, so the second time, I used the forms that would eliminate the plus or minus one.
 August 12th, 2012, 01:19 PM #10 Senior Member   Joined: Jul 2012 Posts: 225 Thanks: 0 Re: Another trigonometric identity Thank you very much.

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