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 October 15th, 2011, 11:11 AM #1 Senior Member   Joined: Jul 2011 Posts: 405 Thanks: 16 Trigonometric sum Prove that $\displaystyle (1)\;\; \tan9 - \tan27- \tan63 + \tan81 =4$ $\displaystyle (5)\;\; \sin^2 12 + \sin^2 21 +\sin^2 39 + \sin^2 48 = 1 + \sin^2 9 + \sin^2 18$ all angles are in degrees. Last edited by skipjack; August 14th, 2019 at 11:15 PM.
 October 15th, 2011, 11:20 AM #2 Senior Member   Joined: Aug 2008 Posts: 113 Thanks: 0 Re: Trigonometric sum tan 9 + cot 9 - (tan 27 + cot 27) = sin 9/cos 9 + cos 9/sin 9 - (sin 27/cos 27 + cos 27/sin 27) = 1/(sin 9 cos 9) - 1/(sin 27 cos 27) = 2/sin 18 - 2/sin 54 = 2[sin 54 - sin 18]/(sin 18 sin 54) = 2[2 cos 36 sin 18]/(sin 18 sin 54) = 4
 October 15th, 2011, 11:46 AM #3 Senior Member   Joined: Aug 2008 Posts: 113 Thanks: 0 Re: Trigonometric sum sin^2 12+ sin^2 21 + sin^2 39 + sin^2 48 = (1/2) [1 - cos 24 + 1 - cos 42 + 1 - cos 78 + 1 - cos 96] = (1/2)[4 - cos 24 - cos 42 - cos 78 - cos 96] = 2 - (1/2)[cos 24 + cos 42 + cos 78 + cos 96] = 2 - (1/2)[cos 24 + cos 96 + cos 42 + cos 78] = 2 - (1/2)[2 cos 60 cos 36 + 2 cos 60 cos 18] = 2 - (1/2)[cos 36 + cos 18] = 1 + (1/2)[1 - cos 18 + 1 - cos 36] = 1 + sin^2 9 + sin^2 18 [color=#FF0000]In fact, it is easier to show that LHS = RHS = 2 - (1/2)[cos 36 + cos 18][/color]

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Tan9 -tan27- cot27 cot9

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