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October 15th, 2011, 11:11 AM   #1
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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.
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October 15th, 2011, 11:20 AM   #2
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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
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October 15th, 2011, 11:46 AM   #3
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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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