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March 28th, 2017, 09:53 AM   #1
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Help me with a very difficult question

(tan15°)×(tan25°)×(tan35°)×(tan85°)
I know the answer, but I do not know the solution
answer: 1
help me please.

Last edited by skipjack; April 24th, 2017 at 08:56 PM.
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April 24th, 2017, 04:26 PM   #2
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Math Focus: Algebra
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Joined: Mar 2017
From: Azerbaijan

Posts: 1
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Help me with a very difficult question
(tan15°)×(tan25°)×(tan35°)×(tan85°)
I know the answer but I do not know the solution
answer: 1
help me please.









It is very possible to get the value of 15
From special angle 45 - 30 and when solve will produce
2 - root 3 but this process will nt work for 25, 35, and 85

Last edited by skipjack; April 24th, 2017 at 08:57 PM.
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April 24th, 2017, 04:32 PM   #3
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Challenging Trigonometry Problems!?


1. If tan 15 tan 25 tan 35 = tan t, and t is
between 0 and 180, find t. (All angle measures
are in degrees.)
2. Find the number of solutions, in degrees, to
the equation
sin^10 x + cos^10 x = (29/16)cos^4 2x, where x is
between 0 and 2007 degrees, inclusive.
3. Find both the maximum and minimum values
of
a) f(x) = sin x + cos x + tan x + csc x + sec x + cot x
b) |f(x)| = |sin x + cos x + tan x + csc x + sec x +
cot x|
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April 24th, 2017, 10:04 PM   #4
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$\displaystyle \begin{align*}\tan15^\circ\tan25^\circ\tan35^\circ \tan85^\circ &= \frac{8\sin15^\circ\sin25^\circ\sin35^\circ\sin85^ \circ}{8\cos15^\circ\cos25^\circ\cos35^\circ\cos85 ^\circ} \\
&= \frac{(\cos10^\circ - \cos40^\circ)(2\cos50^\circ + 1)}{(\cos40^\circ + \cos10^\circ)(2\cos50^\circ - 1)}\\
&= \frac{2\cos10^\circ\cos50^\circ + \cos10^\circ - 2\cos40^\circ\cos50^\circ - \cos40^\circ}{2\cos40^\circ\cos50^\circ - \cos40^\circ + 2\cos10^\circ\cos50^\circ - \cos10^\circ} \\
&= \frac{2\cos10^\circ\cos50^\circ + \cos10^\circ - \cos10^\circ - \cos40^\circ}{\cos10^\circ - \cos40^\circ + 2\cos10^\circ\cos50^\circ - \cos10^\circ} \\
&= 1\end{align*}$
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April 25th, 2017, 05:32 AM   #5
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That right skipjack. Dnt look at it from that corner
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