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cos (-89)+cos (-87) +.......+cos87 +cos 89 =csc 1
tan²(pi/7) +tan²(2pi/7) +tan²(3pi/7) =21
Last edited by tony123 (2007-09-24 08:14:18)
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Let C = cos(1°) + cos(3°) + + cos(87°) + cos(89°)
Note that your sum is equal to 2C since cos is an even function.
Now, cos(89°) = cos(87°+2°) = cos(87°)cos(2°) − sin(87°)sin(2°) = cos(87°)cos(2°) − cos(3°)sin(2°). So:
cos(89°) = cos(87°)cos(2°) − cos(3°)sin(2°)
cos(87°) = cos(85°)cos(2°) − cos(5°)sin(2°)
.
.
.
cos(5°) = cos(3°)cos(2°) − cos(87°)sin(2°)
cos(3°) = cos(1°)cos(2°) − cos(89°)sin(2°)
Now add em up.
C − cos(1°) = [C − cos(89°)]cos(2°) − [C − cos(1°)]sin(2°)
C − cos(1°) = Ccos(2°) − sin(1°)cos(2°) − Csin(2°) + cos(1°)sin(2°) [∵ cos(89°) = sin(1°)]
C[1 + sin(2°) − cos(2°)] = cos(1°) + cos(1°)sin(2°) − sin(1°)cos(2°)
= cos(1°) + sin(2°−1°)
= cos(1°) + sin(1°)
Now 1 + sin(2°) − cos(2°) = 1 + 2sin(1°)cos(1°) − 1 + 2sin[sup]2[/sup](1°) = 2sin(1°)[cos(1°) + sin(1°)]
Hence 2Csin(1°)[cos(1°) + sin(1°)] = cos(1°) + sin(1°)
⇒ 2Csin(1°) = 1
QED
I will do the second problem later.
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It should be
Last edited by JaneFairfax (2007-12-18 12:20:34)
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But with this amazing root of 22000 in radians, you get really close to 4.
Testing time on forum displayed: should read about: 00:23:45
Last edited by John E. Franklin (2007-09-23 16:18:22)
igloo myrtilles fourmis
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Anyway, I havent proved the second equation yet, but this is what Ive done so far.
By exploring with a calcuator, I found that
Recognize the coefficients? Yes! Theyre coefficients of alternate terms in some binomial expansion to the power of 7. The alternating +/− sign also leads one to suspect that complex numbers are involved. Indeed, multiplying the last equation by x,
we see that the above is the real part (taking x to be real) of the binomial expansion of
So if we can show that tan(π⁄7), tan(2π⁄7), tan(3π⁄7) are nonzero real numbers x for which the real part of (x+i)[sup]7[/sup] is 0, we should be well on our way.
PS: As I said, I have not solved the question yet. I was not using a calculator to solve anything, only to explore possibilities which might open the way for a proof. Jesus, man! Why does Tony123 keep posting these questions with no hint whatsoever as to how to proceed? A hint or two would be such a time saver!
Last edited by JaneFairfax (2007-09-24 07:50:03)
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ET VOILÂ!
Since tan[sup]2[/sup](π⁄7), tan[sup]2[/sup](2π⁄7), tan[sup]2[/sup](3π⁄7) are distinct, I conclude that they form the complete solution set to the cubic equation above. Hence their sum is equal to 21. QUOD ERAT DEMONSTRANDUM!!
Last edited by JaneFairfax (2008-04-28 15:01:14)
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Jane, sorry for the confusion.
I was solving for the other side of the equation than you.
I stuck with the four, and stuck with the tangent squareds,
and stuck with the 1:2:3 ratio of terms prior to tangenting them.
I think this is quite relevant.
igloo myrtilles fourmis
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Very well done on that problem Jane, by the way I never knew trigonometric functions could be roots of equations, I always thought they were transcendental.
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.
Last edited by tony123 (2007-09-26 22:14:16)
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JaneFairfax you are jenus
using the addition formula for tangents, we have
.Now obviously, for t =\ pi/7, 2\pi/7, ..., 7\pi/7, we have
. and thus, these are the roots of .Put differently,
. are the roots of .Since
. , we may divide by x to leave an equation of which. are the roots: .Last edited by tony123 (2007-09-26 21:21:38)
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JaneFairfax you are jenus
using the addition formula for tangents, we have
.Now obviously, for t = pi/7, 2pi/7, ..., 7pi/7, we have
. and thus, these are the roots of .Put differently,
. are the roots of .Since
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Since tan[sup]2[/sup](π⁄7), tan[sup]2[/sup](2π⁄7), tan[sup]2[/sup](3π⁄7) are distinct, I conclude that they form the complete solution set to the cubic equation above. Hence their sum is equal to 21.
Using the same trick for
: http://www.artofproblemsolving.com/Foru … p?t=181124Whats Cardanos formula, by the way?
Last edited by JaneFairfax (2007-12-30 18:34:45)
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Same trick used to evaluate
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Im vanity-smitten now. It seems that this trick I discovered all by myself really has a lot of use. Perhaps I should recommend the world mathematical community to name it after me.
http://www.artofproblemsolving.com/Foru … p?t=182995
I used my method to prove that
on my way to showing that
Yes, it seems correct. Ive used a program to compute the product and compare my answer with.
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look to these
http://mathforum.org/library/drmath/view/65389.html
Last edited by tony123 (2008-01-14 07:46:20)
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