prove tri

tony123 · 2007-09-22 05:18:22

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)

JaneFairfax · 2007-09-22 23:44:11

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.

JaneFairfax · 2007-09-23 08:19:46

It should be

Last edited by JaneFairfax (2007-12-18 12:20:34)

John E. Franklin · 2007-09-23 15:23:22

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)

JaneFairfax · 2007-09-23 23:54:12

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)

JaneFairfax · 2007-09-24 01:21:01

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)

John E. Franklin · 2007-09-24 05:18:50

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.

Identity · 2007-09-24 06:26:26

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.

tony123 · 2007-09-26 20:57:02

.

Last edited by tony123 (2007-09-26 22:14:16)

tony123 · 2007-09-26 20:59:02

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)

tony123 · 2007-09-26 21:06:29

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

. ,
we may divide by x to leave an equation of
which .
are the roots:

.

JaneFairfax · 2007-12-30 18:26:08

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=181124

Whats Cardanos formula, by the way?

Last edited by JaneFairfax (2007-12-30 18:34:45)

JaneFairfax · 2008-01-08 06:03:23

Same trick used to evaluate

http://www.artofproblemsolving.com/Foru … p?t=182194

JaneFairfax · 2008-01-13 01:30:43

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.

tony123 · 2008-01-14 07:42:12

look to these
http://mathforum.org/library/drmath/view/65389.html

Last edited by tony123 (2008-01-14 07:46:20)

Math Is Fun Forum

#1 2007-09-22 05:18:22

prove tri

#2 2007-09-22 23:44:11

Re: prove tri

#3 2007-09-23 08:19:46

Re: prove tri

#4 2007-09-23 15:23:22

Re: prove tri

#5 2007-09-23 23:54:12

Re: prove tri

#6 2007-09-24 01:21:01

Re: prove tri

#7 2007-09-24 05:18:50

Re: prove tri

#8 2007-09-24 06:26:26

Re: prove tri

#9 2007-09-26 20:57:02

Re: prove tri

#10 2007-09-26 20:59:02

Re: prove tri

#11 2007-09-26 21:06:29

Re: prove tri

#12 2007-12-30 18:26:08

Re: prove tri

#13 2008-01-08 06:03:23

Re: prove tri

#14 2008-01-13 01:30:43

Re: prove tri

#15 2008-01-14 07:42:12

Re: prove tri

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