Author |
Topic: prove sec40° + sec80° + sec160° =6 (Read 2617 times) |
|
tony123
Junior Member
Posts: 61
|
 |
prove sec40° + sec80° + sec160° =6
« on: Aug 4th, 2007, 9:35am » |
 Quote  Modify
|
prove
sec40° + sec80° + sec160° =6
// title changed --towr
(I'm not good at coming up with titles either, but when the entire problem statement fits there, why not!)
|
| « Last Edit: Aug 5th, 2007, 7:04am by towr » |
 IP Logged |
|
|
|
K Sengupta
Senior Riddler
Gender: 
Posts: 371
|
|
Re: prove
« Reply #1 on: Aug 5th, 2007, 2:01am » |
Quote Modify
|
At the outset, I would like to state that I found the fundamental tenets inclusive of the problem problem very interesting and I am positing a proposed solution in my subsequent post.
However, the problem title appears to be a duplication of a title corresponding to a previous contribution by the same author . In my opinion this is at direct variance with the provisions governing the website.
Accordingly, I would request for moderator intervention to change the problem title.
|
|
IP Logged |
|
|
|
K Sengupta
Senior Riddler
Gender:
Posts: 371
|
|
Re: prove
« Reply #2 on: Aug 5th, 2007, 2:03am » |
Quote Modify
|
Method 1:
Substituting h = 40o, 80o, 160o in turn, we observe that:
cos 3h =-1/2, for each of the foregoing values, and accordingly:
8 cos3h – 6 cos h + 1 =0
Therefore, cos 40, cos 80 and cos 160 are the roots of the cubic equation
8 y3 – 6 y + 1 =0; for y = cos h and so, their respective reciprocals sec 40, sec 80 and sec 160 correspond to the roots of the cubic equation :
y3 – 6y + 8 = 0, and therefore:
The sum of the roots of the cubic is 6.
Consequently:
[EDIT] sec 40o + sec 80o + sec 160o = 6
|
| « Last Edit: Aug 6th, 2007, 8:50am by K Sengupta » |
IP Logged |
|
|
|
ThudnBlunder
wu::riddles Moderator
Uberpuzzler
The dewdrop slides into the shining Sea
Gender:
Posts: 4489
|
|
Re: prove
« Reply #3 on: Aug 5th, 2007, 4:19am » |
Quote Modify
|
on Aug 5th, 2007, 2:01am, K Sengupta wrote:However, the problem title appears to be a duplication of a title corresponding to a previous contribution by the same author . In my opinion this is at direct variance with the provisions governing the website.
[/i] |
|
Yes, it has happened more than once.
As for the problem, it would be more interesting and harder if he asked to solve secx + sec2x + sec4x = 6
The above property of  /9 is not on Mathworld (although Morrie's Law is).
|
|
IP Logged |
THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
|
|
|
JocK
Uberpuzzler
Gender:
Posts: 877
|
|
Re: prove
« Reply #4 on: Aug 5th, 2007, 5:02am » |
Quote Modify
|
Cute problem.
on Aug 5th, 2007, 4:19am, ThudanBlunder wrote:
As for the problem, it would be more interesting and harder if he asked to solve secx + sec2x + sec4x = 6 |
|
Thinking about it, I'd like the formulation:
Prove that the three numbers cos40°, cos80° and cos160° have arithmetic mean 0, harmonic mean 1/2 and geometric mean -1/2.
|
|
IP Logged |
solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
|
|
|
tony123
Junior Member
Posts: 61
|
|
Re: prove sec40° + sec80° + sec160° =6
« Reply #5 on: Aug 5th, 2007, 2:28pm » |
Quote Modify
|
K Sengupta
Method 1:
i think its rong
and
cos 40 + cos 80 + cos 160 not= 6
the problem is
sec40° + sec80° + sec160° =6
|
|
IP Logged |
|
|
|
JocK
Uberpuzzler
Gender:
Posts: 877
|
|
Re: prove sec40° + sec80° + sec160° =6
« Reply #6 on: Aug 5th, 2007, 2:44pm » |
Quote Modify
|
on Aug 5th, 2007, 2:28pm, tony123 wrote:K Sengupta
Method 1:
i think its rong
|
|
Actualy, Sengupta is right: the stated follows from applying Viete's formulas to the cubic equation in cos(x) resulting from expressing 2cos(3x)+1=0 in terms of cos(x). It's just that in the very last line of his post a triple typo ('cos' instead of 'sec') pops up.
|
| « Last Edit: Aug 5th, 2007, 2:44pm by JocK » |
IP Logged |
solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
|
|
|
K Sengupta
Senior Riddler
Gender:
Posts: 371
|
|
Re: prove sec40° + sec80° + sec160° =6
« Reply #7 on: Aug 6th, 2007, 8:32am » |
Quote Modify
|
on Aug 5th, 2007, 2:44pm, JocK wrote:
It's just that in the very last line of his post a triple typo ('cos' instead of 'sec') pops up.
|
|
True.
I was careless and made the typo of triple cos instead of sec.
I confirm having corrected the foregoing anomaly.
|
|
IP Logged |
|
|
|
K Sengupta
Senior Riddler
Gender:
Posts: 371
|
|
Re: prove sec40° + sec80° + sec160° =6
« Reply #8 on: Aug 6th, 2007, 8:56am » |
Quote Modify
|
I furnish hereunder, a second proposed method giving the solution to the problem.
I am not certain if this corresponds to the author's official solution.
Method 2
sec 40 + sec 80 + sec 160
= (cos 40 + cos 80)/(cos 40*cos80) + 1/(cos 160)
= (2*cos 60*cos 20)/(cos 40*cos 80) + 1/ (cos 160)
= (cos 20*cos 160 + cos 40*cos 80)/(cos 40*cos 80*cos 160)
= (cos 180 + cos 140 + cos 120 + cos 40)/ {cos 40*(cos 240 + cos 80)}
= (-1 - 1/2)/{(1/2)(-cos 40 + cos 120 + cos 40)}
= (-3/2)/(-1/4)
= 6
|
|
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print