For Trigonometry Pros
I would like to know how you guys would attempt to solve this problem.
"y = tan(npi/11)x - sin(npi/11) where n is an element of {1, 2, 3, 4, 5}.
Find the product of all the solutions of x for y = 0 in the form 1/a."
This took me 3 hours to solve and I've only completed 11th grade maths.
However, don't let this discourage you as it can be solved in 5 lines of working.
Good luck!
February 15, 2011
16 Comments • Newest first
[quote=davids96]Lol how's it going? So far I have only done easy complex number stuff.[/quote]
It's fun.
However, I bet I only say that due to me doing the MathsQuest textbook, rofl.
[quote=beardoftheyear]@AnasF: really? are you in the IB program taking math HL or further maths SL?[/quote]
Lol, no.
Specialist Maths.
@Garming: Well, if you pay close attention, most calculators will actually have the answer as 0.31249999r.
The actual part of the question is proving cos^5(npi/11). xD
@beardoftheyear: and @Josiah: Yes, we learn De Moivre's theorem in year 11, and I am aware it is one of the methods to solving this problem. As I've stated already, I'm looking for other ways of solving it. When I was first told this question, I was told to only use trigonometry.
I later found out that the trigonometric method was simpler than both De Moivre's theorem and Euler's formula.
well you can use the method which states,
cos nX = 1/2 (z^n + z^-n)
and as they are all complex roots,
it'll turn out that it'll become (1/2)^5 x 1 = 1/32
[quote=AnasF]Yes, but why does
cos(pi/11) x cos(2pi/11) x cos(3pi/11) x cos(4pi/11) x cos(5pi/11) = 1/32?
"Internet said so" doesn't count.[/quote]
I just put it in the calculator... No idea how else to solve it...
[quote=beardoftheyear]de moivre's theorem. i guess they don't teach you that in 11th grade.[/quote]
What this guy said ^^
[quote=Josiah]Complex roots bro.[/quote]
See, this is why I was interested in how people would attempt to solve this question...
I know two ways to solve it, one being multitudes easier than the use of complex numbers...
Was wondering if anyone else could input some insight.
[quote=AnasF]Yes, but why does
cos(pi/11) x cos(2pi/11) x cos(3pi/11) x cos(4pi/11) x cos(5pi/11) = 1/32?
"Internet said so" doesn't count.[/quote]
Complex roots bro.
[quote=Garming]cos(pi/11) x cos(2pi/11) x cos(3pi/11) x cos(4pi/11) x cos(5pi/11) = 1/32?
that's just substituting each element from {1,2,3,4,5} for n?[/quote]
Yes, but why does
cos(pi/11) x cos(2pi/11) x cos(3pi/11) x cos(4pi/11) x cos(5pi/11) = 1/32?
"Internet said so" doesn't count.
Reserving spot
cos(pi/11) x cos(2pi/11) x cos(3pi/11) x cos(4pi/11) x cos(5pi/11) = 1/32?
that's just substituting each element from {1,2,3,4,5} for n?
Excellent.
But why is cos^5(npi/11) = 1/32?
0 = tan(npi/11)x - sin(npi/11)
sin(npi/11) = tan(npi/11)x
sin(npi/11) / tan(npi/11) = x
cos(npi/11) = x [sin(x) / cos(x) = tan(x)]
Substituting and multiplying, I get 1/32... same as above.
[quote=AnasF]You'd think so... But this is only the surface of a cleverly disguised problem. Go ahead and try. [/quote]
What makes you think I'm interested in doing so? I answered your question, and I'm not going to touch it again unless it's for money.
[quote=Omegathorion]I'd probably plug and chug, but it looks like it's set up for a quick and easy solution.[/quote]
You'd think so... But this is only the surface of a cleverly disguised problem. Go ahead and try.
I'd probably plug and chug, but it looks like it's set up for a quick and easy solution.