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wu :: forums    riddles    easy (Moderators: Eigenray, SMQ, towr, Icarus, ThudnBlunder, william wu, Grimbal)    sin 18 « Previous topic | Next topic » Pages: 1  Reply Notify of replies Send Topic Print    Author  Topic: sin 18  (Read 953 times) cool_joh Guest sin 18   « on: Dec 19th, 2007, 5:47am » Quote Modify Remove Find the exact value of sin 18o.   Big hint: draw a 36o-72o-72o triangle. IP Logged Aryabhatta Uberpuzzler     Gender: Posts: 1321 Re: sin 18   « Reply #1 on: Dec 19th, 2007, 12:14pm » Quote Modify Here is one approach: It is enough if we find the roots of x5 = 1 (why?)   Factorizing and removing x=1 factor, we get     x4+x3+x2+x + 1 = 0   Divide by x2 and use the subsitution y = x + 1/x. We get a quadratic in y, whose roots can then be used to give two quadratics in x.   Thus giving us four roots.   I believe the roots come out to be   (-1+- sqrt(5))/4 - i sqrt(10 +- 2sqrt(5))/4   sin18 would be the real part of one of these roots, which I think is (sqrt(5)-1)/4   IP Logged Aryabhatta Uberpuzzler     Gender: Posts: 1321 Re: sin 18   sin18proof.zip « Reply #2 on: Dec 19th, 2007, 4:05pm » Quote Modify And here is a proof using the hint...   I am sure there is a neat geometric proof which didn't involve algebra or finding roots etc.   IP Logged FiBsTeR Senior Riddler     Gender: Posts: 581 Re: sin 18   « Reply #3 on: Dec 19th, 2007, 7:18pm » Quote Modify Aryabhatta has enclosed a similar picture, though it is zipped to avoid spoilers. IP Logged cool_joh Guest Re: sin 18   « Reply #4 on: Dec 19th, 2007, 7:29pm » Quote Modify Remove Sorry, I didn't notice the zip file. I removed my spoiler post « Last Edit: Dec 19th, 2007, 7:42pm by Johan_Gunardi » IP Logged cool_joh Guest Re: sin 18   « Reply #5 on: Dec 19th, 2007, 8:45pm » Quote Modify Remove on Dec 19th, 2007, 12:14pm, Aryabhatta wrote: It is enough if we find the roots of x5 = 1 (why?) I still don't understand. Can anyone give me a hint? IP Logged towr wu::riddles Moderator Uberpuzzler Some people are average, some are just mean.     Gender: Posts: 13730 Re: sin 18   « Reply #6 on: Dec 20th, 2007, 12:37am » Quote Modify on Dec 19th, 2007, 8:45pm, cool_joh wrote: I still don't understand. Can anyone give me a hint? Well, you can get sin/cos 72o out of it; using complex numbers.   (e2/5 pi i)5=e2 pi i = 1, so x = e2/5 pi i is a solution for x5=1 e2/5 pi i = cos 2/5 pi + i sin 2/5 pi = cos 72o + i sin 72o; and cos 72o = sin 18o So taking the real part of the correct root of x5=1 yields the answer. « Last Edit: Dec 20th, 2007, 12:38am by towr » IP Logged Wikipedia, Google, Mathworld, Integer sequence DB cool_joh Guest Re: sin 18   « Reply #7 on: Dec 20th, 2007, 3:13am » Quote Modify Remove Oww. I'm just 13, and I haven't studied that one at school. Thanks anyway! IP Logged mikedagr8 Uberpuzzler A rich man is one who is content; not wealthy.     Gender: Posts: 1105 Re: sin 18   « Reply #8 on: Dec 20th, 2007, 3:16am » Quote Modify on Dec 20th, 2007, 3:13am, cool_joh wrote: Oww. I'm just 13, and I haven't studied that one at school. Thanks anyway! You're 13? IP Logged "It's not that I'm correct, it's that you're just not correct, and so; I am right." - M.P.E. cool_joh Guest Re: sin 18   « Reply #9 on: Dec 20th, 2007, 3:21am » Quote Modify Remove on Dec 20th, 2007, 3:16am, mikedagr8 wrote: You're 13? Is that really surprising? IP Logged mikedagr8 Uberpuzzler A rich man is one who is content; not wealthy.     Gender: Posts: 1105 Re: sin 18   « Reply #10 on: Dec 20th, 2007, 3:25am » Quote Modify on Dec 20th, 2007, 3:21am, cool_joh wrote: Is that really surprising? When you compare your manner to srn347's, who was the same age... Yes. There is a huge difference. I thought you were mid 20's. I only started learning complex and imaginary numbers a few days ago. « Last Edit: Dec 20th, 2007, 3:28am by mikedagr8 » IP Logged "It's not that I'm correct, it's that you're just not correct, and so; I am right." - M.P.E. towr wu::riddles Moderator Uberpuzzler Some people are average, some are just mean.     Gender: Posts: 13730 Re: sin 18   « Reply #11 on: Dec 20th, 2007, 8:39am » Quote Modify on Dec 20th, 2007, 3:13am, cool_joh wrote: Oww. I'm just 13, and I haven't studied that one at school. Thanks anyway! Well, if you're interested to learn something about complex numbers (which, despite their name, simplify many things both in trigonometry and in other areas), feel free to ask.         An example (feel free to ignore if you really aren't interested):   A complex number (z) consists of a real (x) and imaginary part (i*y), the imaginary part is distinguished by a factor i, and importantly, i2=-1   So, e.g. multiplication of two complex numbers goes like (a+i b) * (c+i d) = ac + i2bd + i (ab+cd) = ac - bd + i (ab+cd)   The only other thing we need for now is the Euler formula/identity ei x = cos(x) + i*sin(x)   Knowing just these two things, we can simply derive the double-angle formulas (or any multiple-angle formulas for that matter) ei 2x = cos(2x) + i*sin(2x) ei 2x = (ei x)2       = (cos(x) + i*sin(x))2     = (cos2(x) - sin2(x)) + i (2cos(x)sin(x)) So we get cos(2x) = cos2(x) - sin2(x) and sin(2x) = 2cos(x)sin(x)   So basically, instead of remembering all the double and multiple-angle formulas, just knowing the euler formula is sufficient. IP Logged Wikipedia, Google, Mathworld, Integer sequence DB Pages: 1  Reply Notify of replies Send Topic Print Forum Jump: ----------------------------- riddles -----------------------------=> easy   - medium   - hard   - what am i   - what happened   - microsoft   - cs   - putnam exam (pure math)   - suggestions, help, and FAQ   - general problem-solving / chatting / whatever ----------------------------- general -----------------------------  - guestbook   - truth   - complex analysis   - wanted   - psychology   - chinese « Previous topic | Next topic »

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