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wu :: forums    riddles    medium (Moderators: SMQ, towr, william wu, Eigenray, ThudnBlunder, Grimbal, Icarus)    prove « Previous topic | Next topic » Pages: 1  Reply Notify of replies Send Topic Print    Author  Topic: prove  (Read 542 times) tony123 Junior Member     Posts: 61 prove   « on: Jul 19th, 2007, 9:03am » Quote Modify http://www.mathramz.com/phpbb/latexrender/pictures/3390ffd7b21a77e499662 3f1ddec5d29.png IP Logged pex Uberpuzzler     Gender: Posts: 880 Re: prove   « Reply #1 on: Jul 19th, 2007, 9:13am » Quote Modify Tony - not so long ago, you "were sorry and apologized" for (1) not writing descriptive titles and (2) linking to images which can easily be attached, or even written out.   What happened? IP Logged tony123 Junior Member     Posts: 61 Re: prove   imgcr_E13_C19940-1[1].gif « Reply #2 on: Jul 19th, 2007, 9:44am » Quote Modify why latiks not found any way help me to Attach and thank   IP Logged JohanC Senior Riddler     Posts: 460 Re: prove   4cos36pluscot7dot5.png « Reply #3 on: Jul 19th, 2007, 2:34pm » Quote Modify Hi, Barukh, In my version of Firefox that transparent PNG looks quite ugly, while it renders nicely in IE. I notice you created an embedded link to that external website, which runs the risk that the image will disappear in a few months. IP Logged Barukh Uberpuzzler     Gender: Posts: 2276 Re: prove   « Reply #4 on: Jul 19th, 2007, 10:59pm » Quote Modify on Jul 19th, 2007, 2:34pm, JohanC wrote: I notice you created an embedded link to that external website, which runs the risk that the image will disappear in a few months. Yes, you are right. It was an instant action... I removed the post. IP Logged Barukh Uberpuzzler     Gender: Posts: 2276 Re: prove   « Reply #5 on: Jul 20th, 2007, 6:18am » Quote Modify Amazing!   The whole thing may be derived purely geometrically. IP Logged Barukh Uberpuzzler     Gender: Posts: 2276 Re: prove   Trig_sumof_squares.zip « Reply #6 on: Jul 21st, 2007, 5:57am » Quote Modify All that is needed is on the attached drawing. IP Logged K Sengupta Senior Riddler     Gender: Posts: 371 Re: prove   « Reply #7 on: Jul 21st, 2007, 10:08am » Quote Modify From the identity cos 36 = sin 54, we obtain:   1 – 2x^2 = 3x – 4x^3, where x = sin 18 Or, 4x^3 -2x^2 – 3x + 1 = 0   Or, x_1 = 1; x_2,3 = (-1+/-V5)/4   Since sin t  is increasing in 0<= t<= 90, we must have sin 18 = (V5 – 1)/4     Thus, cos 36 = 1 – 2*(sin 18)^2 = (V5 + 1)/4   Now, cot 7.5   = cot 15/2   = (1+ cos 15)/sin 15   But sin 15 = sin (45-30) = (V6 – V2)/4 (upon simplification); and:   cos 15 = cos (45 – 30) = (V6 + V2)/4   Thus, cot 7.5 = (4+ V6 + V2)/(V6 – V2) = (4(V6 + V2) + (8+ 4V3))/4 = V6 + V2 + 2 + V3   Thus, 4 cos 36 + cot 7.5 = (V5 + 1) + (V6 + V2 + 2 + V3) = V1+ V5 + V6 + V2 + V4 + V3 = V1 + V2 + V3 + V4 + V5 + V6         Q  E  D   IP Logged 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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