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wu :: forums    riddles    medium (Moderators: ThudnBlunder, william wu, towr, Icarus, Grimbal, Eigenray, SMQ)    prove « Previous topic | Next topic » Pages: 1  Reply Notify of replies Send Topic Print    Author  Topic: prove  (Read 1069 times) tony123 Junior Member     Posts: 61 prove   « on: Jun 27th, 2007, 7:34am » Quote Modify prove           tan 70 = tan 20  + 2 tan 40 +4 tan 10 IP Logged FiBsTeR Senior Riddler     Gender: Posts: 581 Re: prove   Tangent_Proof_2.bmp « Reply #1 on: Jun 27th, 2007, 2:29pm » Quote Modify Done.   IP Logged SMQ wu::riddles Moderator Uberpuzzler     Gender: Posts: 2084 Re: prove   « Reply #2 on: Jun 28th, 2007, 4:19am » Quote Modify There may be an easier way, but tan(20) + 2 tan(40) + 4 tan(10) = tan(2(70 - 60)) + 2 tan(70 - 30) + 4 tan(70 - 60), which can be manipulated using the identities tan(a - b) = (tan(a) - tab(b))/(1 + tan(a)tan(b)) and tan(2a) = 2 tan(a)/(1 - tan2(a)) to be entirely in terms of tan(70), tan(60) = sqrt(3), and tan(30) = sqrt(3)/3, then reduced to the desired result.   --SMQ IP Logged --SMQ K Sengupta Senior Riddler     Gender: Posts: 371 Re: prove   « Reply #3 on: Jun 28th, 2007, 8:52pm » Quote Modify The problem can also be alternatively solved in the following manner:   Let tan 70 = p(say) Then, cot 20 = p, so that:   tan 20 = 1/p   Thus, tan 40   = 2*(1/p)/(1- (1/p)^2 ) = 2p/(p^2 - 1)   And, tan 80   = 2*(2p)*(p^2-1)/((p^2-1)^2 – (2p)^2) = 2*(2p)*(p^2-1)/ (p^4 - 6*p^2 + 1) = cot 10   Or, tan 10   = (p^4 -6*p^2 + 1)/(4p(p^2-1))   Thus,   tan 20 + 2*tan 40 + 4*tan 10 = (4*p^2 + p^2 - 1 + p^4 - 6*p^2 +1)/(p(p^2-1) = (p^4 – p^2)/(p(p^2-1)) = p = tan 70                Q            E          D   « Last Edit: Jun 28th, 2007, 9:01pm by K Sengupta » IP Logged K Sengupta Senior Riddler     Gender: Posts: 371 Re: prove   « Reply #4 on: Jun 29th, 2007, 7:32am » Quote Modify Another alternative methodology would turn on referring to the trigonometric identity:   tan x – cot x = -2 cot 2x   From the above identity, we obtain:   tan 20 – cot 20 = - 2cot 40…..(i) 2(tan 40 – cot 40) = - 4cot 80…..(ii)   Also, 4 tan 10 = 4 cot 80 ……(iii)   Adding, (i), (ii), (iii); and simplifying, we have: tan 20 + 2 tan 40 + 4 tan 10 = cot 20 Or, tan 20 + 2 tan 40 + 4 tan 10 = tan 70   « Last Edit: Jun 29th, 2007, 7:32am by K Sengupta » 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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