All the angles are in degrees
6
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Edited version. (Thanks to Brian)
Anyway, I’ll show that instead of
using 45 numbers in the product we can
reduce it to only 15.
The identity to this end is
tan x tan(60+x) tan(60-x) = tan 3x.
So tan 1 tan 59 tan 61 = tan 3
which yields
tan 1 = tan 3 tan 29 tan 31
after dividing by tan 59 tan 61
and taking reciprocals.
(1/tan 59 = tan(90-59) = tan 31.)
Proceeding in the same manner, we get
tan 2 = tan 6 tan 28 tan 32
tan 3 = tan 9 tan 27 tan 33
tan 4 = tan 12 tan 26 tan 34
……
tan 15 = tan 45 tan 15 tan 45
Multiplying the numbers in each column
and cancelling the common factors we get
tan(1)*tan(2)*tan(4) *tan(5)*tan(7)*tan(8)* tan(10)*tan(11)* tan(13)*tan(14) =
tan(18)*tan(21)*tan(24) *tan(27)* tan(30)*tan(33) *tan(36)*tan(39)*tan(42) *
[tan(16)*tan(17)*……*tan(29) *tan(31)*…..* tan(43)*tan(44)]
Buggler: How did you derive your results?
Edit: Brian, thanks for the corrections.
71
So far I have managed to show that Sin 1 (Cos 1 + Sin 1) * Sin 2 (Cos2 + Sin 2) … Sin 45 * (Cos 45 + Sin 45) = Sqrt[180 / 2^134]
Hopefully this might help someone figure out how to finish it off.
Edit 1: You can show from the last thing that (1 + tan 1)(1 + tan2)…(1 + tan 45) = 2^23. Not quite the answer but pretty cool
Steiner, any of these following quantities would be sufficient to prove the answer:
cos1…cos45
sin1…sin45
(1+tan46)…(1+tan89)
(sin1+cos1)…(sin45+cos45)
(sin2+1)(sin4+1)….(sin88+1)
(1+tan^2 1)(1+tan^2 2)..(1+tan^2 45)
What I do know is that there doesn’t seem to be enough information contained in the usual quantities people know about so it seems new information somehow needs to be obtained from another trick. Trig identities probably won’t help.
Oh and as to how I got my formulas above (although I should have checked wiki because they are all known things)
z^(n+1)-1 = (z – 1)(z – w_1)…(z – w_n)
Dividing by z-1 gets us 1 + z + … + z^n = (z – w_1)…(z – w_n)
If you set z = 1, you get that n+1 = (1 – w_1)…(1 – w_n)
Squaring both sides and expanding the right hand side, you get (n+1)^2 = 2^n product (1 – cos (2pi k/(n+1))). ie) n^2 = 2^(n-1) product(1 – cos(2 pi k / n)). Then everything I’ve got so far has followed from this and some trig identities. We need something new like the above method which used a limit to arrive at new data.
@Steiner1745. This looks promising, but won’t you end up with
tan(1)*tan(2)*tan(4) *tan(5)*tan(7)*tan(8)* tan(10)*tan(11)* tan(13)*tan(14) =
tan(18)*tan(21)*tan(24) *tan(27)* tan(30)*tan(33) *tan(36)*tan(39)*tan(42) *
[tan(16)*tan(17)*……*tan(29) *tan(31)*…..* tan(43)*tan(44)] as the equation?
Edit: I’m not sure if I’m just spinning my wheels here, but note that
tan(45 – x) = (1 – tan(x)) / (1 + tan(x)). So then we get
tan(1)*tan(2)* …… *tan(44)*tan(45) =
[(1 – tan(44))/(1 + tan(44))]*[(1 – tan(43))/(1 + tan(43))]* … *[(1 – tan(1))/(1 + tan(1))].
Now using buggler’s result, (which I remember seeing proved before), the product
now becomes [(1 – tan(1))*(1 – tan(2))* ….. *(1 – tan(44))] / (2^23), and thus
2^23 * tan(1)*tan(2)* … *tan(44)*tan(45) = (1 – tan(1))*(1 – tan(2))* … *(1 – tan(44)).
This looks ‘useful’, but I don’t know what else to do with it. 🙂
I found the proof of buggler’s result; gohpihan’s proof was what I remembered:
(a million+tan1)(a million+tan2)(a million+tan3)……………(… = (a million+a million+a million+………40 5 situations)(tan1+tan2+tan3+………..tan45) = 40 5 x 0, considering,in case you proceed (a million+tan1)(a million+tan2)(a million+tan3)……………(… at a factor, you will get Tan 30 levels, that’s comparable to 0. for this reason, the respond is 40 5.
We have: tan(A + B) ≡ (tanA + tanB) / (1 – tanA.tanB),
which, when re-arranged, produces: tanA.tanB ≡ 1 – [(tanA + tanB) / tan(A + B)].
So let’s develop that identity: tanA.tanB ≡ 1 – [(tanA + tanB) / tan(A + B)].
If we multiply both sides by tanC, we get:
tanA.tanB.tanC = tanC{1 – [(tanA + tanB) / tan(A + B)]},
i.e.: tanA.tanB.tanC = tanC – [tanC(tanA + tanB) / tan(A + B)],
i.e.: tanA.tanB.tanC = 1 – [(tanA + tanB) / tan(A + B).tanC].
So, following that pattern, we get:
tanA.tanB.tanC.tanD = 1 – [(tanA + tanB) / tan(A + B).tanC.tanD].
Therefore, if A = 1; B = 2; C = 3 . . . . etc., we eventually get:
tan1.tan2.tan3.tan4 …. tan45 = 1 – [(tan1 + tan2) / tan(1 + 2).tan3.tan4 . . . . . tan45].
A fascinating question!
Wtf rocket science
Source(s): Uneducated me