All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Six tangents (Posted on 2014-08-17)
Show that :

tan(pi/13)*tan(2pi/13)*tan (3pi/13)*tan(4pi/13)*tan(5pi/13)*tan (6pi/13) = 13^1/2.

 No Solution Yet Submitted by Danish Ahmed Khan No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 Solution Comment 8 of 8 |
Let t = tan x.  Let C(x,y) denote the combinatoric function x!/(y!*(x-y)!).  Then the general multi-angle formula for tan(n*x) can be expressed as:
`Sigma{k=0 to floor((n-1)/2)} (-1)^k*C(n,2k+1)*t^(2k+1)------------------------------------------------------  Sigma{k=0 to ceil((n-1)/2)} (-1)^k*C(n,2k)*t^(2k)`

Specifically for this problem tan(13x) then equals:
`13t - 286t^3 + 1287t^5 - 1716t^7 + 715t^9 - 78t^11 + t^13---------------------------------------------------------1 - 78t^2 + 715t^4 - 1716t^6 + 1287t^8 - 286t^10 + 13t^12`

For all the angles x=n*pi/13 with n equals any integer, the value of tan(13x) = 0.  Therefore the values of the tangents of all thirteen angles are the roots of the polynomial:
`13t - 286t^3 + 1287t^5 - 1716t^7 + 715t^9 - 78t^11 + t^13`

Factoring out the trivial root t=0, corresponding to x=0, the product of the remaining 12 roots equals 13.

Note that tan(n*pi/13)=-tan((13-n)pi/13).  Then the product of the 12 roots equals the square of the product of the first six roots - the expression we want to evaluate.

Therefore tan(pi/13) * tan(2pi/13) * tan(3pi/13) * tan(4pi/13) * tan(5pi/13) * tan(6pi/13) = sqrt(13).

This is easily adaptable to prove the more general equation suggested in the comments:
Prod{k=1 to n} tan(k*pi/(2n+1)) = sqrt(2n+1).

 Posted by Brian Smith on 2016-07-04 19:45:10

 Search: Search body:
Forums (0)