Show that :

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

(In reply to re(2): Interesting observation by Larry)

Assuming it's true for all n, then it must be true for some less awkward fraction such as tan(pi/15):

tan(pi/15) tan(2pi/15) tan(3pi/15) tan(4pi/15) tan(5pi/15) tan(6pi/15) tan(7pi/15)

When: tan(pi/15)tan((2pi)/15)tan((4pi)/15)tan((7pi)/15) = 1.

and tan(3pi/15)tan((5pi)/15)tan((6pi)/15) = (5-2*5^(1/2))^(1/2)*3^(1/2)*(5+2*5^(1/2))^(1/2) = 15^(1/2).

So that might be an easier place to start.

   

Edited on August 18, 2014, 5:32 am

Posted by broll on 2014-08-18 05:02:01