Problem:
Find the value of 10cot(cot−13+cot−17+cot−113+cot−121).
Solution:
In order to simplify the notation, let
a=cot−13,b=cot−17,c=cot−113,d=cot−121(1)
e=a+b and f=c+d(2)
Then, in view of (1) and (2) above, the problem is equivalent to finding the value of 10cot(e+f).
It is easy to show that in general,
cot(x+y)=cotx+coty(cotx)(coty)−1(3)
Moreover, even if cot−1 is viewed as a multiple valued function,
cot(cot−1x)=x, for all real x.(4)
Utilizing Equations (1) through (4) above, we find that
cote=cot(a+b)=3+73⋅7−1=2,cotf=cot(c+d)=13+2113⋅21−1=8cot(e+f)=2+82⋅8−1=23
and therefore, 10cot(e+f)=15 is the answer to the problem.
The problems on this page are the property of the MAA's American Mathematics Competitions