Problem:
Let xn=sin2(n∘). What is the mean of x1,x2,x3,…,x90 ?
Answer Choices:
A. 4511
B. 4522
C. 18089
D. 21
E. 18091
Solution:
The required mean is
901n=1∑90sin2(n∘)
Group the summands into 44 pairs plus two additional terms as follows:
sin2(1∘)+sin2(89∘)sin2(2∘)+sin2(88∘)sin2(3∘)+sin2(87∘)⋮sin2(44∘)+sin2(46∘)sin2(45∘)sin2(90∘)=sin2(1∘)+cos2(1∘)=1=sin2(2∘)+cos2(2∘)=1=sin2(3∘)+cos2(3∘)=1=sin2(44∘)+cos2(44∘)=1=21=1
This gives a sum of 4521=291, so the mean is 901⋅291=(E)18091.
The problems on this page are the property of the MAA's American Mathematics Competitions