Problem:
What is the value of
tan216π⋅tan2163π+tan216π⋅tan2165π+tan2163π⋅tan2167π+tan2165π⋅tan2167π?
Answer Choices:
A. 28
B. 68
C. 70
D. 72
E. 84
Solution:
Let θ=16π, and let c=cosθ,s=sinθ, and t=tanθ. Then
e8iθ=e2π⋅i=(c+is)8=0+1i
Expanding the binomial and taking the real part gives
0=1c8−28c6s2+70c4s4−28c2s6+1s8
Divide by c8 to get
1−28t2+70t4−28t6+1t8=0
and let p(t)=1−28t2+70t4−28t6+1t8. In other words, t=tanθ and −t=−tanθ are roots of the polynomial
p(x)=1−28x2+70x4−28x6+1x8
The same argument with θ=16π replaced by θk=16kπ shows that tk=tanθk is also a root of p(x) for k∈{1,3,5,7}. This holds because (ck+isk)8=0±1i when ck=cos(kθ) and sk=sin(kθ). In each case 8k≡8(mod16). Therefore x=tan(kθ) is a root of p(x) whenever
x2∈{tan2θ,tan2(3θ),tan2(5θ),tan2(7θ)}
This accounts for all 8 roots, so it must be that
p(x)=(x2−tan2θ)(x2−tan2(3θ))(x2−tan2(5θ))(x2−tan2(7θ))
Examining the x4 term in the two expressions for p(x) reveals that
tan2θ⋅tan2(3θ)+tan2θ⋅tan2(5θ)+tan2θ⋅tan2(7θ)+tan2(3θ)⋅tan2(5θ)+tan2(3θ)⋅tan2(7θ)+tan2(5θ)⋅tan2(7θ)=70
The identity tanθ⋅tan(2π−θ)=tanθ⋅cotθ=1 implies that
tan2θ⋅tan2(7θ)=tan2(3θ)⋅tan2(5θ)=1
Subtracting those two terms from the previous expression gives
tan2θ⋅tan2(3θ)+tan2θ⋅tan2(5θ)+tan2(3θ)⋅tan2(7θ)+tan2(5θ)⋅tan2(7θ)=(B)68
The problems on this page are the property of the MAA's American Mathematics Competitions
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