Find value of \(\displaystyle{\tan{{\left({\frac{{\pi}}{{{25}}}}\right)}}}\cdot{\tan{{\left({\frac{{{2}\pi}}{{{25}}}}\right)}}}\cdots{\tan{{\left({\frac{{{12}\pi}}{{{25}}}}\right)}}}\) The solution I tried: Assume

Malachi Mullins

Malachi Mullins

Answered question

2022-04-05

Find value of tan(π25)tan(2π25)tan(12π25)
The solution I tried:
Assume
P=tan(π25)tan(2π25)tan(3π25)tan(12π25),
with the help of tan(πθ)=tanθ, then
P=tan(13π25)tan(14π25)tan(15π25)tan(24π25)
which gives
P2={r=1}24tan(rπ25)

Answer & Explanation

Marin Lowe

Marin Lowe

Beginner2022-04-06Added 18 answers

Consider the polynomial
f(t)=12i((1+it)25(1it)25)=t25++25t
When t=tanθ we have
f(tanθ)=12i[(eiθcosθ)25(eiθcosθ)25]=1cos25θsin(25θ)
which vanishes at θ=0,±π25,,±12π25
This implies the 25 roots of 0,±tanπ25,,tan12π25. Apply Vieta's formula to the coefficient of t in f(t) and notice tanπ25,,tan12π25 are positive, we obtain:
k=112(tan2kπ25)=(1)2425k=112tankπ25=25=5

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