I wanted to use this \(\displaystyle{\sin{{\left(\frac{\pi}{{15}}\right)}}}\cdot{\sin{{\left({2}\frac{\pi}{{15}}\right)}}}\ldots{\sin{{\left({7}\frac{\pi}{{15}}\right)}}}=\sqrt{{{15}}}\)

Alan Zavala

Alan Zavala

Answered question

2022-04-07

I wanted to use this
sin(π15)sin(2π15)sin(7π15)=15

Answer & Explanation

posciad48o

posciad48o

Beginner2022-04-08Added 15 answers

We may prove:
cosπ154sin2π15=15sinπ15
by squaring both sides. By setting θ=π15, that leads to:
1322cos(θ)152cos(2θ)+2cos(3θ)+2cos(4θ)=152152cos(2θ)
or to:
cos(θ)+cos(3θ)+cos(4θ)=12
so we just have to prove that cosθ is a root of:
p(x)=16x4+8x316x28x+1
That easily follows from:
Φ30(x)=x8+x7x5x4x3+x+1

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