Evaluating \(\displaystyle\prod^{{{100}}}_{\left\lbrace{k}={1}\right\rbrace}{\left[{1}+{2}{\cos{{\frac{{{2}\pi\cdot{3}^{{k}}}}{{{3}^{{{100}}}+{1}}}}}}\right]}\)

Polchinio3es

Polchinio3es

Answered question

2022-04-03

Evaluating {k=1}100[1+2cos2π3k3100+1]

Answer & Explanation

Mason Knight

Mason Knight

Beginner2022-04-04Added 11 answers

It's unclear if you are asking about
k=1100[1+2cos2π3k3100+1]
or about
k=1100[1+2cos2πk3k3100+1]
I will assume it is the former. At the moment, that is in the body of your question, while the latter is in the title. If it's the former, then using the trig identity you found, you have a telescoping product which then further simplifies nicely:
k=1100sin(π3k+13100+1)sin(π3k3100+1)=sin(π323100+1)sin(π313100+1)sin(π333100+1)sin(π323100+1)sin(π31013100+1)sin(π31003100+1)
=sin(π31013100+1)sin(π313100+1)
=sin(π331003100+1)sin(π33100+1)
=sin(π3(3100+1)33100+1)sin(π33100+1)
=sin(3ππ33100+1)sin(π33100+1)
=sin(π33100+1)sin(π33100+1)
=1
Note near the end that sin(3πX)=sin(X)

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?