\(\displaystyle{\frac{{{1}}}{{{4}}}}{\tan{{\left({\frac{{\pi}}{{{8}}}}\right)}}}+{\frac{{{1}}}{{{8}}}}{\tan{{\left({\frac{{\pi}}{{{16}}}}\right)}}}+{\frac{{{1}}}{{{16}}}}{\tan{{\left({\frac{{\pi}}{{{32}}}}\right)}}}+\cdots\cdots\infty\) Try: \(\displaystyle{\cos{{x}}}\cdot{\cos{{\left(\frac{{x}}{{2}}\right)}}}\cdot{\cos{{\left(\frac{{x}}{{2}^{{2}}}\right)}}}\cdots\cdots{\cos{{\left(\frac{{x}}{{2}^{{{n}-{1}}}}\right)}}}={\frac{{{1}}}{{{2}^{{{n}-{1}}}}}}{\frac{{{\cos{{\left({x}\right)}}}}}{{{\sin{{\left(\frac{{x}}{{2}^{{{n}-{1}}}}\right)}}}}}}\) Could some help me how to

Arianna Villegas

Arianna Villegas

Answered question

2022-03-29

14tan(π8)+18tan(π16)+116tan(π32)+
Try:
cosxcos(x2)cos(x22)cos(x2n1)=12n1cos(x)sin(x2n1)
Could some help me how to solve ahead

Answer & Explanation

Harry Gibson

Harry Gibson

Beginner2022-03-30Added 13 answers

Using the identity:
cotxtanx=2cot2xtanx=cotx2cot2x
one obtains a telescoping sum:
S(x)=n=0tan(x2n)2n=n=0(cot(x2n)2ncot(x2n1)2n1)
=2cot(2x)+limncot(x2n)2n=1x2cot(2x)
Applying this to your case one obtains
14S(π8)=2π12

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