\(\displaystyle{\frac{{{1}}}{{{\sin{{8}}}^{\circ}}}}+{\frac{{{1}}}{{{\sin{{16}}}^{\circ}}}}+\ldots+{\frac{{{1}}}{{{\sin{{4090}}}^{\circ}}}}+{\frac{{{1}}}{{{\sin{{8192}}}^{\circ}}}}={\frac{{{1}}}{{{\sin{\alpha}}}}}\), find \(\displaystyle\alpha\)

jisu61hbke

jisu61hbke

Answered question

2022-03-31

1sin8+1sin16++1sin4090+1sin8192=1sinα, find α

Answer & Explanation

Riya Erickson

Riya Erickson

Beginner2022-04-01Added 12 answers

Using the identity from the Weierstrass Substitution
tan(x2)=sin(x)1+cos(x)
we get
1tan(x2)1tan(x)=1+cos(x)sin(x)cos(x)sin(x)
=1sin(x)
The rest is the same telescoping series
k=0n1sin(2kx)=k=0n[1tan(2k1x)1tan(2kx)]
=1tan(x2)1tan(2nx)
The question has x=8 and n=10, so we get
k=0101sin(2k8)=1tan(4)1tan(8192)
=1tan(4)+1tan(88)
=1tan(4)+tan(2)
=cos(4)sin(4)+sin(4)1+cos(4)
=1sin(4)

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