If a_k := \tan (\sqrt 2+\frac{k\pi}{2011}) , then evaluate \frac{a_1+a_2+\ldots+a_{2011}}{a_1a_2

Joan Thompson

Joan Thompson

Answered question

2022-01-15

If ak=tan(2+kπ2011) , then evaluate a1+a2++a2011a1a2a2011

Answer & Explanation

usumbiix

usumbiix

Beginner2022-01-16Added 33 answers

Using Sum of tangent functions where arguments are in specific arithmetic series,
tan(2m+1)x=(2m+11)t(2m+13)t3++(1)m(2m+12m+1)t2m+1(2m+10)(2m+12)t2+++(1)m(2m+12m)t2m
So, if tan(2m+1)x=tany
(2m+1)x=kπ+y where k is any integer
x=kπ+y2m+1  ;k=1,2,,2m+1
So, the roots of
tany=(2m+11)t(2m+13)t3++(1)m(2m+12m+1)t2m+1(2m+10)(2m+12)t2++(1)m(2m+12m)t2m
(1)mt2m+1tany(1)m(2m+1)t2m++tany=0
are tanx;  x=kπ+y2m+1  ,k=1,2,,2m+1

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