If ak=tan⁡(2+kπ2011) , then evaluate a1+a2+…+a2011a1a2…a2011

Joan Thompson

Joan Thompson

Answered

2022-01-15

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

Answer & Explanation

usumbiix

usumbiix

Expert

2022-01-16Added 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

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

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

Didn't find what you were looking for?