What is \tan(f(1)+f(2)+\cdots+f(21)) where f(x) = \arctan(\frac{1}{x^2+x+1})

Dashawn Robbins

Dashawn Robbins

Answered question

2022-04-25

What is tan(f(1)+f(2)++f(21)) where f(x)=arctan(1x2+x+1)

Answer & Explanation

Leanna Boone

Leanna Boone

Beginner2022-04-26Added 11 answers

Edit: For when you said f(x)=1x2+x+1
A=f(1)++f(21)=25493941612249681421909183335333386891079486272288992946945151
tanA.936697
Edit: For when you put f(x)=tan11x2+x+1=tan1F(x)
A=B+f(21)
tanA=tanB+tanf(21)1tanBtanf(21)
=tanB+F(21)1tanB×F(21)
Keep proceeding down with B=C+f(20)
Jamison Newman

Jamison Newman

Beginner2022-04-27Added 10 answers

My comment shows n=121arctan{1n2+n+1}=π4arctan{122}
This lead your answer.

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?