Question

Evaluate the series
sin^2 0 + sin^2 1 + sin ^2 2+ sin^2 3... sin^2 90

Answer 1

\(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ 90 } ~ {\rm Sin}^2({\rm n})}\)

Answer 2

you can enter a series into wolfram and it will give you the sum, if you are allowed to do that.

Answer 3

My interpretation is off, it should say \(\large\color{black}{ \displaystyle \sum_{ n=0 }^{ \infty } ~ {\rm Sin}^2({\rm n})}\)

Answer 4

http://www.wolframalpha.com/input/?i=sum+from+n%3D0+to+n%3D90+of+sin%5E2%28n%29

Answer 5

thanks @solomonzelman

Answer 6

sure