# Why is \$sumlimits_{k=1}^{leftlfloorfrac n2rightrfloor}sin^2left((2k-1)fracpi nright)=frac n4\$?

I found the relation $$sumlimits_{k=1}^{leftlfloorfrac n2rightrfloor}sin^2left((2k-1)fracpi nright)=frac n4$$But despite my best efforts, I still have no idea as to how to prove it.

Things I’ve tried:

1. Adding $$cos^2$$ terms. Of course, $$sin^2x+cos^2x$$ and $$cos^2x-sin^2x$$ are both simplifiable, so I thought about adding $$cos^2$$ terms to make the sum into $$frac n2$$ and hope that the $$sin^2$$ and $$cos^2$$ terms sum to equal amounts. They don’t in the $$n$$ odd case, so I didn’t know what to do with this approach.
2. Adding more $$sin^2$$ terms: Since $$sin^2 x=sin^2(pi-x)$$, we can add terms to this sum, but honestly, it didn’t make the sum any easier to evaluate.

Any help here?