Question

find the exact value of sin^2(1) + ... + sin^2(90)

Answer 1

oh, ok. haha. i have some questions after. And also, something's up with my OpenStudy, I can't see your previous replies anymore. Or did you do that? Haha

Answer 2

Yeah, I deleted them. No sense in keeping a wrong answer up.

Answer 3

Oh, I thought my connection's acting weird again. Haha. Thanks.

Answer 4

Another mistake...
\[\sum_{n=1}^{90}\sin^2n^\circ=\frac{1}{2}\sum_{n=0}^{90}(1-\cos2n^\circ)=\color{red}{\frac{91}{2}}-\frac{1}{2}\sum_{n=0}^{90}(\cos^2n^\circ-\sin^2n^\circ)\]

Answer 5

is this ok? Haha, I saw you still typing earlier so I thought you weren't done.

Answer 6

It's a work in progress. You can take advantage of symmetry to reduce this sum nicely, but I forget the details of that approach. What I'm currently doing is taking me in circles...

Answer 7

Hey thanks anyway, maybe I'll just skip this for now. I have a lot of Math to cover haha in so little time. Appreciate the help! I got a lot of questions still, from algebra-calculus, if you still have the time to help later, it would be much appreciated! :)

Answer 8

hi

Answer 9

hello

Answer 10

You can see the symmetry in the cosine sum:
\[\frac{91}{2}-\frac{1}{2}\sum_{n=0}^{90}\cos2n^\circ\]with
\[\color{red}{\cos0^\circ}+\color{green}{\cos2^\circ}+\color{blue}{\cos4^\circ}+\cdots+\color{blue}{\cos176^\circ}+\color{green}{\cos178^\circ}+\color{red}{\cos180^\circ}\]This reduces to zero, since \(\cos x^\circ=-\cos(180-x)^\circ\).