數學 AP GP問題~~~pls help!!!

From trigonometry knowledge,
sin 1 = cos (90 - 1) i.e. = cos 89
sin 3 = cos (90 - 3) = cos 87
...
sin 89 = cos (90 -1) =cos 1
etc

Therefore, you can "group" the 1st term with the last term, 2nd term with the 2nd last term etc etc etc

[(sin 1)^2 + (sin89)^2] + [(sin 3)^2 + (sin 87)^2] + ..... (sin45)^2 <--- ( (sin45)^2 will be the odd one out without a "pair")

use equations in AP for the "numbers" in the sin function, (actually not necessary to use AP, it can be easily figure out)
a = 1, d = 2
I = a + (n-1)d (I = nth term)
89 = 1 + (n-1)2
n = 45

Therefore, there are 45 terms. (22 "pairs" and 1 odd one)

using sin^2 X + cos^2 X =1

[(sin 1)^2 + (sin89)^2] + [(sin 3)^2 + (sin 87)^2] + ..... (sin45)^2
will become
1 x 22 + (1/√2)^2 (because [(sin 1)^2 + (sin89)^2] = [(sin 1)^2 + (cos1)^2] = 1)
= 22 + 1/2
= 22.5

[ (sin 1)^2 + (sin89)^2 ]x20=20