✔ 最佳答案
1.
Find the value of cos1 +cos2 +......+cos178 +cos179
Using cos (180-x)=-cosx
We have
cos1 +cos2 +......+cos178 +cos179
=cos1 +cos2 +......+cos90+cos(180-91)+cos(180-92)...cos(180-2) +cos(180-1)
= cos1 +cos2 +......cos88+cos89+cos90-cos89-cos88-.....-cos2-cos1
=0 [since cos90=0]
2.
Find the value of (sin1)^2 +(sin3)^2 +(sin5)^2 +........+(sin87)^2 +(sin89)^2
Using [sin(90-x)]^2=(cosx)^2
We have
(sin1)^2 +(sin3)^2 +(sin5)^2 +........+(sin87)^2 +(sin89)^2
=(sin1)^2 +(sin3)^2 +(sin5)^2 +... (sin 43)^2 + (sin 45)^2 + (sin 47)^2.......+(sin87)^2 +(sin89)^2
=(sin1)^2 +(sin3)^2 +(sin5)^2 +... (sin 43)^2 + (sin 45)^2 + [sin (90-43)]^2.......+[sin(90-3)]^2 +[sin(90-1)]^2
=(sin1)^2 +(sin3)^2 +(sin5)^2 +... (sin 43)^2 + (sin 45)^2 + (cos43)^2.......+ (cos3)^2 +(cos1)^2
=22+ (sin 45)^2
[using (sinx)^2+(cosx)^2=1 and there are total (43-1)/2+1=22 pairs]
=22+1/2
=二十二又二分之一