標題:
F.4 math~~~Trigonometry
1. Find the value of cos1 +cos2 +......+cos178 +cos179 . 2. (sin1)^2 +(sin3)^2 +(sin5)^2 +........+(sin87)^2 +(sin89)^2
最佳解答:
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
其他解答:
1. the value of this Q is 0 solution:cos1+cos179=0,cos2+cos178=0,互相抵銷,,,,,(如此類推,最後剩下cos90,而cos90=0) 2.the value of this Q is 22.5 solution:(sin1)^2+(sin89)^2=1,(sin3)^2+(sin87)^2=1,現在有45個數,每兩個數相加等於1,所以22+(sin45)^2=22+0.5|||||1)cos1 +cos2 +......+cos178 +cos179 =cos1 +cos2 +......+cos(180-2) +cos(180-1) =cos1 +cos2 +......+(-cos2)+(-cos1) =0 2)(sin1)^2 +(sin3)^2 +(sin5)^2 +........+(sin87)^2 +(sin89)^2 =(sin1)^2 +(sin3)^2 +(sin5)^2 +........+(sin90-3)^2 +(sin90-1)^2 =(sin1)^2 +(sin3)^2 +(sin5)^2 +........+(cos3)^2 +(cos1)^2 =22+(sin45)^2 =22.5 2007-05-20 18:36:33 補充： 2)(sin1)^2 +(sin3)^2 +(sin5)^2 +........+(sin87)^2 +(sin89)^2 =(sin1)^2 +(sin3)^2 +(sin5)^2 +........+(sin90-3)^2 +(sin90-1)^2 =(sin1)^2 +(sin3)^2 +(sin5)^2 +........+(cos3)^2 +(cos1)^2=22+(sin45)^2=22.5 -->肯定岩!!!
F.4 math~~~Trigonometry
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發問:1. Find the value of cos1 +cos2 +......+cos178 +cos179 . 2. (sin1)^2 +(sin3)^2 +(sin5)^2 +........+(sin87)^2 +(sin89)^2
最佳解答:
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
其他解答:
1. the value of this Q is 0 solution:cos1+cos179=0,cos2+cos178=0,互相抵銷,,,,,(如此類推,最後剩下cos90,而cos90=0) 2.the value of this Q is 22.5 solution:(sin1)^2+(sin89)^2=1,(sin3)^2+(sin87)^2=1,現在有45個數,每兩個數相加等於1,所以22+(sin45)^2=22+0.5|||||1)cos1 +cos2 +......+cos178 +cos179 =cos1 +cos2 +......+cos(180-2) +cos(180-1) =cos1 +cos2 +......+(-cos2)+(-cos1) =0 2)(sin1)^2 +(sin3)^2 +(sin5)^2 +........+(sin87)^2 +(sin89)^2 =(sin1)^2 +(sin3)^2 +(sin5)^2 +........+(sin90-3)^2 +(sin90-1)^2 =(sin1)^2 +(sin3)^2 +(sin5)^2 +........+(cos3)^2 +(cos1)^2 =22+(sin45)^2 =22.5 2007-05-20 18:36:33 補充： 2)(sin1)^2 +(sin3)^2 +(sin5)^2 +........+(sin87)^2 +(sin89)^2 =(sin1)^2 +(sin3)^2 +(sin5)^2 +........+(sin90-3)^2 +(sin90-1)^2 =(sin1)^2 +(sin3)^2 +(sin5)^2 +........+(cos3)^2 +(cos1)^2=22+(sin45)^2=22.5 -->肯定岩!!!
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