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由 galaxylee 於 星期六 八月 13, 2005 9:42 pm
[數學]求值
由 qeypour 於 星期六 八月 13, 2005 8:48 pm
(1-w)*S=1+w+w^2+w^3+.......+w^48-49w^49
(1-w)*S=[(1-w^49)/(1-w)]-49w^49..............(1)
w^49=1代入(1)得(1-w)*S=-49
S=-49/(1-w)
=-49/(1-cos2p/49-isin2p/49)
=-49/{1-[1-2(sinp/49)^2]-2isinp/49*cosp/49}
=[-49/(2sinp/49)]*[1/(sinp/49-icosp/49)]
=[-49/(2sinp/49)]*[sinp/49+icosp/49]
=-49/2-(49/2)i*cotp/49
所以1+2cosA+3cos2A+........+49cos48A=-49/2
2sinA+3sin2A+..........+49sin48A=(-49/2)*cotp/49
(1-w)*S=[(1-w^49)/(1-w)]-49w^49..............(1)
w^49=1代入(1)得(1-w)*S=-49
S=-49/(1-w)
=-49/(1-cos2p/49-isin2p/49)
=-49/{1-[1-2(sinp/49)^2]-2isinp/49*cosp/49}
=[-49/(2sinp/49)]*[1/(sinp/49-icosp/49)]
=[-49/(2sinp/49)]*[sinp/49+icosp/49]
=-49/2-(49/2)i*cotp/49
所以1+2cosA+3cos2A+........+49cos48A=-49/2
2sinA+3sin2A+..........+49sin48A=(-49/2)*cotp/49
[轉貼]三角函數極式
由 宇智波鼬 於 星期六 八月 13, 2005 8:14 pm
從昌爸轉貼的,因為我也不會,所以貼到yll來求救.
求值式
1+ 2cosA +3cos2A +4cos3A +......+48cos47A +49cos48A =?
某位叫mike的回答,
[缺一條件------>A=(2/49)pi
解答
負的二分之四十九------>(-49/2)
令w=cosA+isinA
Sum=1+2w+3w^2+4w^3+......+48w^47+49w^48
-)w*Sum=w+2w^2+3w^3+4w^4+......+48w^48+49w^49
-----------------------------------------------------
(1-w)*S=1+w+w^2+3w^3+4w^4+......+48w^48-49w^49
S=(-49w^49)/(1-w)......不會解啦!........(煙霧彈......)]
求值式
1+ 2cosA +3cos2A +4cos3A +......+48cos47A +49cos48A =?
某位叫mike的回答,
[缺一條件------>A=(2/49)pi
解答
負的二分之四十九------>(-49/2)
令w=cosA+isinA
Sum=1+2w+3w^2+4w^3+......+48w^47+49w^48
-)w*Sum=w+2w^2+3w^3+4w^4+......+48w^48+49w^49
-----------------------------------------------------
(1-w)*S=1+w+w^2+3w^3+4w^4+......+48w^48-49w^49
S=(-49w^49)/(1-w)......不會解啦!........(煙霧彈......)]