複數運算! 請三角函數高手幫我解答

1)cos(π/12)+isin(π/12)=cos[π/(3×4)]+isin[π/(3×4)]=cos(π/3-π/4)+isin(π/3-π/4)=[cos(π/3)cos(π/4)+sin(π/3)sin(π/4)] +i[sin(π/3)cos(π/4)-sin(π/4)cos(π/3)]=[(1/2)(√2/2)+(√3/2)(√2/2))+i[(√3/2)(√2/2)-(√2/2)(1/2)]=(√2/4+√6/4)+i(√6/4-√2/4)=[(√2+√6)/4]+i[(√6-√2)/4]=[√2(1+√3)/4]+i[√2(√3-1)/4]=[2(1+√3)/(4√2)]+i[2(√3-1)/(4√2)]=(1+√3)/(2√2)+i[(√3-1)/(2√2)] ***cos(π/12)+isin(π/12)=e^(iπ/12) [By Euler’sformula] 2)cos(13π/12)+isin(13π/12)=cos(π+π/12)+isin(π+π/12)=-cos(π/12)-isin(π/12)=-{cos(π/12)+isin(π/12)}=-{[(1+√3)/(2√2)]+i[(√3-1)/(2√2)]}=i[(1-√3)/(2√2)]-(1+√3)/(2√2) ***cos(13π/12)+isin(13π/12)=-cos(π/12)-isin(π/12)=-[cos(π/12)+isin(π/12)]=e^(iπ/12)[ByEuler’s formula]

2011-10-31 18:20:16 補充：
改正：尾二行：

錯：=e^(iπ/12)

對：=-e^(iπ/12)

2011-11-01 11:26:06 補充：
Q1:為什麼是要用3X4!??2X6不行嗎!??

可以。但情況會變複雜（你可試試）

Q2:那接下來為什麼是要用減的!??

如此般可用差角公式計出準確值

Q3:那如果題目是說Cos(5π/6)+iSin(5π/6) 我該怎麼算!??

Cos(5π/6)+iSin(5π/6)=Cos(π-π/6)+iSin(π-π/6)=-Cos(π/6)+iSin(π/6)

可參考 http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Symmetry

恩～!!好謝謝!!
REX大師的意見`~!!

應該要重新了解複數極式的定義跟轉換吧!?

請問Cos(π/12)+iSin(π/12) =(√3+1)/(2√2) +i(√3－1)/(2√2)
能解釋一下嗎??
Sol
Cos(π/12)>0
Cos^2 (π/12)
=[Cos(π /6)+1]/2
=(√3/2+1)/2
=(2√3+4)/8
=(3+2√3+1)/8
=[(√3+1)/(2√2)]^2
So
Cos(π/12)= (√3+1)/(2√2)
Sin(π/12)>0
Sin^2 (π/12)
=[1－Cos(π /6)+1]/2
=(1－√3/2)/2
=(4－2√3)/8
=(3－2√3+1)/8
=[(√3－1)/(2√2)]^2
So
Sin(π/12)= (√3－1)/(2√2)
So
Cos(π/12)+iSin(π /12) =(√3+1)/(2√2) +i(√3－1)/(2√2)
那Cos (13π/12)+iSin(13π /12)的答案為何!??
Cos(13π/12)<0，Sin(13π /12)<0
Cos (13π/12) +iSin(13π /12)=－(√3+1)/(2√2)－i(√3－1)/(2√2)