問幾題高一數學題目..(三角函數)

Note:度字省略
Q1:
原式= sin20*cos20*cos40*cos80/sin(20)
= (1/2) sin40*cos40*cos80/sin20
=(1/2)^2 sin80*cos80/sin20
= 1/8 sin160/sin20 = 1/8

Q2:
原式= sin(2π/5)+sin(4π/5)- sin(4π/5)- sin(2π/5)=0
Note: sinx= - sin(2π-x)

Q3:
原式=(cos4+i sin4)^10*(cos5+isin5)^6*(cos2-i sin2)^10
= (cos40+i sin40)*(cos30+i sin30)*[cos(-20)+i sin(-20)]
= cos(50)+i sin(50)

Q4: 設 z'表z之共軛複數
設 z= sin(π/18)+i cos(π/18)
=> z'= sin(π/18)- i cos(π/18), 且 z*z'= 1
原式= [ (1-z')/(1-z)]^9
=[(z*z'- z')/(1-z)]^9
=[z'(z-1)/(1-z)]^9
=- (z')^9
= - [ sin(π/18)- i cos(π/18)]^9
= - ( cos80- i sin80)^9 (度字省略)
= - ( cos720- i sin720)
= - 1

2009-06-14 23:37:59 補充：
Q3:少一個除號, 更正如下:
原式=(cos4+i sin4)^10*(cos5+isin5)^6 / (cos2-i sin2)^10
= (cos40+i sin40)*(cos30+i sin30) /[cos(-20)+i sin(-20)]
= cos(90)+i sin(90) = i

(1)
cos20度*cos40度*cos80度=?

令P=cos20cos40cos80
則psin20=sin20cos20cos40cos80
=(1/2)sin40cos40cos80
=(1/4)sin80cos80=(1/8)sin160=1/8sin20
psin20=1/8sin20
右sin20不為0
所以p=1/8

(2)
sin2π/5 +sin4π/5 +sin6π/5 +sin8π/5 =?
原式=2sin(3π/5)cos(π/5)+2sin(7π/5)cos(π/5)
=2cos(π/5)[ sin(3π/5)+sin(7π/5) ]
=2cos(π/5)[2sinπcos(2π/5)]
又sinπ=0
故所求=0

(3)
(cos4度+isin4度)^10 *(-sin265度-icos265度)^6 /(cos2度-isin2度)^10
由棣美弗定理
原式=(cos40+isin40)(cos30+isin30)/(cos358+isin358)^10
=(cos70+isin70)/(cos340+isin340)
=cos(-270)+isin(-270)
=i


(4){(1-sinπ/18+i cosπ/18) / (1-sinπ/18-i cosπ/18)}^9 =?
我不會= =
阿......樓上已經有人回答了嗎，真是抱歉沒幫上什麼忙

2009-06-14 23:58:00 補充：
第三題算錯了，應為(cos70+isin70)/[cos(-20)+isin(-20)]
=cos50+isin50才對= =

2009-06-15 00:00:26 補充：
啊阿阿，又錯了，是cos[70-(-20)]+isin[70-(-20)]
=cos90+isin90=i才對啦，原來的才是對的
對不起一直弄錯