三角問題(10分)

1. The person above me is the correct one. It is easy for me. I choose not to answer it.

2. I = sin 12d sin 24d sin 48d sin 84d
= cos 78d cos 66d cos 42d cos 6d
= 1/4 (cos 120d + cos 36d) (cos 72d + cos 60d)
= 1/4 (-cos 60d + cos 36d) (cos 72d + cos 60d)
= 1/16 (-1 + 2 cos 36d) (2 cos 72d + 1)
= 1/16 (4 cos 36d cos 72d + 2 cos 36d - 2 cos 72d - 1)
= 1/16 (4 cos 36d cos 72d + 2 cos 36d + 2 cos 108d - 1)
= 1/16 (4 cos 36d cos 72d + 4 cos 36d cos 72d - 1)
= 1/16 (8 sin 54d sin 18d - 1)

Since sin 5x = 16(sinx)^5 − 20(sinx)^3 + 5sinx (prove it by yourself)
If x = 18d, 5x = 90d
sin 5x = 1
Let y = sin 18d
1 = 16 y^5 − 20 y^3 + 5 y
16 y^5 − 20 y^3 + 5 y - 1 = 0
(y - 1)(16y^4 + 16y^3 - 4y^2 - 4y + 1) = 0
(y - 1)(4y^2 + 2y - 1)^2 = 0
4y^2 + 2y - 1 = 0

I = 1/16 (8 sin 54d sin 18d - 1)
= 1/2 (sin 54d sin 18d) - 1/16
= 1/2 (3 sin 18d - 4 [sin 18d]^3) sin 18d - 1/16
= 1/2 (3y^2 - 4y^4) - 1/16
= 1/2 (3y^2 - 4y^4) - 1/8 + 1/16
= 1/2 (3y^2 - 4y^4 - 1/4) + 1/16
= 1/8 (12y^2 - 16y^4 - 1) + 1/16
= -1/8 (16y^4 - 12y^2 + 1) + 1/16
= -1/8 (4y^2 + 2y - 1)(4y^2 - 2y - 1) + 1/16
= 1/16

1. (sin20)^2 + (cos50)^2 + sin20cos50
=(sin20)^2 + (sin40)^2 + sin20sin40
=(sin20+sin40)^2 - sin20sin40
={2sin[(20+40)/2] cos[(20-40)/2]}^2 -[cos(20-40)+cos(20+40)]/2
=(cos10)^2+ 1/4 - (cos20) /2
=1/2 (1+cos20) + 1/4 - (cos20)/2
=1/2+(cos20)/2 +1/4 - (cos20)/2
=1/2+1/4
=3/4

sorry,錯時想不到如何做第二題,遲些答覆你

2006-11-23 22:39:07 補充：
打錯字........「暫」時