F.4 a.maths2條(複角)

如下:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Aug07/Crazytrigo19.jpg


圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Aug07/Crazytrigo20.jpg

1(a)解關於x的方程ax2+2x-a=0
ax^2+2x-a=0
x=-2±√4+4a^2 /2a
x=-1±√1+a^2

1(b)試以tan(2m)表tan(m)
tan2m=2tanm/1-tan^2 m
tan2m- tan2mtan^2 m=2tanm
(tan2m)tan^2 m+2tanm-(tan2m)=0

1(c)由此,求tan22.5度的值,答案以根式表示
(tan2x22.5)tan222.5+2tan22.5-(tan2x22.5)=0
tan22.5=-1±√1+tan^2 45
tan22.5=-1±√2

2.已知y=sinx+4cosx,其中0度<=x<=360度(360度大過等於x,x大過等於0度),當x=a時,有極大值,當x=b時,y有極小值,求sina,sinb的值
dy/dx=cosx-4sinx
when dy/dx=0,
cosx-4sinx=0
1-4tanx=0
tanx=1/4
x=14 or 194

d^2y/dx^2=-sinx-4sinx
when x=14, d^2y/dx^2<0
x=14, y is the maximum.
So, a=14
sin14=0.242

when x=194, d^2y/dx^2>0
x=14, y is the minimum.
So, b=194
sin194=-0.242

1a. ax^2 + 2x - a = 0
x = { -2 ±√ [2^2 - 4(a)(-a) ] }/ 2a
　= [ -2 ± 2√ (1 + a^2) ] / 2a
　= [-1 ±√ (1 + a^2) ] / a

1b. tan (2m) = 2 tan(m) / [ 1 - tan^2(m) ]
tan (2m) - tan(2m) tan^2(m) = 2 tan(m)
tan (2m) tan^2 (m) + 2 tan(m) - tan (2m) = 0

Let x = tan (m) and a = tan (2m)
By (a), x = [-1 ±√ (1 + a^2) ] / a

Hence, tan(m) = {-1 ±√ [ 1 + tan^2 (2m) } / tan (2m)

1c. By (b), tan 22.5
= [-1 + √ ( 1 + tan45) ] / (tan45)　(∵tan22.5 >0)
= (-1 + √2) / 1
= √2 - 1

2. Let y = r cos (x - α)
i.e. y = r cosx cosα + r sinx sinα

Hence,
r cosα = 4 ... (1)
r sinα = 1 ... (2)

(2) / (1):　tanα = 1/4
α = 14.03624347

When x=a, r cos (x - 14.03624347) is maximum
So, a - 14.03624347 = 0 or 360(rejected ∵x <= 360)
a = 14.03624347
sin a = 0.242535625 = 0.24 (corr. to 3 sig. fig.)

When x=b, r cos (x - 14.03624347) is minimum.
So b - 14.03624347 = 180
b = 194.03624347
sin b = -0.242535625 = -0.24 (corr. to 3 sig. fig.)

2007-08-10 14:36:45 補充：
Sorry, I have corrected to 2 sig. fig. only.The answer should be:sin a = 0.243 (corr. to 3 sig. fig.)and sin b = -0.243 (corr. to 3 sig. fig.)