f.2數 急 (20)

1.cos(40－3θ) = sin(7θ+2)
Sol
cos(40－3θ) = sin(7θ+2)
cos(40－3θ) = cos(90－7θ－2)
40－3θ =360n+/-(90－7θ－2) n is integer
(1) 40－3θ =360n+(90－7θ－2)
4θ =360n+48
θ =90n+12
(2) 40－3θ =360m－(90－7θ－2)
40－3θ =360m-90+7θ+2
－10θ =360m－128
θ =－36m+12.8
θ =36n+12.8
Ans: θ =90n+12 or θ =36n+12.8 n isinteger

2. tan A = 5/3 (GIVEN)FIND 4sinA/(3sinA-cosA):
Sol
4sinA/(3sinA－cosA)
A=4[sinA/cosA]/[(3sinA－cosA)/cosA]
=4tanA/(3sinA/cosA－cosA/cosA)
=4tanA/(3tanA－1)
=[4*5/3]/(3*5/3－1)
=4*5/(15－3)
=20/12
=5/3

wrong method2
tan A = 5/3 =>sinA=5/sqrt(34),cosA=3/sqrt(34)
4sinA/(3sinA－cosA)
=[4*5/sqrt(34)]/[3*5/sqrt(34)－3/sqrt(34)]
=(4*5)/(3*5－3)
=20/12
=5/3
Both have the same answer ,but method 2 is wrong.
The reason is we do"nt know sinA=5/sqrt(34) or sinA=－5/sqrt(34)

1. cos(40°-3θ) = sin(7θ+2°)
cos(40°-3θ) =cos[90°-(7θ+2°)]
(40°-3θ) = [90°-(7θ+2°)]
40°-3θ = 90°-7θ-2°
-3θ+7θ= 90°-2°-40°
4θ= 48°
θ= 12°

2.tan A = 5/3
sinA/cosA=5/3
sinA=5
4sinA
=4(5)
=20

第一題比較特別，因為題目並沒有定出 θ, ( 40° - 3θ) 和　( 7θ+ 2° ) 的範圍，所以在以下過程中：

cos ( 40° - 3θ) = sin [ 90° - ( 40° - 3θ) ] = sin ( 50° + 3θ) = sin ( 7θ+ 2° )

除了 50° + 3θ= 7θ+ 2° 外，亦可以 50° + 3θ= 180° - ( 7θ+ 2° )
4θ= 48° 或 50° + 3θ= 178° - 7θ
θ= 12° 或 θ= 12.8° ( 0° < θ< 90° )

2009-05-24 18:03:20 補充：
而 7θ+ 2° = 7 X 12° + 2° = 86° ( < 90° )
但是，7θ+ 2° = 7 X 12.8° + 2° = 91.6° ( > 90° )


以上的討論只供參考，沒有所謂對錯的問題，因為是否接納 θ= 12.8° , 要看 ( 7θ+ 2° ) 的範圍而定

2009-05-24 19:12:46 補充：
既然是 '急', 就讓我來代 eelyw 解釋：


4 sin x / ( 3 sin x - cos x )
= 4 sin x / [ cos x ( 3 sinx / cos x - 1 ) ] <-----


再清楚看看分母一次：
3 sin x - cos x = cos x ( 3 sinx / cos x - 1 )


如果都唔明，拆右邊的括弧試試 ! !

1. cos(40°-3θ) = sin(7θ+2°)
cos40-3@=sin7@+sin2
cos40-cos3@=cos(90-7@)+cos78
-cos3@-cos(90-7@)=cos78-cos40
cos90=cos38
cos=52

2. 4sinA=4(5)
=20
3sinA-cosA = 3(5)-3
=12

1. cos ( 40 - 3x) = sin [90 - ( 40 - 3x)] = sin ( 50 + 3x)
sin ( 50 + 3x) = sin (7x + 2)
50 + 3x = 7x + 2
48 = 4x
x = 12.
2.
4 sin x/( 3 sin x - cos x)
= 4 sin x/[cos x( 3 sinx /cos x - 1)]
= 4 tan x/( 3 tan x - 1)
= 4(5/3)/[3(5/3) - 1]
= (20/3)/(5 - 1)
= (20/3)/4
= 5/3.