2條A. Maths Compound Angles問題

Q.1(i)
R.H.S. = cos³2A
= (cos²A - sin²A) ³
= (cosA)^6 - 3 (cosA)^4 sin²A + 3 cos²A (sinA)^4 - (sinA)^6
= (cosA)^6 - 3 (cosA)^4 • (1 - cos²A) + 3 (1 - sin²A) • (sinA)^4 - (sinA)^6
= 4 (cosA)^6 - 3 (cosA)^4 + 3 (sinA)^4 - 4 (sinA)^6
= cos³A (4cos³A - 3cosA) + sin³A (3sinA - 4sin³A)
= cos3A cos³A + sin3A sin³A = L.H.S.

Q.1(ii)
L.H.S. = (cos3A + 2cos2A + 7cosA + 6) / (8cosA - 2sinA sin2A)
= [ (4cos³A - 3cosA) + 2 (2cos²A - 1) + 7cosA + 6 ] / [8cosA - 2sinA • (2sinAcosA)]
= (4cos³A + 4cos²A + 4cosA + 4) / [8cosA - 4cosA • (1 - cos²A)]
= [(4cos³A + 4cosA) + (4cos²A + 4)] / (4cos³A + 4cosA)
= 1 + 1/cosA
= 1 + secA

Q.2
x³ + 3(√3)x² - 3x - (√3) = 0 .................<1>
Putting x = tanA,
tan³A + 3(√3)tan²A - 3tanA - (√3) = 0
tan³A - 3tanA = - 3(√3)tan²A + (√3)
tan³A - 3tanA = (√3) (1 - 3tan²A) .......... <2>

Case 1: If 1 - 3tan²A = 0,
Then,
x = tanA = ±(√3)
Putting into <1>,
L.H.S. = x³ - 3x + 3(√3)x² - (√3)
= x • (x² - 3) + 3(√3)x² - (√3)
= x • (3 - 3) + 3(√3) • 3 - (√3)
= 0 + 9(√3) - (√3)
= 8(√3) ≠ 0 = R.H.S.
Thus, no solution.

Case 2: If 1 - 3tan²A ≠ 0
Then,
from <2>,
(tan³A - 3tanA) / (1 - 3tan²A) = (√3)
tan3A = (√3)
3A = -2π/3, π/3, 4π/3 ........ for -π/2 < A < π/2, i.e. -3π/2 < 3A < 3π/2
A = -2π/9, π/9, 4π/9
tanA = tan(-2π/9), tan(π/9), tan(4π/9) ≈ -0.84, +0.36, +5.67 , to 2 decimal places.

Combining the cases,
x = tanA ≈ -0.84, +0.36, +5.67 , to 2 decimal places.

2007-01-17 23:49:57 補充：
A careless mistake was made. Corrected here:Case 2 of Q.2 should be(tan³A - 3tanA) / (1 - 3tan²A) = (√3)- tan3A =√33A = +2π/3, -π/3, -4π/3Thus, x = tanA ≈ +0.84, -0.36, -5.67.Note: -π/2 ＜ A ＜ π/2, as this is the principle domain of "tan A"So, -3π/2 ＜ 3A ＜ 3π/2.

1. (i) sin3Asin³A+cos3Acos³A
= (3 sin A - 4 sin³A)sin³A + (4cos³A - 3cos A)cos³A
= 3 (sin A)^4 - 4(sin A)^6 + 4(cos A)^6 - 3(cos A)^4
= 3[(sin A)^4 - (cos A)^4] - 4[(sin A)^6 - (cos A)^6]
= 3(sin²A + cos²A)(sin²A - cos²A) - 4(sin³A + cos³A)(sin³A - cos³A)
= - 3 cos 2A - 4(sin A + cos A)(sin²A - sin Acos A + cos²A)(sin A - cos A)(sin²A + sin Acos A + cos²A)
= -3 cos 2A - 4(sin A + cos A)(1 - sin A cos A)(sin A - cos A)(1 + sin A cos A)
= -3 cos 2A - 4(sin²A - cos ²A)(1 - sin²Acos²A)
= -3 cos 2A + 4 cos 2A (1 - sin²Acos²A)
= cos 2A[4 - sin²2A - 3]
= cos 2A(1 - sin²2A)
= cos 2A(cos²2A)
= cos³2A.

(ii) (cos3A + 2cos2A + 7cosA + 6)/(8cosA - 2sinAsin2A)
= (4cos³A - 3cosA + 2(2cos²A-1) + 7cos A + 6)/(8 cos A - 4sin²Acos A)
= (4cos³A + 4cos²A + 4cos A + 4)/[4cos A(2 - sin²A)]
= (cos³A + cos²A + cos A + 1)/[cos A(2 - sin²A)]
= [cos A(cos²A + 1) + cos²A + 1]/[cos A(1 + cos²A)]
= (cos²A+1)(cos A+1)/[cos A(1 + cos²A)]
= (cos A + 1)/cos A..........{As cos²A =/= -1 for all real values of A}
= 1 + 1/cos A
= 1 + sec A.

2. x = tan A
x³ + 3(√3)x² - 3x - (√3) = 0
tan³A + 3(√3)tan²A - 3tan A - (√3) = 0
3tan A - tan³A = (√3)(3tan²A - 1)
(3tan A - tan³A)/(1-3tan²A) = -√3
tan 3A = -√3
3A = 120, 300, 480, 660, 840, 1020 degrees
A = 40, 100, 160, 220, 280 or 340 degrees.
x = tan A
= 0.84, -5.67, -0.36, 0.84, -5.67 or -0.36
So x = -5.67, -0.36 or 0.84.



2007-01-17 23:05:51 補充：
In order to transform 3tan A - tan³A = (√3)(3tan²A - 1) to(3tan A - tan³A)/(1-3tan²A) = -√3,1 - 3tan²A =/= 0tan²A =/= 0tan A =/= 0x =/= 0So if the solution(tan A) by solving (3tan A - tan³A)/(1-3tan²A) = -√3 is 0, it has to be rejected.

2007-01-18 00:57:48 補充：
上面補充有錯, sorry.1 - 3tan²A =/= 0應該是tan A =/= ± 1/√3x =/= ±0.58.所以假若最後答案出現x = ±0.58就要reject.