附加數學 三角學 2題

1.

2(cosA)^2 + √3 - 2 = 2sinA - √3 sinA

2[1 - (sinA)^2] + √3 - 2 = (2 - √3)sinA

0 = 2(sinA)^2 + (2 - √3)sinA - √3

0 = (2sinA - √3)(sinA + 1)

2sinA - √3 = 0 或 sinA + 1 = 0

sinA = √3 / 2 或 sinA = -1

A = 60° 或 120° 或 A = 270° [sinA於第1和2象限為正.]


2.

8(sinA)^2 - 9(cosA)^2 = 6sinAcosA

8(sinA)^2 - 6sinAcosA - 9(cosA)^2 = 0

(2sinA - 3cosA)(4sinA + 3cosA) = 0

2sinA - 3cosA = 0 或 4sinA + 3cosA = 0

2sinA = 3cosA 或 4sinA = -3cosA

tanA = 3/2 或 tanA = -3/4 [當cosA為0時, 沒解, 因此cosA不等於0, 全式可除以cosA.]

A = 56.31° 或 236.31° 或 A =143.13° 或 323.13° [tanA於第1和3象限為正, 於第2和4象限為負.]


請留意做法, 學會了下次就可以自己做了, 有不明白可以用電郵再問.

1. 設2 cos² A +√3 - 2 = 2 sin A - √3 sin A ，其中0°≤ A ≤ 360°，設求A的值。
(√3即係開方3)
2 cos² A +√3 - 2 = 2 sin A - √3 sin A
=> 2 (1 -sin² A) + √3 - 2 = (2 - √3) sin A
=> 2 - 2 sin² A + √3 - 2 = (2 - √3) sin A
=> 2 sin² A - √3 + (2 - √3) sin A = 0
=> ( sin A + 1) ( 2 sin A - √3) = 0
Therefore, sin A = -1 or √3/2
When sin A = -1, A = 270°
When sin A = √3/2, A = 60° or 120°
Hence, A = 60°, 120° or 270° (Answer)

2. 設 8 sin² A - 9 cos² A = 6 sin A cos A。求 tan A的值
8 sin² A - 9 cos² A = 6 sin A cos A
=> 8 tan A - 9 cot A = 6 ..... (Divide both side by sin A cos A)
=> 8 tan² A - 9 = 6 tan A ..... (Multiply both sides by tan A)
=> 8 tan² A - 6 tan A - 9 = 0
=> (2 tan A - 3) (4 tan A + 3) = 0
Therefore tan A = 3/2 = 1.5 or -3/4 = -0.75 (Answer)

若0° ≤ A ≤ 360°，試求A的值。
When tan A = 3/2, A ~= 56° 18' 36" or 236° 18' 36"
When tan A = -3/4, A ~= 143° 07' 48" or 323° 07' 48"
Hence, A ~= 56° 18' 36", 236° 18' 36", 143° 07' 48" & 323° 07' 48" (Answer)

1. 2 cos^2 A +√3 - 2 = 2 sin A - √3 sin A
=> 2(1 - sin^2 A) +√3 - 2 = 2 sin A - √3 sin A
=> 2sin^2 A + (2 - √3)sin A - √3 = 0
=> delta of the above quadratic equation = (2 - √3)^2 - 4x2x(- √3)
= 4 - 4√3 + 3 + 8√3
= 4 + 4√3 + 3
= (2 + √3)^2
Then, sin A = [-(2 - √3)+√delta]/(2x2) or [-(2 - √3)-√delta]/(2x2)
= [-(2 - √3)+(2+ √3)]/(2x2) or [-(2 - √3)-(2+ √3)]/(2x2)
= √3/2 or -1
For sin A = √3/2, A = 60° or 120° for 0°<=A<=360°
For sin A = -1, A = 270° for 0°<=A<=360°
In conclusion, A = 60°, 120° or 270°

2. 8 sin^2 A - 9 cos^2 A = 6 sin A cos A
=> 8 sin^2 A - 6 sin A cos A - 9 cos^2 A = 0
=> (2 sin A - 3 cos A)(4 sin A + 3 cos A) = 0
=> 2 sin A - 3 cos A = 0 or 4 sin A + 3 cos A = 0
=> tan A = 3/2 or tan A = -3/4 (tan A = sin A / cos A)
=> A = 56.31° or 236.31° or A = 143.13° or 323.13°

1.設2 cos^2 A +√3 - 2 = 2 sin A - √3 sin A ，其中0°<=A<=360°，設求A的值。
(√3即係開方3)
2 cos^2 A +√3 - 2 = 2 sin A - √3 sin A
2 (1 – sin 2 A) + Ö3 – 2 = 2 sin A - √3 sin A
0 = 2 sin 2 A + 2 sin A - Ö3 sin A – 2
0 = 2 sin 2 A + (2 - Ö3) sin A – 2
???
0 = (sin A – 1) (sin A + 2) ????????????, any mistake in the question?
sin A – 1 = 0 or sin A + 2 = 0
sin A = 1 or sin A = -2 (rej)
A = 90°

2.設8 sin^2 A - 9 cos^2 A = 6 sin A cos A。
求tan A的值
若0°<=A<=360°，試求A的值

8 sin^2 A - 9 cos^2 A = 6 sin A cos A
8 tan A sin A – 9 cos A = 6 sin A
8 tan A – 9/tan A = 6
8 tan 2 A – 9 = 6 tan A
8 tan 2 A – 6 tan A – 9 = 0
(2 tan A + 3) (4 tan A – 3) = 0
2 tan A + 3 = 0 or 4 tan A – 3 = 0
tan A = - 3/2 or tan A = ¾
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