plz come and help

2006-11-13 12:37 am
Solve the following equations for 0 degree < (or) = x < (or) = 360 degree
a. 2 cos ^4 A - 2 sin ^4 A + cos 2 A = 1
b. sin 5 A / sin A - cos 5 A / cos A = 2
c. sin ^2 A + 1/2 sin ^2 2 A = 1

回答 (1)

2006-11-13 12:54 am
✔ 最佳答案
a.
2 cos ^4 A - 2 sin ^4 A + cos (2A) = 1
2 (cos ^4 A - sin ^4 A) + cos (2A) = 1
2 (cos² A - sin²A)(cos²A + sin²A) + cos (2A) = 1 [difference of two squares]
2 cos (2A) (1) + cos (2A) = 1 [Because cos² A - sin²A = cos(2A) and cos²A + sin²A = 1]
3 cos(2A) = 1
cos (2A) = 1/3
2A = 70.5288, 289.4712, 430.5288, 649.4712 [Because 0≦2A≦720]
A = 35.26, 144.74, 215.26, 324.74 (3 d. p.)

b.
sin (5A) / sin A - cos (5A) / cos A = 2
[sin (5A)cosA - cos(5A)sinA]/(sinA cosA) = 2 [Common denominator]
sin(5A-A)/(sinA cosA) = 2 [Compound Angle formula]
sin(4A)/(sinA cosA) = 2
2sin(2A)cos(2A)/(sinA cosA) = 2 [Double Angle formula]
4sinA cosA cos(2A)/(sinA cosA) = 2 [Double Angle formula]
4 cos(2A) = 2 [Cancelling common factors in numerator and denominator]
cos(2A) = 1/2
2A = 60, 300, 420, 660 [Because 0≦2A≦720]
A = 30, 150, 210, 330

c.
sin²A + 1/2 sin²(2A) = 1
sin²A + 1/2 [sin(2A)]² = 1
sin²A + 1/2 (2 sinA cosA)² = 1 [Double Angle formula]
sin²A + 1/2 (4 sin²A cos²A) = 1
sin²A + 2 sin²A cos²A = 1
sin²A + 2 sin²A (1 - sin²A) = 1 [Because cos²A + sin²A = 1]
-2 sin^4 A + 3 sin²A = 1
2 sin^4 A - 3 sin²A +1 = 0
(2 sin² A - 1)(sin²A - 1) = 0
sin²A = 1/2 or sin²A = 1
sinA = √(1/2) or sin A = -√(1/2) or sinA = 1 or sinA = -1
A = 45, 135 or A = 225, 315 or A = 90 or A = 270
Rearranging, A = 45, 90, 135, 225, 270, 315.
參考: Myself


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