omckyyo的部落格

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plz come and help

發問:

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

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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.

其他解答:7638E7481407D16B