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標題:
f.4 a maths
發問:
tanA and cotA are the roots of the equation x^2-(4√3/3)x + 1=0 a)Without calculating the values of tanA and cotA, find the value of tanA+cotA. b)By the result of(a), find the value of sin2A. c)Find the general solution of A.
最佳解答:
(a) tanA + cotA = sum of root = -b / a = 4√3 / 3 (b) tanA + cotA = sinA/cosA + cosA/sinA tanA + cotA = (sin^2A + cos^2A) / sinAcosA 4√3 / 3 = 2 / 2sinAcosA 4√3 / 3 = 2 / sin2A sin2A = 6 / 4√3 = 3 / 2√3 = √3 / 2 (c) sin2A = √3 / 2 2A = 180n + (-1)^n * 60 A = 90n + (-1)^n * 30
其他解答:
The answer is 圖片參考：http://hk.geocities.com/namsm4e/P68.jpg|||||a) Use sum of roots tanA + cotA = 4√3/3 ******************** b) tanA + cotA = 4√3/3 sinA/cosA + cosA/sinA = 4√3/3 (sin2A + cos2A)/(sinAcosA) = 4√3/3 1/(sinAcosA) = 4√3/3 sinAcosA = 3/(4√3) = √3/4 2sinAcosA = √3/2 sin2A = √3/2 ************************************* c) sin2A = √3/2 2A = n(180o) ± 60o A = n(90o) ± 30o 2007-03-27 00:15:28 補充： Sorry, Part c is wrong !!!It should be n(90o) + (-1)^n (30o)
f.4 a maths
發問:
tanA and cotA are the roots of the equation x^2-(4√3/3)x + 1=0 a)Without calculating the values of tanA and cotA, find the value of tanA+cotA. b)By the result of(a), find the value of sin2A. c)Find the general solution of A.
最佳解答:
(a) tanA + cotA = sum of root = -b / a = 4√3 / 3 (b) tanA + cotA = sinA/cosA + cosA/sinA tanA + cotA = (sin^2A + cos^2A) / sinAcosA 4√3 / 3 = 2 / 2sinAcosA 4√3 / 3 = 2 / sin2A sin2A = 6 / 4√3 = 3 / 2√3 = √3 / 2 (c) sin2A = √3 / 2 2A = 180n + (-1)^n * 60 A = 90n + (-1)^n * 30
其他解答:
The answer is 圖片參考：http://hk.geocities.com/namsm4e/P68.jpg|||||a) Use sum of roots tanA + cotA = 4√3/3 ******************** b) tanA + cotA = 4√3/3 sinA/cosA + cosA/sinA = 4√3/3 (sin2A + cos2A)/(sinAcosA) = 4√3/3 1/(sinAcosA) = 4√3/3 sinAcosA = 3/(4√3) = √3/4 2sinAcosA = √3/2 sin2A = √3/2 ************************************* c) sin2A = √3/2 2A = n(180o) ± 60o A = n(90o) ± 30o 2007-03-27 00:15:28 補充： Sorry, Part c is wrong !!!It should be n(90o) + (-1)^n (30o)
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