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.
f.4 a maths
2007-03-27 7:34 am
回答 (3)
2007-03-27 8:10 am
✔ 最佳答案
(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
2007-03-27 8:21 am
2007-03-27 8:13 am
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)
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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