F.4 [A MATHS]

1) sinA + sin2A + sin3A = 0
2sin[(3A + A) / 2]cos[(3A - A) / 2] + sin2A = 0
2sin2AcosA + sin2A = 0
sin2A(2cosA + 1) = 0
sin2A = 0 or cosA = -0.5
2A = nπ + (-1)^n (0) or A = mπ + 2π/3 or mπ - 2π/3
A = 0.5nπ or mπ + 2π/3 or mπ - 2π/3, where m,n are integers

2) 4cos(A/2)cos(3A/2) = 1
4*0.5[cos(4A/2) + cos(2A/2)] = 1
cos2A + cosA = 0.5
2(cosA)^2 - 1 + cosA - 0.5 = 0
cosA = (-1 + √13)/4 or cosA = (-1 - √13)/4 (rej)
A = nπ + 0.861 or nπ - 0.861, where n are integers

3)(sinA)^3 - (cosA)^3 = sinA - cosA
(sinA - cosA)((sinA)^2 + sinAcosA + (cosA)^2) = sinA - cosA
sinA - cosA = 0 or ((sinA)^2 + sinAcosA + (cosA)^2) = 1
tanA = 1 or sinAcosA = 0
tanA = 1 or sinA = 0 or cosA = 0
A = mπ + π/4 or nπ or pπ + π/2 or pπ - π/2

2007-03-11 23:53:28 補充：
where m, n, p are integers for the answer of Q3

1)
sinA + sin2A + sin3A = 0
(sinA + sin 3A) + sin2A = 0
(2sin2AcosA) + sin2A = 0
sin2A(2cosA+1) = 0
sin2A = 0 or cosA = -1/2

sin2A = 0
2A = 180n
A = 90n (degree)

cosA = -1/2
A = 360n +/- 120

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2)
4cosA/2 cos3A/2 = 1
4[1/2(cos2A+cosA)] = 1
2(cos2A + cosA) = 1
2[2(cosA)^2 - 1 + cosA] = 1
4(cosA)^2 + 2cosA - 3 = 0
cosA = (-sqrt13-1)/4 ....rejected
or
cosA = (sqrt13-1)/4
A = 360n +/- 49.35 (degree)

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3)
sin^3 A - cos^3 A = sinA-cosA
(sinA)^3 - sinA = (cosA)^3 - cosA
sinA[(sinA)^2 - 1] = cosA[(cosA)^2 - 1]
sinA[-(cosA)^2] = cosA[-(sinA)^2]
sinA/cosA = (sinA)^2/(cosA)^2
tanA = (tanA)^2
(tanA)^2 - tanA = 0
tanA(tanA - 1) = 0
tanA = 0 or tanA = 1

tanA = 0
A = 180n (degree)

tanA = 1
A = 180n + 45 (degree)