附加數學(請問)

1: Evaluate cos^2 1+cos^2 2 +cos^2 3 +...+ cos^2 88+ cos^2 89
= (cos^2 1 + cos^2 89) + (cos^2 2 + cos^2 88) + ... + (cos^2 44 + cos^2 46) + cos^2 45
= (sin^2 89 +cos^2 89) + (sin^2 88 + cos^2 88) + ... + (cos^2 46 + cos^2 46) + 1/2
= 89/2
2: Find the solutions between 0 and 360 of the equatons:
(a)cos^2θ = (1 + sinθ) / (1 + 2sinθ)
1- sin^2θ = (1 + sinθ) / (1 + 2sinθ)
(1 + sinθ)(1 - sinθ) = (1 + sinθ) / (1 + 2sinθ)
(1 + sinθ)(1 - sinθ) - (1 + sinθ) / (1 + 2sinθ) = 0
(1 + sinθ)[(1 - sinθ) - 1 / (1 + 2sinθ)] = 0
(1 + sinθ)[(1 - sinθ + 2sinθ - 2sin^2θ - 1) / (1 + 2sinθ)] = 0
(1 + sinθ)(sinθ - 2sin^2θ) = 0
sinθ(1 + sinθ)(1 - 2sinθ) = 0
sinθ = 0 or sinθ = -1 or sinθ = 1/2
θ = 0, 180, 360 OR θ = 270 OR θ = 30, 150
(b) 2sinθ + 2√2 sinθcosθ - √2cosθ = 1
2sinθ + 2√2 sinθcosθ - √2cosθ - 1 = 0
2sinθ - 1 + 2√2sinθcosθ - √2cosθ = 0
(2sinθ - 1) + √2cosθ(2sinθ - 1) = 0
(2sinθ - 1)(1 + √2cosθ) = 0
sinθ = 1/2 or cosθ = -1/√2 = -√2/2
θ = 30 or 150 OR θ = 135 or 225
3:This is a tetrahedron with AB=AC=BC=PB=PC=10cm,and AP=8cm
Find
(a) Let the angle between plane PBC and ABC be X
Let the mid-point of BC be D
Let the vertical line from P touches plane ABC at O
AD = (10)(√3/2) = 5√3
PD = (10)(√3/2) = 5√3
cosX = (PD^2 + AD^2 - AP^2) / [2(PD)(AD)]
cosX = (75 + 75 - 64)/150
cosX = 86/150
X = 55.02 degrees
(b) Let the angle between plane PAB and ABC be Y
Let E be a point on AB such that AB is perpendicular to PE
cos(PAB) = (AP^2 + AB^2 - PB^2) / [2(AP)(AB)]
= (64 + 100 - 100) / 160
= 64/160
cos(PAB) = 2/5
PAB = 66.42 degrees
PE = PAsinPAB = 8sinPAB = 7.332
PO = PDsinX = 5√3sinX = 7.096
sinY = PO/PE = 7.096/7.332
Y = 75.4 degrees

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