2題A. Maths Compound Angles問題

1)
As A+B+C = 180°,
A = 180° - (B+C)
sinA = sin (B+C) = sinBcosC + cosBsinC
sin²A = sin²Bcos²C + cos²Bsin²C + 2sinBcosBsinCcosC
and
cosA = -cos(B+C) = sinBsinC - cosBcosC

LHS = (k²sin²B + k²sin²C - k²sin²A)/2k²sinBsinC
= (sin²B + sin²C - sin²A)/2sinBsinC
= (sin²B+sin²C-sin²Bcos²C-cos²Bsin²C-2sinBcosBsinCcosC)/2sinBsinC
= [sin²B(1-cos²C)+sin²C(1-cos²B)-2sinBcosBsinCcosC]/2sinBsinC
= (sin²Bsin²C+sin²Csin²B-2sinBcosBsinCcosC)/2sinBsinC
= (2sin²Bsin²C-2sinBcosBsinCcosC)/2sinBsinC
= sinBsinC(sinBsinC-cosBcosC)/sinBsinC
= sinBsinC-cosBcosC
= cosA
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2a)
cos(x+β) = Acos(x-β)
cos(x-β) + cos(x+β) = cos(x-β) + Acos(x-β) = [1+A]cos(x-β)
2cosxcosβ = [1+A]cos(x-β) ---------- (1)

cos(x+β) = Acos(x-β)
cos(x-β) - cos(x+β) = cos(x-β) - Acos(x-β) = [1-A]cos(x-β)
2sinxsinβ = [1-A]cos(x-β) ------------- (2)

(2) / (1)
sinxsinβ/cosxcosβ = (1-A)/(1+A)
tanxtanβ = (1-A)/(1+A)
tan x = [(1-A)/(1+A)]*(1/tanβ)

2b)
Consider β = 30° and A = 1/2
By a, tan x = [(1-1/2)/(1+1/2)]*(1/tan 30°) = 1/sqrt 3
x = 30° or 210°

1.
(b²+c²-a²) - 2bc cosA
= k²sin²B + k²sin²C - k² + k²cos²A - 2k²sinBsinC cosA
= k²sin²B + k²sin²C - k² + k²cos²(B+C) - 2k²sinBsinC cos(B+C)
= k²sin²B + k²sin²C - k² + k²(cos(B)cos(C) - sin(B)sin(C))² - 2k²sinBsinC(cos(B)cos(C) - sin(B)sin(C))
= k²sin²B + k²sin²C - k² + k²cos²(B)cos²(C) - k²sin²(B)sin²(C)
= k²sin²B - k²sin²(B)sin²(C) + k²sin²C - k² + k²cos²(B)cos²(C)
= k²sin²Bcos²C - k²cos²C + k²cos²(B)cos²(C)
= k²sin²Bcos²C + k²cos²(B)cos²(C) - k²cos²C
= k²cos²C - k²cos²C
= 0

2. (a)
(1-A)/(1+A)
= (cos(x-β) - cos(x+β)) / (cos(x-β) + cos(x+β))
= 2 sin x sin β/ 2 cos x cos β
= tan x tan β

2. (b)
cos(x+30°) = [cos(x-30°)]/2
From (a), tan x = 1/3 . 1/tan 30° = 1/3^(0.5)
so x = 30° or 210°