✔ 最佳答案
Let me finish Q2 and Q3 for you:
2) Looking into ab2cos A = bc2cos B and applying the sine formula, we have:
c/sin C = b/sin B
c2/sin2 C = b2/sin2 B
c2 = b2sin2 C/sin2 B
Also, a/b = sin A/sin B
∴ab2cos A = bc2cos B
ab2cos A = bcos B(b2sin2 C/sin2 B)
acos A = bsin2 Ccos B/sin2 B
a/b = (sin2 Ccos B)/(sin2 Bcos A)
sin A/sin B = (sin2 Ccos B)/(sin2 Bcos A)
sin A cos A sin B = sin2 C cos B
sin A(1/2)[sin (B+A) + sin (B-A)] = sin C(1/2)[sin (C+B) + sin (C-B)]
sin Asin (π-C) + sin Asin (B-A) = sin Csin (π-A) + sin Csin (C-B)
sin Asin C + sin Asin (B-A) = sin Csin A + sin Csin (C-B)
sin Asin (B-A) = sin Csin (C-B)
-2{cos[A+(B-A)]/2}{cos[A-(B-A)]/2} = -2{cos[C+(C-B)]/2}{cos[C-(C-B)]/2}
cos(B/2)cos[A-(B/2)] = cos(B/2)cos[C-(B/2)]
∴cos(B/2) = 0 (rejected since B < π) or
cos[A-(B/2)] = cos[C-(B/2)]
For 0 < A,B,C < π, A-(B/2) = C-(B/2)
∴ A = C
Redoing the same deductions for another pair of equality, we can obtain A = B or C = B and hence △ABC is an equilateral triangle.
3) Let me start with a(b cosC - c cosB) first and explain in the latter part.
a(b cosC - c cosB) = ab cos C - ac cos B
In △ABC, by the cosine formula:
b2 = a2 + c2 - 2ac cos B ... (1)
c2 = a2 + b2 - 2ab cos C ... (2)
(1) - (2) yields:
b2 - c2 = c2 - 2ac cos B - b2 + 2ab cos C
2(b2 - c2) = 2ab cos C - 2ac cos B
∴ab cos C - ac cos B = b2 - c2
(You may use a simple right-angled triangle with sides 3,4 and 5 to check against my result and the one you stated in the question.)
