F4 A Maths - prove

(a) In the triangle ABC, by using Cosine Rule, we have
cos C = (a^2 + b^2 - c^2)/2ab and cos B = (a^2 + c^2 - b^2)/2ac

So we have
L.H.S.
= b cos C - c cos B
= b[(a^2 + b^2 - c^2)/2ab] - c[(a^2 + c^2 - b^2)/2ac]
= (a^2 + b^2 - c^2)/2a - (a^2 + c^2 - b^2)/2a
= [(a^2 + b^2 - c^2) - (a^2 + c^2 - b^2)]/2a
= (a^2 + b^2 - c^2 - a^2 - c^2 + b^2)/2a
= (b^2 - c^2)/a
= R.H.S.

(b) L.H.S.
= sin^2 A + sin^2 B - sin^2 C
= sin^2 A - sin^2 C + sin^2 B
= (sin A + sin C)(sin A - sinC) + sin^2 B
= [2 sin 0.5(A + C) cos 0.5(A - C)][2 cos 0.5(A + C) sin 0.5(A - C)]
+ sin^2 B ------ (by compound angle formulae)

= [2 sin 0.5(A + C) cos 0.5(A + C)][2 cos 0.5(A - C) sin 0.5(A - C)]
+ sin^2 {180 - (A + C)] ------ (because A + B + C = 180)

= sin (A + C) sin (A - C) + sin^2 (A + C)
= sin (A + C) [sin (A - C) + sin (A + C)]
= sin (180 - B) [2 sin 0.5(A - C + A + C) cos 0.5(A - C - A - C)] ------ (by compound angle formulae)
= sin B [2 sin A cos (-C)]
= 2 sin A sin B cos C ------ (because cos (-C) = cos C)
= R.H.S.