F.5 trigonometry\\\\\\\\\\

(a) (sin A + cos A)2 = sin2 A + cos2 A + 2 sin A cos A = 1 + sin 2A

Similarly:

(sin B + cos B)2 = 1 + sin 2B

(sin C + cos C)2 = 1 + sin 2C

(sin A + cos A)2 - (sin B + cos B)2 - (sin C + cos C)2 = (1 + sin 2A) - (1 + sin 2B) - (1 + sin 2C)

= sin 2A - sin 2B - sin 2C - 1

(b1) (sin A + cos A)2 - (sin B + cos B)2 - (sin C + cos C)2 = sin 2A - sin 2B - sin 2C - 1

= 2 sin (A - B) cos (A + B) - 2 sin C cos C - 1

= - 2 sin (A - B) cos C - 2 sin C cos C - 1

= - 2 cos C [sin (A - B) + sin C] - 1

= - 4 cos C sin [(A - B + C)/2] cos [(A - B - C)/2] - 1

= - 4 cos C sin (90 - B) cos (A - 90) - 1

= - 4 sin A cos B cos C - 1

(b) A = 90:

(sin A + cos A)2 - (sin B + cos B)2 - (sin C + cos C)2 = - 4 sin A cos B cos C - 1

1 - (sin B + cos B)2 - (sin C + cos C)2 = - 4 cos B cos C - 1

- 4 cos B cos C = 2 - (sin B + cos B)2 - (sin C + cos C)2

0 <= (sin B + cos B)2 , (sin C + cos C)2 <= 2

Hence

2 >= - 4 cos B cos C >= -2

-1/2 <= cos B cos C <= 1/2