This question is rather challenging for O level students, but it's fun :D
A, B and C are angles in a triangle.
Show that
(i) cosA + cosB + cosC = 1 + 4sin(A/2) sin(B/2) sin(C/2)
(ii) sin2A - sin2B + sin2C = 4 cos A sin B cos C
(iii) sin A cos B cos C + cos A sin B cos C + cos A cos B sin C = sin A sin B sin C.
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Answer:
Some notes:
Look at your formula sheet. The following identities are what we have to use
1) sin A + sin B = 2 sin ½(A+B) cos ½(A-B)
2) sin A - sin B = 2 cos ½(A+B) sin ½(A-B)
3) cos A + cos B = 2 cos ½(A+B) cos ½(A-B)
4) cos A - cos B = - 2 sin ½(A+B) sin ½(A-B)
Also, the following identity is used
sin (90° - x) = cos x
cos (90° - x) = sin x
A, B and C are angles in a triangle => A + B + C = 180°
(i) RHS = 1 + 4sin(A/2) sin(B/2) sin(C/2)
Note: A/2 = 90° - (B + C)/2
B/2 = 90° - (A + C)/2
C/2 = 90° - (A + B)/2
Hence, RHS = 1 + 4 sin {90° - (B + C)/2} sin {90° - (A + C)/2} sin {90° - (A + B)/2}
= 1 + 4 cos (B + C)/2 cos (A + C)/2 cos (A + B)/2
= 1 + 2 cos (B + C)/2 [ cos (2A+B+C)/2 + cos (C-B)/2 ]
= 1 + cos (A+B+C) + cos (-2A/2) + cos (2C/2) + cos (2B/2)
= 1 - 1 + cos A + cos B + cos C
= cos A + cos B + cos C
= LHS (proved)
(ii) LHS = sin2A - sin2B + sin2C
= 2 cos (A+B) sin (A-B) + 2 sin C cos C
= 2 cos (180° - C) sin (A-B) + 2 sin (180° - (A+B) ) cos C
= - 2 cos C ( sin (A-B) - sin (A+B) )
= - 2 cos C ( 2 cos A sin (-B) )
= 4 cos A sin B cos C (shown)
(iii) sin(A+B+C) = sin 180° = 0
Using the identity (or expanding yourself)
sin(A+B+C) = sin A cos B cos C + cos A sin B cos C + cos A cos B sin C - sin A sin B sin C
Thus, sin A cos B cos C + cos A sin B cos C + cos A cos B sin C = sin A sin B sin C (shown)
