Buktikan rumus-rumus trigonometri ini! sin α sin β = 1/2 [cos (α - β) - cos (α + β)] sin α cos β = 1/2 [sin (α - β) + sin (α + β)] cos α cos β = 1/2 [cos (α - β

1) Add vertically the first and second equations to get:
2 sin A cos B = sin (A + B) + sin (A - B)
Divide by 2 in both sides,
sin A cos B = 1/2[sin (A + B) + sin (A - B)]

2) If we loop out the way in first solution, in other words, we substract it, hence we get
2 cos A sin B = sin (A + B) - sin (A - B)
Divide by 2 in both sides,
cos A sin B = 1/2 [sin (A + B) - sin (A - B)]

3) Add vertically the third and fourth equations,
2 cos A cos B = cos (A + B) + cos (A - B)
cos A cos B = 1/2[cos (A+B) + cos (A-B)]

4) Substract vertically the third and fourth equations,
-2 sin A sin B = cos (A + B) - cos (A - B)
sin A sin B = -1/2 [cos (A+B) - cos (A-B)]

Hence all the cases simply proved by using sum and difference formulas

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