sinAsinB = 1/2[cos(A-B)-cos(A+B)]?

How can i prove this union?

sinA * sinB = 1/2 * [cos(A – B) – cos(A + B)]

RS = 1/2 * [cosA cosB + sinA sinB – (cosA cosB – sinA sinB)]

= 1/2 * (cosA cosB + sinA sinB – cosA cosB + sinA sinB)

= 1/2 * (2 * sinA sinB)

= sinA sinB = LS

1/2[cos(A-B)-cos(A+B)]
=1/2[(cosAcosB+sinAsinB)-(cosAcosB-sinAsinB)]
=1/2[2sinAsinB]
=sinAsinB

cos(A-B)=cos(A)*cos(B)+sin(A)*sin(B)
cos(A+B)=cos(A)*cos(B)-sin(A)*sin(B)
cos(A-B)-cos(A+B)= cos(A)*cos(B)+sin(A)*sin(B)-
(cos(A)*cos(B)-sin(A)*sin(B))=
2*sin(A)*sin(B)
So sin(A)*sin(B)=0.5(cos(A-B)-cos(A+B))

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