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

by

= sinA sinB = LS

= 1/2 * (2 * sinA sinB)

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))

= sinA sinB = LS

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

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))

This post is last updated on hrtanswers.com at Date : 1st of September – 2022

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