( root2 )cosA=cosB+cos3B...........(1)(root2 )sinA=sinB-sin3B................(2)
Squaring & adding (1)&(2)& simplifying the eq. in terms of cos B we get
3cos4B+cos2B-2=0
Neglecting the -ve value of cos2B we get cos2B=2/3
therefore by using suitable formulae we get
sin 4B=4(root2)/9
As solved by me earlier
sin(A-B)=sin4B/4root2
=1/9
Am i correct?
Moderation Team
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