Biến đổi thành tổng: $A=4sin3asin2asina$ $B=4sin3asin2acosa$

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1. $A=4\sin2a(\sin3a.\sin a)$

   $=4.\sin2a.\dfrac{-1}{2}\Big(\cos4a-\cos2a\Big)$

   $=-2\sin2a.(\cos4a-\cos2a)$

   $=-2\sin2a.\cos4a+2\sin2a.\cos2a$

   $=-2.\dfrac{1}{2}\Big(\sin6a+\sin(-2a)\Big)+\sin4a$

   $=-\sin6a+\sin2a+\sin4a$

   $=\sin2a+\sin4a-\sin6a$

   $B=4\sin2a.(\sin3a.\cos a)$

   $=4\sin2a.\dfrac{1}{2}\Big(\sin4a+\sin2a\Big)$

   $=2\sin4a.\sin2a+2\sin^22a$

   $=-(\cos6a-\cos2a)+2.\dfrac{1-\cos4a}{2}$

   $=-\cos6a+\cos2a+2-2\cos4a$

   $=\cos2a-2\cos4a-\cos6a+2$

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