Sin10asin8a+sin 8asin6a-sin4asin2a

Question

Sin10asin8a+sin 8asin6a-sin4asin2a

Answer 1

Sin8a*(sin10a+sin6a)-sin4a*sin2a=sin8a*2sin8acos2a-sin4asin2a=
=2sin²8acos2a-2sin²2acos2a=2cos2a*(sin²8a-sin²2a)=
=2cos2a*[(1-cos16a)/2-(1-cos4a)/2]=2cos2a*1/2(1-cos16a+1-cos4a)=
=cos2a(2-(cos16a+cos4a))=cos2a*(2-2cos10acos6a)=
=2cos2a*(1-cos10acos6a)