Asked by HIRDESH

SinAsin2A+sin3Asin6A/sinAcos2A+sin3Acos6A
Answered by Steve
using your product-to-sum formulas,
sinA sin2A = 1/2 (cosA-cos3A)
sin3A sin6A = 1/2 (cos3A-cos9A)
and so on ...

so your fraction now is

(cosA-cos3A)+(cos3A-cos9A)
-------------------------------
(cos3A-sinA)+(sin9A-sin3A)

= (cosA-cos9A)/(sin9A-sinA)

now go back the other way using sun-to-product formulas, and simplify
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