SinAsin2A+sin3Asin6A-sin4Asin5A=0
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First of all the question is not properly written like what is asked to do. If it's asked to prove that then follow the below steps:
SinAsin2A+sin3Asin6A-sin4Asin5A
= 0.5(cos(2A-A) - cos(2A+A)) + 0.5(cos(6A-3A) - cos(6A+3A)) - 0.5(cos(5A-4A) - cos(5A+4A))
= 0.5cosA - 0.5cos3A + 0.5cos3A - 0.5cos9A - 0.5cosA + 0.5cos9A
= 0 (as all gets cancelled).
SinAsin2A+sin3Asin6A-sin4Asin5A
= 0.5(cos(2A-A) - cos(2A+A)) + 0.5(cos(6A-3A) - cos(6A+3A)) - 0.5(cos(5A-4A) - cos(5A+4A))
= 0.5cosA - 0.5cos3A + 0.5cos3A - 0.5cos9A - 0.5cosA + 0.5cos9A
= 0 (as all gets cancelled).
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