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Problem:sin(8x)*cos(3x) - sin(6x)*cos(4x) = sin(2x)*cos(4x)Where I got to:\( sin(8x)*cos(3x) - sin(6x)*cos(4x) = {1 \over 2} * [ sin(11x) + sin(5x) - sin(10x) - sin(2x)]\)Via formula:\( sinx*cosy = {1 \over 2} * [ cos(x-y) - cos(x+y)]\)Any help is greatly appreciated, thank You in advance!

Answer»

sin 8x cos 3x - sin 6x cos 4x

= 2 sin 4x cos 4x cos 3x - 2sin 3x cos 3x cos 4x

(\(\because\) sin c- sin D = 2 cos\((\frac{C+D}2)\) sin \((\frac{C-D}2)\))

= 4 cos 3x cos 4x cos 7x/2 sin x/2



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