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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! |
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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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