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Use the formula: \[\sin(ax)\sin(bx)=\frac{1}{2}(\cos((a-b)x)-\cos((a+b)x)))\] So: \[\sin(9x)\sin(4x)=\frac{1}{2}(\cos((9-4)x)-\cos((9+4)x))=\frac{1}{2}(\cos(5x)-\cos(13x))\]
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