# Sin4x-sin8x using a sum to product to rewrite the expression

2 answers

## Answers

sin(4x) - sin(8x)sin(2(2x)) - sin(2(4x))2sin(2x)cos(2x) - 2sin(4x)cos(4x)2[sin(x)cos(x)][cos²(x) - sin²(x)] - [2sin(2(2x))cos(2(2x))]{2[sin(x)cos³(x) - sin³(x)cos(x)]} - [2[2sin(2x)cos(2x)][cos²(x) - sin²(x)]][2sin(x)cos³(x) - 2sin³(x)cos(x)] - {[2[2[sin(x)cos(x)][cos²(x) - sin²(x)]][cos²(x) - sin²(x)]}[2sin(x)cos³(x) - 2sin³(x)cos(x)] - [2[2[sin(x)cos(x)][cos⁴(x) - 2cos²(x)sin²(x) - sin⁴(x)]]]][2sin(x)cos³(x) - 2sin³(x)cos(x)] - [2[2[sin(x)cos⁵(x) - 2sin³(x)cos³(x) + sin⁵(x)cos(x)]]]2sin(x)cos³(x) - 2sin³(x)cos(x) - 4sin(x)cos⁵(x) + 8sin³(x)cos³(x) - 4sin⁵(x)cos(x)

Hello,sin(a)-sin(b)=2cos((a+b)/2)sin((a-b)/2)sin 4x-sin 8x=-(sin 8x-sin 4x)=-2cos 6x sin2x

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