in terms of complex numbers
let z=cosθ+isinθ
then
2cosθ=z+z12isinθ=z−z1
then
(2cosθ)3(2isinθ)4=(z+z1)3(z−z1)4128cos3θsin4θ=(z−z1)(z2−z21)3128cos3θsin4θ=(z−z1)(z6−3z2+z23−z61)128cos3θsin4θ=(z7+z71)−(z5+z51)−3(z3+z31)+3(z+z1)
substituting back:
128cos3θsin4θ=2cos7θ−2cos5θ−3(2cos3θ)+3(2cosθ)
dividing through by 128
cos3θsin4θ=641[cos7θ−cos5θ−3cos3θ+3cosθ]
cos3θsin4θ=641[3cosθ−3cos3θ−cos5θ+cos7θ]
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