Answer to Question #114401 in Complex Analysis for Mcsheos

Question #114401
express cos^(3)tetasin^(4)teta in terms of multiple angle
1
Expert's answer
2020-05-08T19:00:49-0400

in terms of complex numbers

let z=cosθ+isinθz=\cos\theta+i\sin\theta

then

2cosθ=z+1z2isinθ=z1z2\cos\theta=z+\frac{1}{z}\\2i\sin\theta=z-\frac{1}{z}

then

(2cosθ)3(2isinθ)4=(z+1z)3(z1z)4128cos3θsin4θ=(z1z)(z21z2)3128cos3θsin4θ=(z1z)(z63z2+3z21z6)128cos3θsin4θ=(z7+1z7)(z5+1z5)3(z3+1z3)+3(z+1z)(2\cos\theta)^{3}(2i\sin\theta)^{4}=(z+\frac{1}{z})^{3}(z-\frac{1}{z})^{4}\\128\cos^{3}\theta\sin^{4}\theta=(z-\frac{1}{z})(z^{2}-\frac{1}{z^{2}})^{3}\\128\cos^{3}\theta\sin^{4}\theta=(z-\frac{1}{z})(z^{6}-3z^{2}+\frac{3}{z^{2}}-\frac{1}{z^{6}})\\128\cos^{3}\theta\sin^{4}\theta=(z^{7}+\frac{1}{z^{7}})-(z^{5}+\frac{1}{z^{5}})-3(z^{3}+\frac{1}{z^{3}})+3(z+\frac{1}{z})

substituting back:

128cos3θsin4θ=2cos7θ2cos5θ3(2cos3θ)+3(2cosθ)128\cos^{3}\theta\sin^{4}\theta=2\cos7\theta-2\cos5\theta-3(2\cos3\theta)+3(2\cos\theta)


dividing through by 128


cos3θsin4θ=164[cos7θcos5θ3cos3θ+3cosθ]\\\cos^{3}\theta\sin^{4}\theta=\frac{1}{64}[\cos7\theta-\cos5\theta-3\cos3\theta+3\cos\theta]


cos3θsin4θ=164[3cosθ3cos3θcos5θ+cos7θ]\cos^{3}\theta\sin^{4}\theta=\frac{1}{64}[3\cos\theta-3\cos3\theta-\cos5\theta+\cos7\theta]


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