Answer to Question #114401 in Complex Analysis for Mcsheos

Question

express cos^(3)tetasin^(4)teta in terms of multiple angle

Answer 1

in terms of complex numbers

let $z=\cos\theta+i\sin\theta$

then

$2\cos\theta=z+\frac{1}{z}\\2i\sin\theta=z-\frac{1}{z}$

then

$(2\cos\theta)^{3}(2i\sin\theta)^{4}=(z+\frac{1}{z})^{3}(z-\frac{1}{z})^{4}\\128\cos^{3}\theta\sin^{4}\theta=(z-\frac{1}{z})(z^{2}-\frac{1}{z^{2}})^{3}\\128\cos^{3}\theta\sin^{4}\theta=(z-\frac{1}{z})(z^{6}-3z^{2}+\frac{3}{z^{2}}-\frac{1}{z^{6}})\\128\cos^{3}\theta\sin^{4}\theta=(z^{7}+\frac{1}{z^{7}})-(z^{5}+\frac{1}{z^{5}})-3(z^{3}+\frac{1}{z^{3}})+3(z+\frac{1}{z})$

substituting back:

$128\cos^{3}\theta\sin^{4}\theta=2\cos7\theta-2\cos5\theta-3(2\cos3\theta)+3(2\cos\theta)$

dividing through by 128

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

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