Question Number 143222 by ZiYangLee last updated on 11/Jun/21
$$\mathrm{If}\:{z}=\mathrm{cos}\:\theta+{i}\:\mathrm{sin}\:\theta,\:\mathrm{by}\:\mathrm{expand} \\ $$$$\left({z}+\frac{\mathrm{1}}{{z}}\right)^{\mathrm{4}} \left({z}−\frac{\mathrm{1}}{{z}}\right)^{\mathrm{4}} \mathrm{or}\:\mathrm{other}\:\mathrm{method}, \\ $$$$\mathrm{prove}\:\mathrm{128}\:\mathrm{sin}^{\mathrm{4}} \theta\mathrm{cos}^{\mathrm{4}} \theta=\mathrm{cos}\:\mathrm{8}\theta−\mathrm{4cos}\:\mathrm{4}\theta+\mathrm{3}. \\ $$
Answered by Olaf_Thorendsen last updated on 11/Jun/21
![If z = cosθ+isinθ = e^(iθ) z+(1/z) = e^(iθ) +e^(−iθ) = 2cosθ z−(1/z) = e^(iθ) −e^(−iθ) = 2isinθ Z = (z+(1/z))^4 (z−(1/z))^4 = 16cos^4 θ×16sin^4 θ Z = 256sin^4 θcos^4 θ (1) Z = (z+(1/z))^4 (z−(1/z))^4 Z = (z^2 +(1/z^2 )+2)^2 (z^2 +(1/z^2 )−2)^2 Z = (e^(2iθ) +e^(−2iθ) +2)^2 (e^(2iθ) +e^(−2iθ) −2)^2 Z = (2cos2θ+2)^2 (2cos2θ−2)^2 Z = 16(cos2θ+1)^2 (cos2θ−1)^2 Z = 16(1−cos^2 2θ)^2 Z = 16(((1−cos4θ)/2))^2 Z = 4[((1+cos8θ)/2)−2cos4θ+1] Z = 2cos8θ−8cos4θ+6 (2) (1) and (2) : 128sin^4 θcos^4 θ = cos8θ−4cos4θ+3](https://www.tinkutara.com/question/Q143224.png)
$$\mathrm{If}\:{z}\:=\:\mathrm{cos}\theta+\mathrm{isin}\theta\:=\:{e}^{{i}\theta} \\ $$$${z}+\frac{\mathrm{1}}{{z}}\:=\:{e}^{{i}\theta} +{e}^{−{i}\theta} \:=\:\mathrm{2cos}\theta \\ $$$${z}−\frac{\mathrm{1}}{{z}}\:=\:{e}^{{i}\theta} −{e}^{−{i}\theta} \:=\:\mathrm{2}{i}\mathrm{sin}\theta \\ $$$$ \\ $$$$\mathrm{Z}\:=\:\left({z}+\frac{\mathrm{1}}{{z}}\right)^{\mathrm{4}} \left({z}−\frac{\mathrm{1}}{{z}}\right)^{\mathrm{4}} =\:\mathrm{16cos}^{\mathrm{4}} \theta×\mathrm{16sin}^{\mathrm{4}} \theta \\ $$$$\mathrm{Z}\:=\:\mathrm{256sin}^{\mathrm{4}} \theta\mathrm{cos}^{\mathrm{4}} \theta\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$ \\ $$$$\mathrm{Z}\:=\:\left({z}+\frac{\mathrm{1}}{{z}}\right)^{\mathrm{4}} \left({z}−\frac{\mathrm{1}}{{z}}\right)^{\mathrm{4}} \\ $$$$\mathrm{Z}\:=\:\left({z}^{\mathrm{2}} +\frac{\mathrm{1}}{{z}^{\mathrm{2}} }+\mathrm{2}\right)^{\mathrm{2}} \left({z}^{\mathrm{2}} +\frac{\mathrm{1}}{{z}^{\mathrm{2}} }−\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\mathrm{Z}\:=\:\left({e}^{\mathrm{2}{i}\theta} +{e}^{−\mathrm{2}{i}\theta} +\mathrm{2}\right)^{\mathrm{2}} \left({e}^{\mathrm{2}{i}\theta} +{e}^{−\mathrm{2}{i}\theta} −\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\mathrm{Z}\:=\:\left(\mathrm{2cos2}\theta+\mathrm{2}\right)^{\mathrm{2}} \left(\mathrm{2cos2}\theta−\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\mathrm{Z}\:=\:\mathrm{16}\left(\mathrm{cos2}\theta+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{cos2}\theta−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{Z}\:=\:\mathrm{16}\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \mathrm{2}\theta\right)^{\mathrm{2}} \\ $$$$\mathrm{Z}\:=\:\mathrm{16}\left(\frac{\mathrm{1}−\mathrm{cos4}\theta}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\mathrm{Z}\:=\:\mathrm{4}\left[\frac{\mathrm{1}+\mathrm{cos8}\theta}{\mathrm{2}}−\mathrm{2cos4}\theta+\mathrm{1}\right] \\ $$$$\mathrm{Z}\:=\:\mathrm{2cos8}\theta−\mathrm{8cos4}\theta+\mathrm{6}\:\:\:\left(\mathrm{2}\right) \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:\mathrm{and}\:\left(\mathrm{2}\right)\:: \\ $$$$\mathrm{128sin}^{\mathrm{4}} \theta\mathrm{cos}^{\mathrm{4}} \theta\:=\:\mathrm{cos8}\theta−\mathrm{4cos4}\theta+\mathrm{3} \\ $$