Question Number 28164 by abdo imad last updated on 21/Jan/18
$${simplify}\: \\ $$$${A}={cos}^{\mathrm{4}} \theta\:+{cos}^{\mathrm{4}} \left(\theta+\frac{\pi}{\mathrm{4}}\right)\:+{cos}^{\mathrm{4}} \left(\theta\:+\frac{\mathrm{2}\pi}{\mathrm{4}}\right)\:+{cos}^{\mathrm{4}} \left(\theta\:+\frac{\mathrm{3}\pi}{\mathrm{4}}\right). \\ $$
Answered by ajfour last updated on 21/Jan/18
![A=cos^4 θ+cos^4 (θ+(π/2))+ cos^4 (θ+(π/4))+cos^4 (θ+(π/4)+(π/2)) =(cos^4 θ+sin^4 θ)+ cos^4 (θ+(π/4))+sin^4 (θ+(π/4)) =1−2sin^2 θcos^2 θ+ 1−2sin^2 (θ+(π/4))cos^2 (θ+(π/4)) =2−(1/4)[2sin^2 2θ+2sin^2 (θ+(π/4))] =2−(1/4)[1−cos 2θ+1−cos (2θ+(π/2))] A=(3/2)+(((cos 2θ−sin 2θ)/4)) .](https://www.tinkutara.com/question/Q28175.png)
$${A}=\mathrm{cos}\:^{\mathrm{4}} \theta+\mathrm{cos}\:^{\mathrm{4}} \left(\theta+\frac{\pi}{\mathrm{2}}\right)+ \\ $$$$\:\:\:\:\:\:\:\mathrm{cos}\:^{\mathrm{4}} \left(\theta+\frac{\pi}{\mathrm{4}}\right)+\mathrm{cos}\:^{\mathrm{4}} \left(\theta+\frac{\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{2}}\right) \\ $$$$\:\:\:=\left(\mathrm{cos}\:^{\mathrm{4}} \theta+\mathrm{sin}\:^{\mathrm{4}} \theta\right)+ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{cos}\:^{\mathrm{4}} \left(\theta+\frac{\pi}{\mathrm{4}}\right)+\mathrm{sin}\:^{\mathrm{4}} \left(\theta+\frac{\pi}{\mathrm{4}}\right) \\ $$$$\:\:\:=\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \theta\mathrm{cos}\:^{\mathrm{2}} \theta+ \\ $$$$\:\:\:\:\:\:\:\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \left(\theta+\frac{\pi}{\mathrm{4}}\right)\mathrm{cos}\:^{\mathrm{2}} \left(\theta+\frac{\pi}{\mathrm{4}}\right) \\ $$$$\:\:=\mathrm{2}−\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{2sin}\:^{\mathrm{2}} \mathrm{2}\theta+\mathrm{2sin}\:^{\mathrm{2}} \left(\theta+\frac{\pi}{\mathrm{4}}\right)\right] \\ $$$$=\mathrm{2}−\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta+\mathrm{1}−\mathrm{cos}\:\left(\mathrm{2}\theta+\frac{\pi}{\mathrm{2}}\right)\right] \\ $$$${A}=\frac{\mathrm{3}}{\mathrm{2}}+\left(\frac{\mathrm{cos}\:\mathrm{2}\theta−\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{4}}\right)\:. \\ $$