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Question Number 48360 by maxmathsup by imad last updated on 22/Nov/18

calculate cos^4 ((π/8))+cos^4 (((3π)/8)) +cos^4 (((5π)/8)) +cos^4 (((7π)/8))

$${calculate}\:{cos}^{\mathrm{4}} \left(\frac{\pi}{\mathrm{8}}\right)+{cos}^{\mathrm{4}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)\:+{cos}^{\mathrm{4}} \left(\frac{\mathrm{5}\pi}{\mathrm{8}}\right)\:+{cos}^{\mathrm{4}} \left(\frac{\mathrm{7}\pi}{\mathrm{8}}\right) \\ $$

Commented by Abdo msup. last updated on 23/Nov/18

S=cos^4 ((π/8))+cos^4 (π−(π/8))+cos^4 (((3π)/8))+cos^4 (π−((3π)/8)) =2cos^4 ((π/8))+2cos^4 (((3π)/8)) =2cos^4 ((π/8))+2cos^4 ((π/2) −(π/8))=2(cos^4 ((π/8))+sin^4 ((π/8))) =2{ (cos^2 (π/8) +sin^2 (π/8))^2 −2cos^2 (π/8)sin^2 (π/8)} =2{1−(1/2)sin^2 ((π/4)) =2{1−(1/2)(1/2)}=2(3/4) =(3/2) .

$${S}={cos}^{\mathrm{4}} \left(\frac{\pi}{\mathrm{8}}\right)+{cos}^{\mathrm{4}} \left(\pi−\frac{\pi}{\mathrm{8}}\right)+{cos}^{\mathrm{4}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)+{cos}^{\mathrm{4}} \left(\pi−\frac{\mathrm{3}\pi}{\mathrm{8}}\right) \\ $$$$=\mathrm{2}{cos}^{\mathrm{4}} \left(\frac{\pi}{\mathrm{8}}\right)+\mathrm{2}{cos}^{\mathrm{4}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right) \\ $$$$=\mathrm{2}{cos}^{\mathrm{4}} \left(\frac{\pi}{\mathrm{8}}\right)+\mathrm{2}{cos}^{\mathrm{4}} \left(\frac{\pi}{\mathrm{2}}\:−\frac{\pi}{\mathrm{8}}\right)=\mathrm{2}\left({cos}^{\mathrm{4}} \left(\frac{\pi}{\mathrm{8}}\right)+{sin}^{\mathrm{4}} \left(\frac{\pi}{\mathrm{8}}\right)\right) \\ $$$$=\mathrm{2}\left\{\:\left({cos}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}\:+{sin}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}\right)^{\mathrm{2}} −\mathrm{2}{cos}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}{sin}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}\right\} \\ $$$$=\mathrm{2}\left\{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}\right)\:=\mathrm{2}\left\{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\frac{\mathrm{1}}{\mathrm{2}}\right\}=\mathrm{2}\frac{\mathrm{3}}{\mathrm{4}}\:=\frac{\mathrm{3}}{\mathrm{2}}\:.\right. \\ $$$$ \\ $$

Answered by behi83417@gmail.com last updated on 22/Nov/18

c^4 ((π/8))=c^4 (((7π)/8))=(((1+c(π/4))/2))^2 =(((1+((√2)/2))/2))^2 =((3+2(√2))/8) c^4 (((3π)/8))=c^4 (((5π)/8))=(((1+c((3π)/4))/2))^2 =((3−2(√2))/8) ⇒S=2×((3+2(√2))/8)+2×((3−2(√2))/8)=(3/2).

$${c}^{\mathrm{4}} \left(\frac{\pi}{\mathrm{8}}\right)={c}^{\mathrm{4}} \left(\frac{\mathrm{7}\pi}{\mathrm{8}}\right)=\left(\frac{\mathrm{1}+{c}\frac{\pi}{\mathrm{4}}}{\mathrm{2}}\right)^{\mathrm{2}} =\left(\frac{\mathrm{1}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{8}} \\ $$$${c}^{\mathrm{4}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)={c}^{\mathrm{4}} \left(\frac{\mathrm{5}\pi}{\mathrm{8}}\right)=\left(\frac{\mathrm{1}+{c}\frac{\mathrm{3}\pi}{\mathrm{4}}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{8}} \\ $$$$\Rightarrow{S}=\mathrm{2}×\frac{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{8}}+\mathrm{2}×\frac{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{8}}=\frac{\mathrm{3}}{\mathrm{2}}. \\ $$

Answered by hknkrc46 last updated on 24/Dec/18

α+β=π ⇒cos^(2k) α=cos^(2k) β (k∈N) α=(π/8) , β=((7π)/8) , k=2 ⇒cos^4 ((π/8))=cos^4 (((7π)/8)) ⇒cos^4 ((π/8))+cos^4 (((7π)/8))=2cos^4 ((π/8)) α=((3π)/8) , β=((5π)/8) , k=2 ⇒cos^4 (((3π)/8))=cos^4 (((5π)/8)) ⇒cos^4 (((3π)/8))+cos^4 (((5π)/8))=2cos^4 (((3π)/8)) 2cos^4 ((π/8))+2cos^4 (((3π)/8)) =2[cos^4 ((π/8))+cos^4 (((3π)/8))] =2[cos^4 ((π/8))+cos^4 ((π/2)−(π/8))] =2[cos^4 ((π/8))+sin^4 ((π/8))] =2[(cos^2 ((π/8))+sin^2 ((π/8)))^2 −((4cos^2 ((π/8))sin^2 ((π/8)))/2)] =2[1^2 −(((2cos ((π/8))sin ((π/8)))^2 )/2)] =2[1−((sin^2 ((π/4)))/2)]=2−sin^2 ((π/4)) =2−(((√2)/2))^2 =2−(1/2)=(3/2)

$$\:\alpha+\beta=\pi \\ $$$$\Rightarrow\mathrm{cos}\:^{\mathrm{2}{k}} \alpha=\mathrm{cos}\:^{\mathrm{2}{k}} \beta\:\left({k}\in\mathbb{N}\right) \\ $$$$\alpha=\frac{\pi}{\mathrm{8}}\:,\:\beta=\frac{\mathrm{7}\pi}{\mathrm{8}}\:,\:{k}=\mathrm{2} \\ $$$$\Rightarrow\mathrm{cos}\:^{\mathrm{4}} \left(\frac{\pi}{\mathrm{8}}\right)=\mathrm{cos}\:^{\mathrm{4}} \left(\frac{\mathrm{7}\pi}{\mathrm{8}}\right) \\ $$$$\Rightarrow\mathrm{cos}\:^{\mathrm{4}} \left(\frac{\pi}{\mathrm{8}}\right)+\mathrm{cos}\:^{\mathrm{4}} \left(\frac{\mathrm{7}\pi}{\mathrm{8}}\right)=\mathrm{2cos}\:^{\mathrm{4}} \left(\frac{\pi}{\mathrm{8}}\right) \\ $$$$\alpha=\frac{\mathrm{3}\pi}{\mathrm{8}}\:,\:\beta=\frac{\mathrm{5}\pi}{\mathrm{8}}\:,\:{k}=\mathrm{2} \\ $$$$\Rightarrow\mathrm{cos}\:^{\mathrm{4}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)=\mathrm{cos}\:^{\mathrm{4}} \left(\frac{\mathrm{5}\pi}{\mathrm{8}}\right) \\ $$$$\Rightarrow\mathrm{cos}\:^{\mathrm{4}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)+\mathrm{cos}\:^{\mathrm{4}} \left(\frac{\mathrm{5}\pi}{\mathrm{8}}\right)=\mathrm{2cos}\:^{\mathrm{4}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right) \\ $$$$\mathrm{2cos}\:^{\mathrm{4}} \left(\frac{\pi}{\mathrm{8}}\right)+\mathrm{2cos}\:^{\mathrm{4}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right) \\ $$$$=\mathrm{2}\left[\mathrm{cos}\:^{\mathrm{4}} \left(\frac{\pi}{\mathrm{8}}\right)+\mathrm{cos}\:^{\mathrm{4}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)\right] \\ $$$$=\mathrm{2}\left[\mathrm{cos}\:^{\mathrm{4}} \left(\frac{\pi}{\mathrm{8}}\right)+\mathrm{cos}\:^{\mathrm{4}} \left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{8}}\right)\right] \\ $$$$=\mathrm{2}\left[\mathrm{cos}\:^{\mathrm{4}} \left(\frac{\pi}{\mathrm{8}}\right)+\mathrm{sin}\:^{\mathrm{4}} \left(\frac{\pi}{\mathrm{8}}\right)\right] \\ $$$$=\mathrm{2}\left[\left(\mathrm{cos}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)+\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)\right)^{\mathrm{2}} −\frac{\mathrm{4cos}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)}{\mathrm{2}}\right] \\ $$$$=\mathrm{2}\left[\mathrm{1}^{\mathrm{2}} −\frac{\left(\mathrm{2cos}\:\left(\frac{\pi}{\mathrm{8}}\right)\mathrm{sin}\:\left(\frac{\pi}{\mathrm{8}}\right)\right)^{\mathrm{2}} }{\mathrm{2}}\right] \\ $$$$=\mathrm{2}\left[\mathrm{1}−\frac{\mathrm{sin}^{\mathrm{2}} \:\left(\frac{\pi}{\mathrm{4}}\right)}{\mathrm{2}}\right]=\mathrm{2}−\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}\right) \\ $$$$=\mathrm{2}−\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$

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