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Question Number 45830 by jashim last updated on 17/Oct/18

Given A= sin^2 θ + cos^4 θ, then for all  real θ

$$\mathrm{Given}\:{A}=\:\mathrm{sin}^{\mathrm{2}} \theta\:+\:\mathrm{cos}^{\mathrm{4}} \theta,\:\mathrm{then}\:\mathrm{for}\:\mathrm{all} \\ $$$$\mathrm{real}\:\theta \\ $$

Commented by maxmathsup by imad last updated on 17/Oct/18

the Q is absent let suppse simplification  A =((1−cos(2θ))/2) +(((1+cos(2θ))/2))^2 =((1−cos(2θ))/2) +((1+2cos(2θ)+cos^2 (2θ))/4)  =((2−2cos(2θ) +1+2cos(2θ) +cos^2 (2θ))/4) =((3+cos^2 (2θ))/4)  =((3 +((1+cos(4θ))/2))/4) =((6+1+cos(4θ))/8) =(7/8) +(1/8)cos(4θ) .

$${the}\:{Q}\:{is}\:{absent}\:{let}\:{suppse}\:{simplification} \\ $$$${A}\:=\frac{\mathrm{1}−{cos}\left(\mathrm{2}\theta\right)}{\mathrm{2}}\:+\left(\frac{\mathrm{1}+{cos}\left(\mathrm{2}\theta\right)}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{1}−{cos}\left(\mathrm{2}\theta\right)}{\mathrm{2}}\:+\frac{\mathrm{1}+\mathrm{2}{cos}\left(\mathrm{2}\theta\right)+{cos}^{\mathrm{2}} \left(\mathrm{2}\theta\right)}{\mathrm{4}} \\ $$$$=\frac{\mathrm{2}−\mathrm{2}{cos}\left(\mathrm{2}\theta\right)\:+\mathrm{1}+\mathrm{2}{cos}\left(\mathrm{2}\theta\right)\:+{cos}^{\mathrm{2}} \left(\mathrm{2}\theta\right)}{\mathrm{4}}\:=\frac{\mathrm{3}+{cos}^{\mathrm{2}} \left(\mathrm{2}\theta\right)}{\mathrm{4}} \\ $$$$=\frac{\mathrm{3}\:+\frac{\mathrm{1}+{cos}\left(\mathrm{4}\theta\right)}{\mathrm{2}}}{\mathrm{4}}\:=\frac{\mathrm{6}+\mathrm{1}+{cos}\left(\mathrm{4}\theta\right)}{\mathrm{8}}\:=\frac{\mathrm{7}}{\mathrm{8}}\:+\frac{\mathrm{1}}{\mathrm{8}}{cos}\left(\mathrm{4}\theta\right)\:. \\ $$

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