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Question Number 17080 by Kunal kumar shukla last updated on 30/Jun/17
sin^4 θ/2+cos^4 θ/2≥1/2
$$\mathrm{sin}^{\mathrm{4}} \theta/\mathrm{2}+\mathrm{cos}\:^{\mathrm{4}} \theta/\mathrm{2}\geqslant\mathrm{1}/\mathrm{2} \\ $$
Answered by virus last updated on 30/Jun/17
1−2sin^2 (θ/2)cos^2 (θ/2)≥1/2 2−4sin^2 (θ/2)cos^2 (θ/2)≥1 2−sin^2 θ≥1 1−sin^2 θ≥0 cos^2 θ≥0 ∴θ∈(2n+1)π/2
$$\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \left(\theta/\mathrm{2}\right)\mathrm{cos}\:^{\mathrm{2}} \left(\theta/\mathrm{2}\right)\geqslant\mathrm{1}/\mathrm{2} \\ $$$$\mathrm{2}−\mathrm{4sin}\:^{\mathrm{2}} \left(\theta/\mathrm{2}\right)\mathrm{cos}\:^{\mathrm{2}} \left(\theta/\mathrm{2}\right)\geqslant\mathrm{1} \\ $$$$\mathrm{2}−\mathrm{sin}\:^{\mathrm{2}} \theta\geqslant\mathrm{1} \\ $$$$\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \theta\geqslant\mathrm{0} \\ $$$$\mathrm{cos}\:^{\mathrm{2}} \theta\geqslant\mathrm{0} \\ $$$$\therefore\theta\in\left(\mathrm{2}{n}+\mathrm{1}\right)\pi/\mathrm{2} \\ $$
Answered by mrW1 last updated on 02/Jul/17
sin^4 (θ/2)+cos^4 (θ/2) =sin^4 (θ/2)+(1−sin^2 (θ/2))^2 =sin^4 (θ/2)+1−2sin^2 (θ/2)+sin^4 (θ/2) =2(sin^4 (θ/2)−sin^2 (θ/2))+1 =2(sin^4 (θ/2)−sin^2 (θ/2)+(1/4))+(1/2) =(1/2)+2(sin^2 (θ/2)−(1/2))^2 =(1/2)+2(((1−cos θ)/2)−(1/2))^2 =(1/2)+((cos^2 θ)/2) =(1/2)(1+cos^2 θ) ≥(1/2)
$$\mathrm{sin}^{\mathrm{4}} \:\frac{\theta}{\mathrm{2}}+\mathrm{cos}^{\mathrm{4}} \:\frac{\theta}{\mathrm{2}} \\ $$$$=\mathrm{sin}^{\mathrm{4}} \:\frac{\theta}{\mathrm{2}}+\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$=\mathrm{sin}^{\mathrm{4}} \:\frac{\theta}{\mathrm{2}}+\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}+\mathrm{sin}^{\mathrm{4}} \:\frac{\theta}{\mathrm{2}} \\ $$$$=\mathrm{2}\left(\mathrm{sin}^{\mathrm{4}} \:\frac{\theta}{\mathrm{2}}−\mathrm{sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}\right)+\mathrm{1} \\ $$$$=\mathrm{2}\left(\mathrm{sin}^{\mathrm{4}} \:\frac{\theta}{\mathrm{2}}−\mathrm{sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}\right)+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{2}\left(\mathrm{sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{2}\left(\frac{\mathrm{1}−\mathrm{cos}\:\theta}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{cos}^{\mathrm{2}} \:\theta}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:\theta\right) \\ $$$$\geqslant\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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