Theresa Archer

2022-06-30

How to find the smallest value of ${\mathrm{sin}}^{4}\left(\alpha \right)+{\mathrm{cos}}^{4}\left(\alpha \right)$ ?

America Barrera

Expert

Because actually
${\mathrm{sin}}^{4}\left(\alpha \right)+{\mathrm{cos}}^{4}\left(\alpha \right)=\left({\mathrm{sin}}^{2}\left(\alpha \right)+{\mathrm{cos}}^{2}\left(\alpha \right){\right)}^{2}-2{\mathrm{sin}}^{2}\left(\alpha \right){\mathrm{cos}}^{2}\left(\alpha \right)=1-\frac{1}{2}{\mathrm{sin}}^{2}\left(2\alpha \right),$

Ayanna Trujillo

Expert

${\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x=\left({\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x{\right)}^{2}-2{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x=1-\frac{\left(\mathrm{sin}2x{\right)}^{2}}{2}$
Now $0\le {\mathrm{sin}}^{2}2x\le 1$
Alternatively using $\mathrm{cos}2x=1-2{\mathrm{sin}}^{2}x=2{\mathrm{cos}}^{2}x-1$,
$\left(2{\mathrm{sin}}^{2}x{\right)}^{2}+\left(2{\mathrm{cos}}^{2}x{\right)}^{2}=\left(1-\mathrm{cos}2x{\right)}^{2}+\left(1+\mathrm{cos}2x{\right)}^{2}=2\left(1+{\mathrm{cos}}^{2}2x\right)=2+1+\mathrm{cos}4x$
Now $-1\le \mathrm{cos}4x\le 1$

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