Pamela Meyer

2022-01-17

Help me prove the exponential inequality
${2}^{{\mathrm{sin}}^{2}x}+{2}^{{\mathrm{cos}}^{2}x}\le 3$
Ive

Bernard Lacey

Expert

Since ${2}^{{\mathrm{sin}}^{2}x}\ge 1$ and ${2}^{{\mathrm{cos}}^{2}x}\ge 1$, then
${2}^{{\mathrm{sin}}^{2}x}+{2}^{{\mathrm{cos}}^{2}x}=3-\left({2}^{{\mathrm{sin}}^{2}x}-1\right)\cdot \left({2}^{{\mathrm{cos}}^{2}x}-1\right)\le 3$
Moreover,
${2}^{{\mathrm{sin}}^{2}x}+{2}^{{\mathrm{cos}}^{2}x}={\left[{\left(\sqrt{2}\right)}^{{\mathrm{sin}}^{2}x}-{\left(\sqrt{2}\right)}^{{\mathrm{cos}}^{2}x}\right]}^{2}+2\sqrt{2}\ge 2\sqrt{2}$.

RizerMix

Expert

Rearranging the equation we have that
${2}^{t}+{2}^{1-t}=3⇒{2}^{2t}-3\cdot {2}^{t}+2=0$
which after factoring gets us
$\left({2}^{t}-1\right)\left({2}^{t}-2\right)=0$
This is a parabola with roots at 1 and 2 that opens upward. Therefore we have that
$\left({2}^{t}-1\right)\left({2}^{t}-2\right)\le 0⇔1\le 2t\le 2⇔0\le t\le 1$
which immediately gives us our desired result.

alenahelenash

Expert