Key to "Help me prove the exponential inequality 2sin2x+2cos2x≤3 Ive"

Name: Help me prove the exponential inequality 2sin2x+2cos2x≤3 Ive
Uploaded: 2022-02-23T17:22:57+00:00
Description: Help me prove the exponential inequality 2sin2x+2cos2x≤3 Ive

Pamela Meyer Answered2022-01-17

Help me prove the exponential inequality

2^{\sin^{2} x} + 2^{\cos^{2} x} \leq 3

Ive

Answer & Explanation

Bernard Lacey

Expert

2022-01-16Added 30 answers

Since

2^{\sin^{2} x} \geq 1

and

2^{\cos^{2} x} \geq 1

, then

2^{\sin^{2} x} + 2^{\cos^{2} x} = 3 - (2^{\sin^{2} x} - 1) \cdot (2^{\cos^{2} x} - 1) \leq 3

Moreover,

2^{\sin^{2} x} + 2^{\cos^{2} x} = {[{(\sqrt{2})}^{\sin^{2} x} - {(\sqrt{2})}^{\cos^{2} x}]}^{2} + 2 \sqrt{2} \geq 2 \sqrt{2}

RizerMix

Expert

2022-01-19Added 437 answers

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