what is the lowest value of2^sin^2x+2^cos^2x

Question

anonymous · Answer

are you familiar with min/maxing functions using derivatives?

$oumya · Answer

no

anonymous · Answer

oh man...sorry then I can't help you.

ganeshie8 · Answer

Say,  $\large  2^{\sin^2x}  + 2^{\cos^2x} =  t$  is the min value



$\large  2^{\sin^2x}  + 2^{1-\sin^2x} =  t$


$\large  \left(2^{\sin^2x} \right)^2    + 2 = \left(2^{\sin^2x}\right)  t$


plugin  $\large    2^{\sin^2x}    =  y$


$\large  y^2    -yt + 2 =  0$

Since $y$ is always real (positive), D > 0 :

$\large    t^2 -  4(1)(2) \ge 0    $

$\large    t \ge \sqrt{8}    $

anonymous · Answer

The function y = 2cos^2 x + 2 sin^2 x is always positive, Then, the lowest value of y is zero.
We can prove by another way.
y = 2 (sin^2 x + cos^2 x) = 2[(sin^2 x + cos^2 x)^2 - 2sin x*cos x)
y is minimum when (2sin x*cos x = sin 2x) is max, meaning sin 2x = 1, then y = 0.
It occurs when 2x = Pi/2 -> x = Pi/4.