<s>[INST] A question of trigonometry on how to find minimum value. 
Find The minimum value of $$2^{\sin^2 \alpha} + 2^{\cos^2 \alpha}.$$

I can easily get the maximum value but minimum value is kinda tricky. Please help.
 [/INST] $x:= \sin^2(\alpha) $; $ 1-x = \cos^2(\alpha)$ .
Then: $ f(x):= 2^x + \frac{2}{2^x} $,  $0 \le x \le1$.
$z := 2^x$ ; 
$ g(z): =  z + \frac{2}{z} , 1 \le z \le 2$.
AM GM inequality:
$ (1/2) ( z + \frac{2}{z}) \ge (z \frac{2}{z})^{1/2} =$
$ \sqrt{2}$.
$g(z) = z + \frac{2}{z} \ge 2 \sqrt{2}$.
Equality for $z = √2$.
Back substitution: $2^x = 2^{1/2}$.
Minimum at $x= 1/2$, I.e 
$x = \sin^2(\alpha) = 1/2 = \cos^2(\alpha)$ , 
hence $\alpha = 45°$.
$\min(f(x)) = f(x =1/2) = 2√2$.
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