<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] $f'(\alpha)=2\log 2 \sin \alpha \cos \alpha \left[2^{\sin ^2\alpha}- 2^{\cos ^2\alpha}\right]$
$f'(\alpha)=0 \to 2\sin\alpha\cos\alpha=0$ or $2^{\sin ^2\alpha}- 2^{\cos ^2\alpha}=0$
$ \sin 2\alpha=0\to 2\alpha=k\pi\to\alpha=\dfrac{k\pi}{2}$
$2^{\sin ^2\alpha}- 2^{\cos ^2\alpha}=0\to 2^{\sin ^2\alpha}= 2^{\cos ^2\alpha}$
$\sin^2\alpha=\cos^2\alpha\to |\sin\alpha|=|\cos\alpha|\to \alpha=\dfrac{\pi}{4}+k\dfrac{\pi}{2}$
Now it's easy to see that $\alpha=\dfrac{\pi}{4}$ etc leads to the minimum 
$2^{\sin^2 \alpha} + 2^{\cos^2 \alpha}=2^{\frac12}+2^{\frac12}=2\sqrt 2$
hope this helps
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