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.
 A: $f(x)=2^x$ is a convex function.
Thus, by Jensen:
$$2^{\sin^2\alpha}+2^{\cos^2\alpha}\geq2\cdot2^{\frac{\sin^2\alpha+\cos^2\alpha}{2}}=2\sqrt2.$$
The equality occurs for $\alpha=\beta=45^{\circ}$, which says that we got a minimal value.
Done!
Also, we can use $(x+y)^2\geq4xy$, which is $(x-y)^2\geq0$:
$$2^{\sin^2\alpha}+2^{\cos^2\alpha}=\sqrt{\left(2^{\sin^2\alpha}+2^{\cos^2\alpha}\right)^2}\geq$$
$$\geq\sqrt{4\cdot2^{\sin^2\alpha}\cdot2^{\cos^2\alpha}}=\sqrt{4\cdot2^{\sin^2\alpha+\cos^2\alpha}}=\sqrt8=2\sqrt2$$
A: HINT: By $AM-GM$ we have $$\frac{2^{\sin(x)^2}+2^{\cos(x)^2}}{2}\geq \sqrt{2^{\sin(x)^2+\cos(x)^2}}=...$$
A: Hint: Write $\cos(\alpha)^2=1-\sin(\alpha)^2$, so that
\begin{align}
2^{\sin(\alpha)^2}+2^{\cos(\alpha)^2}&=2^{\sin(\alpha)^2}+2^{1-\sin(\alpha)^2}
\\&=2^{\sin(\alpha)^2}+\frac{2}{2^{\sin(\alpha)^2}}
\end{align}
With $t={\sin(\alpha)^2}$, can you minimize the expression above?
A: $f'(\alpha)=2\log 2 \sin \alpha \cos \alpha \left[2^{\sin ^2\alpha}- 2^{\cos ^2\alpha}\right]$
$f'(\alpha)=0 \to 2\sin\alpha\cos\alpha=0$ 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
A: $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 $\alpha = 45°$.
$\min(f(x)) = f(x =1/2) = 2√2$.
