How to find the maximum and minimum value of $2^{\sin x}+2^{\cos x}$ My try: Let $y$ =  $2^{\sin x}+2^{\cos x}$


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*Applying AM GM inequality I get 
$y$ $> 2.2^{(\sin x+\cos x)/2}$. 
Now, the highest value of R.H.S is $2^{\frac{\left(2+\sqrt{2}\right)}{2}}$.


Should this mean that $y$ is always greater than $2^{\frac{ \left ( 2+\sqrt{2}\right ) }{2}}$?
But this is not true (we can see in the graph).


*Calculus method:
$dy/dx$ = $\ln\left(2\right){\cdot}{2}^{\sin\left(x\right)}\cos\left(x\right){-\ln\left(2\right){\cdot}{2}^{\cos\left(x\right)}\sin\left(x\right)}$


When $dy/dx$ =0,
$\tan x = 2^{\sin x- \cos x}$ and I am stuck here.
https://www.desmos.com/calculator/p3zfvkq2mn

Let $f(u,v)=2^u+2^v;\;$ we want to find the extrema of $f$ on the circle $u^2+v^2=1$.
Using Lagrange multipliers, $\;\;2^u\ln2=\lambda(2u)$ and $2^v\ln 2=\lambda(2v)\;$ so $\;\displaystyle\frac{2^u\ln2}{2u}=\frac{2^v\ln2}{2v}$.
Then $\displaystyle\frac{2^u}{u}=\frac{2^v}{v}$ and therefore $u=v$, 
$\;\;\;$since the function $\displaystyle g(t)=\frac{2^t}{t}$ is 1-1 on $[-1,0)$ and $(0,1]$
$\;\;\;$because $\displaystyle g^{\prime}(t)=\frac{2^t(t\ln 2-1)}{t^2}<0$ for $t\in(-1,0)\cup(0,1)$.
Then $u^2+v^2=1\implies 2u^2=1\implies u=v=\pm\frac{\sqrt{2}}{2}$.
Thus $\displaystyle f\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)=2^{1+\frac{\sqrt{2}}{2}}$ is the maximum, and $\displaystyle f\left(-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)=2^{1-\frac{\sqrt{2}}{2}}$ is the minimum.
