My try: Let $y$ = $2^{\sin x}+2^{\cos x}$

  1. 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).

  1. 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.


  • 2
    $\begingroup$ We have $y\ge 2\cdot 2^{(\sin x+\cos x)/2}$, with equality when $2^{\sin x}=2^{\cos x}$. Minimize $\sin x+\cos x$, and you will have minimized $y$. (We will want for example $x=3\pi/4$.) So you have dealt with the minimum. Now tackle the maximum. $\endgroup$ – André Nicolas Sep 24 '15 at 17:52
  • $\begingroup$ @AndréNicolas I think it is the best process so far. For max $\sin x = \cos x=1/\sqrt2$ And for min $\sin x = \cos x=-1/\sqrt2$ $\endgroup$ – Archisman Panigrahi Sep 24 '15 at 18:13
  • $\begingroup$ I have written a fairly detailed outline for the minimum, based on your AM/GM calculation. That does not directly deal with the maximum. $\endgroup$ – André Nicolas Sep 24 '15 at 18:29
  • $\begingroup$ Well, $y$ is maximum when $\sin x +\cos x$ is maximum so it deals with maximum too. $\endgroup$ – Archisman Panigrahi Sep 25 '15 at 13:17

Let $f(x)=2^{\sin x}+2^{\cos x},\;$ so $f^{\prime}(x)=(2^{\sin x}\cos x-2^{\cos x}\sin x)(\ln 2)$.

$\textbf{A)}$ To find the maximum of $f(x)$, it is enough to consider $f$ on $[0,\frac{\pi}{2}]$.

$\;\;\;$Then $f(0)=3$ and $f(\frac{\pi}{2})=3$, and

$\;\;\;f^{\prime}(x)=0\iff 2^{\sin x}\cos x=2^{\cos x}\sin x\iff\frac{2^{\sin x}}{\sin x}=\frac{2^{\cos x}}{\cos x}\iff x=\frac{\pi}{4}$,

$\;\;\;$since if $\displaystyle g(t)=\frac{2^t}{t},\;$ $\displaystyle g^{\prime}(t)=\frac{2^t(t\ln 2-1)}{t^2}<0$ for $0< t<1$ so $g$ is 1-1 on $(0,1]$.

$\;\;\;$Since $f(\frac{\pi}{4})=2\cdot2^{\sqrt{2}/2}=2^{1+\frac{\sqrt{2}}{2}}>3,\;$ $f(\frac{\pi}{4})=2^{1+\frac{\sqrt{2}}{2}}$ is the maximum value.

$\textbf{B)}$ Similarly, to find the minimum of $f(x)$ it is enough to consider $f$ on $[\pi, \frac{3\pi}{2}]$.

$\;\;\;$Then $f(\pi)=\frac{3}{2}$ and $f(\frac{3\pi}{2})=\frac{3}{2}$, and $f^{\prime}(x)=0\iff\frac{2^{\sin x}}{\sin x}=\frac{2^{\cos x}}{\cos x}\iff x=\frac{5\pi}{4}$

$\;\;\;$ since $\displaystyle g(t)=\frac{2^t}{t}$ is 1-1 on $[-1,0)$ as above.

$\;\;\;$Since $f(\frac{5\pi}{4})=2\cdot2^{-\frac{\sqrt{2}}{2}}=2^{1-\frac{\sqrt{2}}{2}}<\frac{3}{2},\;\;$ $f(\frac{5\pi}{4})=2^{1-\frac{\sqrt{2}}{2}}$ is the minimum value.

  • $\begingroup$ Using the fact that $2^z/z$ is 1-to-1 for $0<|z|\le 1 which is readily shown by differentiating. $\endgroup$ – DanielWainfleet Sep 25 '15 at 10:32

Using AM GM inequality $$2^{\sin x}+2^{\cos x}\ge2\cdot2^{\frac{\sin x+\cos x}2}$$

Now the the minimum value i.e., equality occurs if $2^{\sin x}=2^{\cos x}\iff \sin x=\cos x$

$\implies\tan x=1\implies\sin x=\cos x=\pm\dfrac1{\sqrt2}$

So, the minimum value $$=2^{1-\frac{1+1}{\sqrt2}}$$

Set $y=\pi+x$ to find the maximum value

  • $\begingroup$ But this should lead to minimum value, right? However this is the maximum value $\endgroup$ – Archisman Panigrahi Sep 24 '15 at 17:53
  • $\begingroup$ To find the maximum value, you might want to study $y^2$ (which attains max/min at the same points as $y$) and then use the AM GM inequality on the product term. $\endgroup$ – mickep Sep 24 '15 at 17:55
  • $\begingroup$ @ArchismanPanigrahi, This the minimum value. Now check the updated answer $\endgroup$ – lab bhattacharjee Sep 24 '15 at 17:57
  • $\begingroup$ @labbhattacharjee $ y=π+x$ ? $\endgroup$ – Archisman Panigrahi Sep 24 '15 at 18:00
  • $\begingroup$ @ArchismanPanigrahi, As that makes $\sin y=-\sin x,\cos y=-\cos x$ $\endgroup$ – lab bhattacharjee Sep 24 '15 at 18:03

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.


Write $x:={\pi\over4}+t$, and put ${\displaystyle{1\over\sqrt{2}}\log 2}=:\lambda\doteq0.490$. Then $$2^{\cos x}+2^{\sin x}=2 e^{\lambda\cos t}\>\cosh(\lambda\sin t)=: g(t)\ .$$ Since $g$ is a smooth $2\pi$-periodic function its global extrema are at the zeros of its derivative $$g'(t)=2\lambda e^{\lambda\cos t}\bigl(-\sin t\>\cosh(\lambda\sin t)+\cos t\>\sinh(\lambda\sin t) \bigr)\ .$$ When $\cos t=0$ one necessarily has $g'(t)\ne0$. We may therefore consider the equation $$\tan t=\tanh(\lambda \sin t)\ .\tag{1}$$ For $0< t<{\pi\over2}$ one has $$\tanh(\lambda \sin t)<\lambda\sin t<t<\tan t\ .$$ Symmetry considerations then imply that $t=0$ and $t=\pi$ are the only solutions of $(1)$ mod $2\pi$. Therefore $$\bigl\{g(0), g(\pi)\bigr\}=\{2e^\lambda, 2e^{-\lambda}\}=\bigl\{2\cdot 2^{1/\sqrt{2}}, \> 2\cdot 2^{-1/\sqrt{2}}\bigr\}$$ are the maximal and the minimal value of $g$, and whence of $2^{\cos x}+2^{\sin x}$, on ${\mathbb R}$.


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