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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.


marked as duplicate by lab bhattacharjee trigonometry Aug 19 '17 at 1:13

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$f(x)=2^x$ 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.


Also, we can use $(x+y)^2\geq4xy$, which is $(x-y)^2\geq0$: $$2^{\sin^2\alpha}+2^{\cos^2\alpha}=\sqrt{\left(2^{\sin^2\alpha}+2^{\cos^2\alpha}\right)^2}\geq$$ $$\geq\sqrt{4\cdot2^{\sin^2\alpha}\cdot2^{\cos^2\alpha}}=\sqrt{4\cdot2^{\sin^2\alpha+\cos^2\alpha}}=\sqrt8=2\sqrt2$$

  • $\begingroup$ could you give me a more "beginner-type" solution? $\endgroup$ – Lokesh Sangewar Aug 17 '17 at 15:09
  • $\begingroup$ @Lokesh Sangewar My solution for beginner. Draw graph of $f(x)=2^x$ and two points $A\left(\sin^2\alpha,2^{\sin^2\alpha}\right)$ and $B\left(\cos^2\alpha,2^{\cos^2\alpha}\right)$. Then graph of $f$ placed behind the segment $AB$. Now, the inequality is obviuous! $\endgroup$ – Michael Rozenberg Aug 17 '17 at 15:15
  • 1
    $\begingroup$ @Lokesh Sangewar I added something. See now. $\endgroup$ – Michael Rozenberg Aug 17 '17 at 15:21

HINT: By $AM-GM$ we have $$\frac{2^{\sin(x)^2}+2^{\cos(x)^2}}{2}\geq \sqrt{2^{\sin(x)^2+\cos(x)^2}}=...$$


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?

  • $\begingroup$ How should I minimize it? $\endgroup$ – Lokesh Sangewar Aug 17 '17 at 15:07
  • $\begingroup$ We know $t\in[0,1]$, so $2^t\in[1,2]$. Let $x=2^t$. The question is hence equivalent to minimizing $x+\frac2x$. This can be done in several different ways -- you can use calculus, AM-GM inequality, and if you can picture the graph of $f(x)=x+\frac2x$, you can even do it via quadratic equations (when does a quadratic polynomial have a single real root?). $\endgroup$ – Fimpellizieri Aug 17 '17 at 15:10

$f'(\alpha)=2\log 2 \sin \alpha \cos \alpha \left[2^{\sin ^2\alpha}- 2^{\cos ^2\alpha}\right]$

$f'(\alpha)=0 \to 2\sin\alpha\cos\alpha=0$ 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


$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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